It follows from the definition of $B_{k}$ that $H_{w}^{(n)}=(2k+1)^{n}$ for
every $w\in V_{n}$, $n\in\mathbb{N}_{0}$. As in Subsection 6.1, we can show
that the entries of ${G^{\prime}}^{(n,m)}$ are determined by the formula:
${g^{\prime}}_{vw}^{(n,m)}=K_{v-w}^{(m)}$ for $v\in\\{w-mk,\ldots,w+mk\\}$,
and ${g^{\prime}}_{vw}^{(n,m)}=0$, otherwise.

Hence, the $v$-th row of ${G^{\prime}}^{(n,m)}$ has the form

${\overline{g}^{\prime}}_{v}^{(n,m)}=\langle\ldots,0,K_{mk}^{(m)},K_{mk-1}^{(m)},\ldots,K_{-mk}^{(m)},0,\ldots\rangle,$

where $K_{mk}^{(m)}$ is the $(v-mk)$-th entry.

Further, we have

$g_{vw}^{(n,m)}={g^{\prime}}_{vw}^{(n,m)}\cdot\frac{H_{w}^{(n)}}{H_{v}^{(n+m)}}=\frac{1}{{(2k+1)}^{m}}\cdot{g^{\prime}}_{vw}^{(n,m)},$

and

${\overline{g}}_{v}^{(n,m)}=\frac{1}{{(2k+1)}^{m}}\cdot\langle\ldots,0,K_{mk}^{(m)},\ldots,K_{-mk}^{(m)},0,\ldots\rangle.$

######  Proposition 6.4.

There is no probability tail invariant measure on the generalized Bratteli
diagram $B_{k}$.

###### Proof.

We will show that the set of all limit vectors of
${\overline{g}}_{v}^{(n,m)}$’s is a zero vector
$\overline{0}=\langle\ldots,0,0,0,\ldots\rangle$.

Note that $K_{0}^{(m)}$ is the largest number among ${K_{i}^{(m)}}$’s where
$i=mk,\ldots,-mk$. We prove that
$\lim_{m\rightarrow\infty}\dfrac{K_{0}^{(m)}}{{(2k+1)}^{m}}=0$. Indeed, since

(6.8)
$\frac{K_{0}^{(m+1)}}{{(2k+1)}^{m+1}}=\frac{1}{2k+1}\cdot\frac{{[K}_{k}^{(m)}+\ldots+K_{0}^{(m)}+\ldots+K_{-k}^{(m)}]}{(2k+1)^{m}}\leq\frac{K_{0}^{(m)}}{{(2k+1)}^{m}},$

the limit $g_{0}=\lim_{m\rightarrow\infty}\frac{K_{0}^{(m)}}{{(2\cdot
k+1)}^{m}}$ exists. We will show that $g_{0}=0$.

Choosing subsequences of $m$’s, we can define
$g_{i}=\lim_{m\rightarrow\infty}\dfrac{K_{i}^{(m)}}{(2k+1)^{m}}$, where
$i=k,\ldots,1,-1,\ldots,-k$. Obviously, $0\leq g_{i}\leq g_{0}$. It follows
from (6.8) that

$\frac{2k}{2k+1}g_{0}=\frac{1}{2k+1}\cdot[g_{k}+\ldots+g_{1}+g_{-1}+\ldots+g_{-k}].$

This implies that $g_{i}=g_{0}$ for $i=k,\ldots,1,-1,\ldots,-k$.

Finally, we have

$(2k+1)g_{0}=\lim_{m\rightarrow\infty}\frac{{[K}_{k}^{(m)}+\ldots+K_{0}^{(m)}+\ldots+K_{-k}^{(m)}]}{(2k+1)^{m}}\leq
1.$

This implies that $g_{0}\ \leq\frac{1}{2k+1}$.

In a similar way we can prove that $g_{0}\leq\frac{1}{{(2\cdot k+1)}^{2}}$ and
so on. Hence $g_{0}=0$, and the proposition is proved. ∎

##  7\. Examples

In this section, we consider several examples of generalized Bratteli
diagrams. Our goal is to show how the methods, developed in Section 4, can be
applied to the study of probability tail invariant measures.

###  7.1.  Bratteli diagrams and substitutions

The shift dynamical systems associated with substitutions on a finite alphabet
have been studied by many authors; we mention here only several of them
[Fog02], [Que10], [DHS99], [DP22], [Put18], etc. The primary interest is
usually focused on minimal substitution dynamical systems. It was shown that
substitution dynamical systems are completely described by stationary Bratteli
diagrams. In [BKM09], the authors constructed Bratteli diagrams for aperiodic
substitution dynamics.

In recent years, substitution dynamical systems have been considered on a
countable or even compact alphabet, see [Fer06], [BJS24], [Man23], [MnRW22].
The problem of finding finite (or sigma-finite) invariant measures for
substitution dynamical systems is highly non-trivial in this case. In [DFMV23]
investigated the existence of invariant probability measures for substitutions
on countable alphabet. In particular, they found a sufficient condition under
which there are no invariant probability measures. We show below that this
sufficient condition is, in fact, a consequence of Theorem 4.9.

Let $A$ be a countable set (an alphabet), $A^{*}$ be the set of all finite
words on $A$, and $A^{\mathbb{N}_{0}}$ be the set of infinite words on $A$. A
substitution is a map $\sigma\colon A\rightarrow A^{*}$ such that for every
$a\in A$, the finite word $\sigma(a)$ is not empty. We can extend a map
$\sigma$ to $A^{*}$ and $A^{\mathbb{N}_{0}}$ by concatenation:
$\sigma(a_{0},a_{1},\ldots)=\sigma(a_{0})\sigma(a_{1})\ldots$. In particular,
we can define words $\sigma^{n}(a),\ a\in A,\ n\in\mathbb{N}$, by setting
$\sigma^{1}(a)=\sigma(a)=u_{0}u_{1}\ldots{u}_{k}$,
$\sigma^{n+1}(a)=\sigma^{n}(u_{0})\sigma^{n}(u_{1}),\ldots,\sigma^{n}(u_{k})$.
A substitution $\sigma$ determines a shift dynamical system $(X_{\sigma},S)$,
where $X_{\sigma}$ is the set of all sequences $u\in A^{\mathbb{N}_{0}}$ (or
$u\in A^{\mathbb{Z}}$) such that any finite subword of $u$ occurs in
$\sigma^{n}(a)$ for some $a\in A$ and $n\in\mathbb{N}$ and $S$ is the shift on
$X_{\sigma}$. To avoid unnecessary complications, we will assume that
$|{\sigma}^{n}(a)|\rightarrow\infty$ as $n\to\infty$ for any $a\in A$ where
$|\sigma^{n}(a)|$ denotes the length of $\sigma^{n}(a)$.

Now we define a Bratteli diagram $B_{\sigma}=(V,E)$ associated with a
substitution $\sigma$. Set $V_{n}=A$ $n\in\mathbb{N}_{0}$. For $a\in V_{1}$,
the set $r^{-1}(a)$ consists of single edges $e(u_{i},a)$ connecting $a$ and
the vertices $u_{0},\ldots,u_{k}$ where
$(u_{0}u_{1},\ldots{u}_{k})=\sigma(a)$.

The Bratteli diagram $B_{\sigma}=(V,E)$ is stationary and its incidence matrix
$M=(M_{ij})$, $i,j=0,1,2,\ldots$ is the matrix associated to $\sigma$, i.e.,
$M_{ij}$ is the number of occurrences of the letter $j$ in the word
$\sigma(i)$. In general, the dynamics of the Bratteli diagram
$B_{\sigma}=(V,E)$ is not a good model for the shift dynamical system
($X_{\sigma},S)$. However, if we assume that every pair $(uv)$ (where $u$ is
the last letter of some $\sigma(i)$ and $v$ is the first letter of some
$\sigma(j)$) appears inside some $\sigma(k)$, then both dynamical systems
$(X_{\sigma},S)$ and the Vershik map $(X_{B},\varphi_{B})$ are isomorphic (for
a more detailed discussion of this relation see [BKM09]).

In [DFMV23], the authors proved the following theorem.

######  Theorem 7.1.

Let $\sigma:\mathbb{N}_{0}\to\mathbb{N}_{0}$ be a bounded length substitution
such that $\sigma$ has a periodic point $u$ and the matrix $M=M_{\sigma}$ of
the substitution is irreducible and aperiodic. If $M$ satisfies

(7.1)
$\lim_{n\to\infty}\sup_{i\in\mathbb{N}_{0}}\frac{M^{(n)}_{ij}}{\sum_{k\geq
0}M^{(n)}_{ik}}=0\qquad\forall j\in N_{0}$

then the dynamical system $(X_{\sigma},S)$ has no finite invariant measure.

######  Proposition 7.2.

Condition (7.1) implies that the Bratteli diagram $B_{\sigma}=(V,E)$ has no
tail invariant probability measure.

###### Proof.

We use our standard notation of matrices related to a Bratteli diagram. Note
that $M=F^{\prime}$ in our notation. Because $B_{\sigma}$ is a stationary
Bratteli diagram, we have ${G^{\prime}}^{(n,m)}={F^{\prime}}^{m}$ and, for the
stochastic matrix $G^{(n,m)}$, we can write

$\overline{g}_{i}^{(n,m)}=\left\langle\frac{{f^{\prime}}_{ij}^{(m)}\cdot
H_{j}^{(n)}}{\sum_{k=0}^{\infty}{f^{\prime}}_{ik}^{(m)}\cdot H_{k}^{(n)}}\ :\
j=0,1,2,\ldots\right\rangle.$

Then using (7.1) we have, for each $j\in\mathbb{N}_{0}$,

$g_{ij}^{(n,m)}=\frac{{f^{\prime}}_{ij}^{(m)}\cdot
H_{j}^{(n)}}{\sum_{k=0}^{\infty}{f^{\prime}}_{ik}^{(m)}\cdot
H_{k}^{(n)}}\leq\frac{{f^{\prime}}_{ij}^{(m)}\cdot
H_{j}^{(n)}}{\sum_{k=0}^{\infty}{f^{\prime}}_{ik}^{(m)}}\rightarrow 0$

Thus, $\lim_{m\to\infty}{\overline{g}}_{i}^{(n,m)}=0$ for any sequence
$\\{i_{m}\\}$ and therefore $\mathrm{\Delta}^{(n,cl,\infty)}=\\{\langle
0,0,\ldots\rangle\\}$ for each $n\in\mathbb{N}_{0}$. ∎

###  7.2. Reducible Bratteli diagrams with infinitely many odometers

We consider here a class of reducible non-stationary generalized Bratteli
diagrams $B=B_{IO}$ consisting of infinitely many odometers connected by
single edges. This class of diagrams was first considered in [BKK24], where
the authors used the procedure of measure extension from a subdiagram to
obtain results concerning the number of ergodic tail invariant measures. Here
we recall the obtained results and show how apply methods developed in Section
4 to these diagrams. For more results concerning tail invariant measures and
Vershik maps for reducible generalized Bratteli diagrams with infinitely many
odometers see [BKK24].

Let the generalized Bratteli diagram $B=B_{IO}$ be defined by the sequence of
incidence matrices

(7.2)
${F}_{n}^{{}^{\prime}}=\left[\begin{array}[]{llllllll}a_{n}^{(1)}&1&0&0&\ldots&\ldots&\ldots&\ldots\\\
0&a_{n}^{(2)}&1&0&0&\ldots&\ldots&\ldots\\\
0&0&a_{n}^{(3)}&1&0&0&\ldots&\ldots\\\ 0&0&0&a_{n}^{(4)}&1&0&0&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\end{array}\right],\qquad n\in\mathbb{N}_{0},$

where the natural numbers $a_{n}^{(i)}\geq 2$ for all $n,i$. The index $n$
points out at the $n$-th level of the diagram $B$, and $i$ indicates the
odometer supported by the $i$-vertex in each level, see Figure 2 where a part
of the diagram between levels $V_{n}$ and $V_{n+1}$ is shown ($a_{n}^{(i)}$
indicates the number of vertical edges).

Figure 2. The diagram $B_{IO}$.

The diagram $B_{IO}$ has a natural set of elementary vertex subdiagrams
$\overline{B}(i)$ consisting of vertical odometers where $i$ runs the set
$\mathbb{N}$. There are exactly $a_{n}^{(i)}$ edges connecting the vertices
$i\in V_{n}$ and $i\in V_{n+1}$. The subdiagram $\overline{B}(i)$ of $B$
admits a unique tail invariant probability measure $\overline{\mu}(i)$ on the
path space $X_{\overline{B}(i)}$ such that for a cylinder set
$[\overline{e}]=[e_{1},\ldots,e_{n}],s(e_{j})=r(e_{j})=i$,

$\overline{\mu}(i)([\overline{e}])=\frac{1}{a^{(i)}_{1}\ \cdots\
a^{(i)}_{n}}.$

The measure extension procedure applied to $\overline{B}(i)$ gives us the
measure $\widehat{\overline{\mu}}(i)$ on the tail invariant set
$\widehat{X}_{\overline{B}(i)}$. It follows from Theorem 3.13 that

$\widehat{\overline{\mu}}(i)(\widehat{X}_{\overline{B}(i)})<\infty\
\Longleftrightarrow\ \sum_{n=1}^{\infty}\frac{H^{(n)}_{i+1}}{a^{(i)}_{1}\
\cdots\ a^{(i)}_{n}}<\infty.$

Thus, it follows from the construction of $B_{IO}$ that there are infinitely
many ergodic measures $\widehat{\overline{\mu}}(i)$ on the path space $X_{B}$.
Some of them may be finite the others are infinite. We will give an example
below, for more examples see [BKK24]. Moreover, the measures
$\widehat{\overline{\mu}}(i)$ and $\widehat{\overline{\mu}}(j)$ are mutually
singular ($i\neq j)$ because they are supported by non-intersecting tail
invariant sets $\widehat{X}_{\overline{B}(i)}$ and
$\widehat{X}_{\overline{B}(j)}$. Our goal is to show that there are no other
ergodic measures.

######  Remark 7.3.

Let $\theta$ be a finite tail invariant measure on
$\widehat{X}_{\overline{B}(i)}$. Then there is a constant $C$ such that
$\theta=C\widehat{\overline{\mu}}(i)$.

Indeed, let $C=\theta(X_{\overline{B}(i)})$, then
$\overline{\theta}:=\theta|_{X_{\overline{B}(i)}}=C\overline{\mu}(i)$. By tail
invariance,

$\theta([e_{0},\ldots,e_{n}])=\frac{C}{a^{(i)}_{1}\ \cdots\
a^{(i)}_{n}}=C\overline{\mu}(i)([e_{0},\ldots,e_{n}]).$

######  Theorem 7.4.

Let $\mathcal{M}$ be the family of measures $\widehat{\overline{\mu}}(i)$ such
that $\widehat{\overline{\mu}}(i)(\widehat{X}_{\overline{B}(i)})<\infty$.
Then, after normalization, $\mathcal{M}$ coincides with the set of all ergodic
probability tail invariant measures on the path space $X_{B}$ of the diagram
$B$.

######  Remark 7.5.

(i) The proof of Theorem 7.4 can be found in [BKK24], and it contains a
version of the famous Rokhlin theorem about a canonical system of measures
associated with a measurable partition. Note that $X_{B}$ is partitioned into
sets $\widehat{X}_{\overline{B}(i)}$ for $i=1,2,\ldots$ Thus, for any
probability ergodic invariant measure $\mu$ on $X_{B}$, we have
$\mu(\widehat{X}_{\overline{B}(i)})=1$ for some $i$. Hence by Remark 7.3, we
have

$\mu=\frac{\widehat{\overline{\mu}}(i)}{\widehat{\overline{\mu}}(i)\left(\widehat{X}_{\overline{B}(i)}\right)}.$

(ii) The result of Theorem 7.4 can be obtained in the case when the vertical
odometers are replaced with simple stationary standard Bratteli diagrams
$\overline{B}_{i}$ with the incidence matrix $\overline{F}_{i}$. As for
odometers, we will have a unique ergodic probability measure
$\overline{\mu}_{i}$ on the path space $X_{\overline{B}_{i}}$. The measure
$\overline{\mu}_{i}$ is completely determined by the values on cylinder sets
$\overline{\mu}_{i}([\overline{e}])=\frac{\xi_{v}}{\lambda^{n}}$ where
$\xi=(\xi_{v})$ is the Perron-Frobenius eigenvector,
$\xi\overline{F}_{i}=\lambda\xi$ and $r(\overline{e})=v\in V_{n}$, see, e.g.
[BKMS10]. Assuming that the extension
$\widehat{\overline{\mu}}_{i}(\widehat{X}_{\overline{B}_{i}})$ is finite, we
get that this measure is a unique ergodic measure (up to a constant). The same
arguments as in the proof of Theorem 7.4 can be repeated.

(iii) For the Bratteli diagram $B_{IO}$, the set of limit points
$L(\\{\overline{g}_{v}:v\in V_{n+m},m\in\mathbb{N}\\})$ coincides with
$\\{\lim_{m\to\infty}\overline{g}^{(n,m)}_{i_{m}}\\}$. Since the incidence
matrices of the diagram are upper triangular, we get that the limit of
$\overline{g}^{(n,m)}_{i_{m}}$ is zero when $i_{m}\rightarrow\infty$ as
$m\rightarrow\infty$. This means that we should only consider the sequences
$\\{i_{m}\\}$ such that $i_{m}=i$. Denote by $\nu(i)$ a probability measure
determined by the limit $\lim_{m\to\infty}\overline{g}^{(n,m)}_{i}$. Then it
can be shown that the following fact holds:

Claim. If for some $i$ the extension of the measure $\overline{\mu}(i)$ is
finite, then $\nu(i)=c_{i}\widehat{\overline{\mu}}(i)$ for a constant $c_{i}$.

From Theorem 7.4 it follows, that $B_{IO}$ can have not more than countably
many probability ergodic invariant measures. We formulate here a statement
that was proved in [BKK24].

######  Theorem 7.6.

[BKK24] A stationary generalized Bratteli diagram with infinitely many
odometers can have only finitely many probability ergodic invariant measures.
In particular, one can find diagrams which (i) have a unique probability
ergodic invariant measure, (ii) have no probability invariant measure, but
possess an infinite $\sigma$-finite invariant measure that takes finite values
on all cylinder sets, and (iii) have no invariant measure that takes finite
values on all cylinder sets. A non-stationary generalized Bratteli diagram
with infinitely many odometers can have countably infinitely many probability
ergodic invariant measures.

Consider a non-stationary generalized Bratteli diagram $B$ defined by a
sequence of natural numbers $\\{a_{i}:i\in\mathbb{N}_{0}\\}$. Without loss of
generality, we can assume that $a_{n}\geq 2$ for all $n$. The diagram $B$
consists of an infinite sequence of non-stationary odometers connected with
the neighboring odometer by single edges. More precisely, let
$V_{n}=\mathbb{N}$ and the incidence matrices $F^{{}^{\prime}}_{n}$ has the
form

(7.3)
${F}_{n}^{{}^{\prime}}=\left[\begin{array}[]{llllllll}a_{n}&1&0&0&\ldots&\ldots&\ldots&\ldots\\\
0&a_{n}&1&0&0&\ldots&\ldots&\ldots\\\ 0&0&a_{n}&1&0&0&\ldots&\ldots\\\
0&0&0&a_{n}&1&0&0&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\end{array}\right],\qquad n\in\mathbb{N}_{0}.$

Fix $i\geq 1$, set $W_{n}=\left\\{i\right\\}$, $n=1,2,\ldots$ and define the
subdiagram ${\overline{B}}_{(i)}=(\overline{W},{\overline{E}})$ as above with
the only difference that the set $\overline{E}_{n}$ is formed now by $a_{n}$
edges connecting the vertices $i\in V_{n}$ and $i\in V_{n+1}$. The unique tail
invariant probability measure $\overline{\mu}=\overline{\mu}(i)$ on
${\overline{B}}_{(i)}=(\overline{W},{\overline{E}})$ is given by the formula
$\overline{\mu}([\overline{e}])=\dfrac{1}{a_{0}\ \cdots\ a_{n}}$ where
$r(\overline{e})=i\in V_{n+1}$.

By definition of the diagram, $H^{(n)}=H^{(n)}_{i}$ for all $n$ and $i$ (as
usual, we set $H^{(0)}_{i}=1$). Then
$H^{(n+1)}=H_{i}^{(n+1)}=(a_{n}+1)H^{(n)}$ which implies that

$H^{(n+1)}=(a_{0}+1)\ \cdots\ (a_{n}+1),\quad n\in\mathbb{N}_{0}.$

The proof of the following statement can be found in [BKK24].

######  Proposition 7.7.

Let the sequence $(a_{n})$ be such that $\sum_{n}a_{n}^{-1}<\infty$. Then, for
every $i$, the extension $\widehat{\overline{\mu}}(i)$ is finite. The set of
all ergodic finite tail invariant measures on the path space of the diagram
$B$ is formed by $\\{\widehat{\overline{\mu}}(i)\ :i\in\mathbb{N}\\}$.

The next statement is, in some sense, the converse to Proposition 7.7.

######  Proposition 7.8.

Let the Bratteli diagram $B$ be defined by the sequence of incidence matrices
$F^{\prime}_{n}$, see (7.3). If the series $\sum_{n\geq 1}a_{n}^{-1}$
diverges, then the diagram does not admit finite tail invariant measures.

###### Proof.

The following formulas hold for the diagram $B$ (they can be easily proved by
induction taking into account that the height of a tower does not depend on
the vertex):

$H^{(n)}_{i}=(a_{1}+1)\ \cdots\ (a_{n-1}+1),\quad i\in\mathbb{N}.$

The entries of the matrix ${G^{\prime}}^{(n,m)}$ can be found as follows:

${g^{\prime}}^{(n,m)}_{ii}=a_{n}\ \cdots\ a_{n+m-1}$

and, for $j=i+1,\ldots i+m$,

${g^{\prime}}^{(n,m)}_{ij}=(a_{n}\ \cdots\
a_{n+m-1})\left[\sum_{s_{1}<s_{2}<\cdots<s_{j-i}}\frac{1}{a_{s_{1}}\ \cdots\
a_{s_{j-i}}}\right]$

where $n\leq s_{1}<\cdots<s_{j-i}\leq n+m-1$. Clearly,
${g^{\prime}}^{(n,m)}_{ij}=0$ for other $i,j$. Now we can find the entries of
the stochastic matrix ${G}^{(n,m)}$: if $j=i+1,\ldots,i+m$, then

${g}^{(n,m)}_{ij}={g^{\prime}}^{(n,m)}_{ij}\frac{H^{(n)}_{j}}{H^{(n+m)}_{i}}=\left[\frac{a_{n}}{1+a_{n}}\
\cdots\
\frac{a_{n+m-1}}{1+a_{n+m-1}}\right]\cdot\left[\sum_{s_{1}<s_{2}<\cdots<s_{j-i}}\frac{1}{a_{s_{1}}\
\cdots\ a_{s_{j-i}}}\right]$

and

${g}^{(n,m)}_{ii}=\frac{a_{n}}{1+a_{n}}\ \cdots\
\frac{a_{n+m-1}}{1+a_{n+m-1}}$

Assume now that the set of tail invariant probability measures $M_{1}(B)$ is
not empty. Then, by Theorem 4.9, there is some $i$ such that the sequence of
vectors $\\{{\overline{g}}_{i}^{(n,m)}\\}$ has the limit as $m\to\infty$. This
means that

$\lim_{m\to\infty}{g}^{(n,m)}_{ii}=\lim_{m\to\infty}\frac{a_{n}}{1+a_{n}}\
\cdots\ \frac{a_{n+m-1}}{1+a_{n+m-1}}$

exists. The latter is equivalent to the convergence of the series $\sum_{n\geq
1}a_{n}^{-1}$. This is a contradiction. ∎

The corresponding stochastic matrices $F_{n}$’s have the form

${F}_{n}=\left[\begin{array}[]{llllllll}\frac{a_{n}}{a_{n}+1}&\frac{1}{a_{n}+1}&0&0&\ldots&\ldots&\ldots&\ldots\\\
0&\frac{a_{n}}{a_{n}+1}&\frac{1}{a_{n}+1}&0&0&\ldots&\ldots&\ldots\\\
0&0&\frac{a_{n}}{a_{n}+1}&\frac{1}{a_{n}+1}&0&0&\ldots&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\end{array}\right],$

i.e., $f_{ii}^{(n)}=\frac{a_{n}}{(a_{n}+1)}$,
$f^{(n)}_{i(i+1)}=\frac{1}{(a_{n}+1)}$, ${f}^{(n)}_{ij}=0$ for $j\neq i,i+1$,
$i\in\mathbb{N}$. Let $\overline{g}_{i}=\left\langle f^{(n)}_{ij}:\
j\in\mathbb{N}\right\rangle$ be the $i$-th row of the matrix $F_{n}$. Then
$|{\overline{g}}_{i}|\leq\frac{1}{2^{a(i)}}+\frac{1}{2^{a(i+1)}}\to 0$ as
$i\to\infty$ where $a(i)$ is an enumeration of vertices in $V_{n}$. It follows
from Theorem 2.10 that the linear maps
$F_{n}^{T}\colon\Delta_{1}\to\Delta_{1}$ are continuous.

We observe that the above statement is true in a more general situation when
the entries of an incidence matrix $F$ under its main diagonal are zeros.

##  8\. Uncountably many ergodic probability tail invariant measures

This section studies a class of reducible generalized Bratteli diagrams
$B_{\infty}$ whose incidence matrices are triangular. It turns out that such
diagrams have uncountably many ergodic probability tail invariant measures. We
also consider subdiagrams of $B_{\infty}$, standard and generalized, and
answer the questions about internal tail invariant measures on such
subdiagrams and the finiteness of their extensions.

###  8.1. Triangular generalized Bratteli diagram $B_{\infty}$

We define the diagram $B_{\infty}=(V,E)$ by taking $V_{n}=\mathbb{N}$ for all
$n\in\mathbb{N}$ so that each vertex $v\in V_{n}$ can be written as $(n,i)$.
For $v=(n,i)\in V_{n}$, an edge $e(v,u)$ where $u\in V_{n+1}$, exists whenever
$u=(n+1,j),\ j=i,i+1,\ldots$. The incidence matrices $F^{\prime}_{n}$ are the
same for all levels, $F^{\prime}_{n}=F^{\prime}$, where

$F^{\prime}=\left[\begin{array}[]{ccccccc}1&0&0&0&\ldots&\ldots&\ldots\\\
1&1&0&0&0&\ldots&\ldots\\\ 1&1&1&0&0&0&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\ \end{array}\right].$

The stochastic matrix $F$ obtained from the matrix $F^{\prime}$ has a form

$F=\left[\begin{array}[]{llllllll}1&0&0&0&\ldots&\ldots&\ldots&\ldots\\\
\frac{1}{2}&\frac{1}{2}&0&0&0&\ldots&\ldots&\ldots\\\
\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0&0&\ldots&\ldots&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
\end{array}\right].$

We claim that the matrix $F$ satisfies the assumption of Theorem 2.10. Indeed,
the $i$-th row
${\overline{g}}_{i}=\left\langle\frac{1}{i},\ldots,\frac{1}{i},0,\ldots\right\rangle$
of $F$ satisfies the relation:
$|{\overline{g}}_{i}|\leq\left[\frac{1}{2^{a(1)}}+\frac{1}{2^{a(2)}}+\ldots\right]\cdot\frac{1}{i}\leq\frac{2}{i}\to
0$, as $i\to\infty$ where $a(\cdot)$ is an enumeration of the vertices. Thus,
${G^{\prime}}^{(n,m)}={F^{\prime}}^{m}=\\{f_{ij}^{(m)}i,j\in\mathbb{N}\\}$ for
all $n\geq 1$, and these matrices generate continuous mappings.

To find the matrices ${F^{\prime}}^{n}=\left({f^{\prime}}_{ij}^{n}\
:i,j=1,2,\ldots\right)$, we will use the numbers the $S_{i}^{(k)},\
k\in\mathbb{N}_{0},\ i\in\mathbb{N}$, which are defined as follows:

(8.1) $S_{i}^{(0)}=1,\quad S_{i}^{(k+1)}=S_{1}^{(k)}+\ldots+S_{i}^{(k)}$

The numbers $S_{i}^{(k)}$ are used in the Cesaro summability method, and it is
known [Har92] that they are

(8.2) $S_{i}^{(k)}=\begin{pmatrix}i+k-1\\\ k\\\ \end{pmatrix}.$

Using the induction it is not hard to prove that
${f^{\prime}}_{ij}^{(n)}=S_{i-j+1}^{(n-1)}$ whenever $j=1,\ldots,i$ and
${f^{\prime}}_{ij}^{(n)}=0$, otherwise ($i=1,2,\ldots,\ n=1,2,\ldots$). Thus,

(8.3)
${F^{\prime}}^{(m)}=\left[\begin{array}[]{ccccccc}S_{1}^{(m-1)}&0&0&0&\ldots&0&\ldots\\\
S_{2}^{(m-1)}&S_{1}^{(m-1)}&0&0&\ldots&0&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
S_{k}^{(m-1)}&S_{k-1}^{(m-1)}&S_{k-2}^{(m-1)}&\ldots&\ldots&S_{1}^{(m-1)}&\ldots\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\ \end{array}\right],$

It follows from (8.3) that

(8.4) $H_{i}^{(n)}=S_{i}^{(n-1)}=\begin{pmatrix}i+n-2\\\ n-1\\\
\end{pmatrix},\quad n,i\in\mathbb{N},$

and we find that

$g_{ij}^{(n,m)}=\frac{{f^{\prime}}_{ij}^{m}H_{j}^{(n)}}{\sum_{s=1}^{i}{{f^{\prime}}_{is}^{(m)}H_{s}^{(n)}}}=\frac{S_{i-j+1}^{(m-1)}H_{j}^{(n)}}{\sum_{s=1}^{i}{S_{i-s+1}^{(m-1)}H_{s}^{(n)}}}$

if $j=1,\ldots i$ and $g_{ij}^{(n,m)}=0$ otherwise.

To describe the set $\Delta^{(n,\infty,cl)}$, we use the method developed in
Theorem 4.9. For this, we have first to find the set of all limit points
$\overline{y}^{(n,\infty)}$ of the vectors

$\overline{y}_{i}^{(n,m)}=\left\langle{y}_{ij}^{(n,m)}=\frac{{f^{\prime}}_{ij}^{(m)}}{\sum_{s=1}^{\infty}{f^{\prime}}_{is}^{(m)}}\
:j=1,2,\ldots\right\rangle,\quad i\in\mathbb{N},$

when $m\to\infty$.

######  Lemma 8.1.

The set of limits of the vectors ${\overline{{y}}}_{i}^{(n,m)}$ is formed by
the vectors

$\overline{{y}}_{a}^{(n,\infty)}=\lim_{m\to\infty}\overline{{y}}_{i_{m}}^{(n,m)}=\left\langle\frac{1}{(a+1)},\frac{a}{{(a+1)}^{2}},\frac{a^{2}}{{(a+1)}^{3}},\ldots\right\rangle,\quad
0\leq a<\infty.$

###### Proof.

Recall that we have ${f^{\prime}}_{ij}^{(m)}=S_{i-j+1}^{(m-1)}$ if
$j=1,\ldots,i$ and ${f^{\prime}}_{ij}^{(m)}=0$ if $j>i$. Moreover,

$\sum_{s=1}^{\infty}{f^{\prime}}_{is}^{(m)}=\sum_{s=1}^{i}S_{i-s+1}^{(m-1)}=S_{i}^{(m)}.$

Let $\overline{{y}}=\lim_{m\to\infty}{\overline{{y}}}_{i}^{(n,m)},\ i=i_{m}$.
Taking a subsequence of $\\{m\\}$, we can assume that
$\dfrac{i_{m}}{m}\rightarrow a$ for some $0\leq a<\infty$. Fix $j\geq 1$.
Since $i_{m}\rightarrow\infty$, we can assume that $i_{m}>j$. Compute for
$j=2,\ldots,i$

${y}_{ij}^{(n,m)}=\frac{S_{i-j+1}^{(m-1)}}{S_{i}^{(m)}}=\frac{\begin{pmatrix}i+m-j-1\\\
m-1\\\ \end{pmatrix}\ }{\begin{pmatrix}i+m-1\\\ m\\\
\end{pmatrix}}=\left[\frac{(i+m-j-1)!}{(m-1)!\cdot(i-j)!}\right]\cdot\left[\frac{m!\cdot(i-1)!}{(i+m-1)!}\right]$
$=\frac{m\cdot(i-j+1)\cdot\ldots\cdot(i-1)}{(i+m-j)\cdot\ldots\cdot(i+m-1)}=\frac{\left[\frac{(i-j+1)}{m}\cdot\ldots\cdot\frac{(i-1)}{m}\right]}{\left[\frac{(i+m-j)}{m}\cdot\ldots\cdot\frac{(i+m-1)}{m}\right]}$

If $j=1$, we have ${y}_{i1}^{(n,m)}=\dfrac{m}{i+m-1}$. Taking the limit in the
formulas for ${y}_{ij}^{(n,m)}$ as $m\to\infty$, we get that

$\lim_{m\to\infty}{y}_{ij}^{(n,m)}=\lim_{m\to\infty}{\frac{\left[\frac{(i-j+1)}{m}\cdot\ldots\cdot\frac{(i-1)}{m}\right]}{\left[\frac{(i+m-j)}{m}\cdot\ldots\cdot\frac{(i+m-1)}{m}\right]}}=\frac{a^{j-1}}{{(a+1)}^{j}}.$

If the sequence $\\{i_{m}\\}$ is bounded (we can assume $i_{m}=i$), then
$a=0$, and in this case ${\overline{p}}_{0}^{(n,\infty)}=\langle
1,0,0,\ldots\rangle$.

This proves the lemma. ∎

Following the method described in Section 4, we need to find the limit set
$L^{(n)}(\\{\overline{g}_{v}\\})$ in terms of the vectors
$\overline{{y}}_{a}^{(n,\infty)}$.

######  Lemma 8.2.

The vectors $\overline{{y}}_{a}^{(n,\infty)}$ belong to the set $P^{(n)}$ and
satisfy conditions (4.17) (defined in Theorem 4.9) and (4.18). That is

(8.5) $H(n,a):=\sum_{u\in V_{n}}{y}_{u}^{(n,\infty)}H_{u}^{(n)}<\infty$

and

(8.6) $\lim_{m\to\infty}\sum_{u\in
V_{n}}\left[\frac{{g^{\prime}}^{(m,n)}_{vu}}{\sum_{w\in
V_{n}}{g^{\prime}}^{(m,n)}_{vw}}\right]H_{u}^{(n)}=H(n,a).$

###### Proof.

We first show that, for all $n\geq 1$,

(8.7) $H(n,a)=(1+a)^{n-1}.$

Note that by Lemma 8.1

$H(n,a)=\sum_{j=1}^{\infty}\frac{a^{j-1}}{(1+a)^{j}}H_{j}^{(n)}.$

For $n=1$, $H_{j}^{(1)}=1$ and

$H(1,a)=\frac{1}{1+a}\sum_{j=1}^{\infty}\left(\frac{a}{1+a}\right)^{j}=1.$

In the following computation, we show that $H(n,a)$ is finite. This fact
proves that (8.5) holds.

$\displaystyle H(n,a)=$ $\displaystyle\
\sum_{j=1}^{\infty}\frac{a^{j-1}}{(1+a)^{j}}H_{j}^{(n)}=\sum_{j=1}^{\infty}\frac{a^{j-1}}{(1+a)^{j}}S_{j}^{(n-1)}$
$\displaystyle=$ $\displaystyle\
\sum_{j=1}^{\infty}\frac{a^{j-1}}{(a+1)^{j}}\left(\begin{array}[]{c}j+n-2\\\
n-1\\\
\end{array}\right)=\frac{1}{(n-1)!}\cdot\sum_{j=1}^{\infty}\frac{a^{j-1}}{(a+1)^{j}}\frac{(j+n-2)!}{(j-1)!}$
$\displaystyle=$ $\displaystyle\
\frac{1}{(n-1)!}\cdot\sum_{j=1}^{\infty}\frac{a^{j-1}}{(a+1)^{j}}\cdot\left[j(j+1)\cdot\ldots\cdot(j+n-2)\right]<\infty.$

Next, we prove that (8.7) holds.

$\displaystyle H(n+1,a)=$ $\displaystyle\
\sum_{j=1}^{\infty}\frac{a^{j-1}}{(1+a)^{j}}H_{j}^{(n+1)}=\sum_{j=1}^{\infty}\frac{a^{j-1}}{(1+a)^{j}}\sum_{l=1}^{j}H_{l}^{(n)}$
$\displaystyle=$ $\displaystyle\
\sum_{l=1}^{\infty}H_{l}^{(n)}\sum_{j=l}^{\infty}\frac{a^{j-1}}{(1+a)^{j}}=\sum_{l=1}^{\infty}H_{l}^{(n)}\left(\frac{a}{1+a}\right)^{l-1}$
$\displaystyle=$ $\displaystyle\ (1+a)H(n,a),$

and (8.7) follows.

It remains to show that condition (8.6) is satisfied. It was proved in Lemma
8.1 that

(8.8)
$\sum_{j=1}^{i}{y}_{ij}^{(n,m)}H_{j}^{(n)}={y}_{i,1}^{(n,m)}+\sum_{j=2}^{i}H_{j}^{(n)}\frac{\frac{i-j+1}{m}\
\cdots\ \frac{i-1}{m}}{\frac{i+m-j}{m}\ \cdots\ \frac{i+m-1}{m}}$

Fix $k\geq 1$ and take $i=i_{m}>k$ as $m\to\infty$. Then

$\sum_{j=1}^{i}{y}_{ij}^{(n,m)}H_{j}^{(n)}>\frac{m}{i+m-1}+\sum_{j=2}^{k}H_{j}^{(n)}\frac{\frac{i-j+1}{m}\
\cdots\ \frac{i-1}{m}}{\frac{i+m-j}{m}\ \cdots\ \frac{i+m-1}{m}}$

As $m\to\infty$ (remember that $\dfrac{i_{m}}{m}\to a$), we get from the above
inequality that for every $k\in\mathbb{N}$

$\liminf_{m\to\infty}\sum_{j=1}^{i}{p_{ij}^{(n,m)}H_{j}^{(n)}}\geq\sum_{j=1}^{k}\frac{a^{j-1}}{(a+1)^{j}}H_{j}^{(n)}.$

Hence

(8.9)
$\liminf_{m\to\infty}\sum_{j=1}^{i}{{y}_{ij}^{(n,m)}H_{j}^{(n)}}\geq\sum_{j=1}^{\infty}\frac{a^{j-1}}{(a+1)^{j}}H_{j}^{(n)}=H(n,a)=(a+1)^{n-1}.$

Take $b>a$ (then $\frac{a}{a+1}<\frac{b}{b+1}$). It follows from
$\frac{i_{m}}{m}\to a$ that, for sufficiently large $m$,

$\frac{\frac{i-l}{m}}{\frac{i-l}{m}+1}<\frac{b}{b+1},\quad l=0,1,\ldots j-1,\
i=i_{m}.$

Therefore (8.8) can be estimated from above

$\sum_{j=1}^{i}{y}_{ij}^{(n,m)}H_{j}^{(n)}<\frac{m}{i+m-1}+\sum_{j=2}^{i}H_{j}^{(n)}\frac{m}{m+i-j}\left(\frac{b}{1+b}\right)^{j-1}$

Taking the limit when $m\to\infty$, we obtain the inequality

$\limsup_{m\to\infty}\
\sum_{j=1}^{i}{y}_{ij}^{(n,m)}H_{j}^{(n)}\leq\frac{1}{1+a}+\sum_{j=2}^{\infty}H_{j}^{(n)}\frac{1}{1+a}\left(\frac{b}{1+b}\right)^{j-1}.$

Since this relation holds for all $b>a$, we can deduce that

(8.10) $\displaystyle\limsup_{m\to\infty}\
\sum_{j=1}^{i}{y}_{ij}^{(n,m)}H_{j}^{(n)}\leq$ $\displaystyle\
\frac{1}{1+a}+\lim_{b\to
a}\left(\sum_{j=2}^{\infty}H_{j}^{(n)}\frac{1}{1+a}\left(\frac{b}{1+b}\right)^{j-1}\right)$
$\displaystyle=$ $\displaystyle\
\frac{1}{1+a}+\sum_{j=2}^{\infty}H_{j}^{(n)}\frac{a^{j-1}}{(1+a)^{j}}$
$\displaystyle=$ $\displaystyle\ H(n,a)$ $\displaystyle=$ $\displaystyle\
(a+1)^{n-1}.$

It follows from (8.9) and (8.10) that

$\lim_{m\to\infty}\
\sum_{j=1}^{i}{y}_{ij}^{(n,m)}H_{j}^{(n)}=(a+1)^{n-1}=\sum_{j=1}^{\infty}y_{ij}^{(n,m)}H_{j}^{(n)},$

and (8.6) is proved. ∎

Let the vectors $\overline{q}_{{a}}^{(n,\infty)}=\langle
q_{{a,i}}^{(n,\infty)}:\ i\in V_{n}\rangle$ be defined as in Theorem 4.9, see
(4.19), and let $\overline{p}_{a}^{(n,\infty)}=\left\langle
p_{a,i}^{(n,\infty)}\ |\ i\in V_{n}\right\rangle$ be such that
$q_{a,i}^{(n,\infty)}=H_{i}^{(n)}p_{a,i}^{(n,\infty)},\ n,i=1,2,\ldots$.

Then we can find the entries of $\overline{q}_{a}^{(n,\infty)}$:

$\displaystyle q_{a,i}^{(n,\infty)}=$ $\displaystyle\frac{1}{\sum_{j\geq
1}p^{(n,\infty)}_{j}H_{j}^{(n)}}\cdot p^{(n,\infty)}_{i}H_{i}^{(n)}$
$\displaystyle=$ $\displaystyle\
\frac{1}{(a+1)^{n-1}}\frac{a^{i-1}}{(a+1)^{i}}S_{i}^{(n-1)}$ $\displaystyle=$
$\displaystyle\ \frac{a^{i-1}}{(a+1)^{n+i-1}}\left(\begin{array}[]{c}n+i-2\\\
n-1\end{array}\right),\qquad i\geq 1.$

Thus, we proved that

(8.11) $\overline{q}_{a}^{(n,\infty)}=\left\langle\frac{1}{{(a+1)}^{n}},\
\frac{a}{{(a+1)}^{n+1}}\left(\begin{array}[]{c}n\\\
1\end{array}\right),\frac{a^{2}}{{(a+1)}^{n+2}}\left(\begin{array}[]{c}n+1\\\
2\end{array}\right),\ \ldots\right\rangle$

According to Theorem 4.9, we get that

$L^{(n)}(\\{\overline{g}_{i}^{(n,m)}:i\in
V_{n+m},m\in\mathbb{N}\\})=\\{\overline{q}_{a}^{(n,\infty)}:\ 0\leq
a<\infty\\}.$

It follows from (8.11) that

$\overline{p}_{a}^{(n,\infty)}=\left\langle\frac{1}{(a+1)^{n}},\frac{a}{(a+1)^{n+1}},\frac{a^{2}}{(a+1)^{n+2}},\ldots\right\rangle=\left\langle\frac{a^{(i-1)}}{H(n,a)\cdot(a+1)^{i}}\
|\ i\in\mathbb{N}\right\rangle,$

because $H_{i}^{(n)}=\left(\begin{array}[]{c}i+n-2\\\
n-1\end{array}\right)=\left(\begin{array}[]{c}i+n-2\\\ i-1\end{array}\right),\
n,i\in\mathbb{N}.$

######  Proposition 8.3.

For every $0\leq a<\infty$, the sequence of vectors
$\\{\overline{q}_{{a}}^{(n,\infty)}:\ n\in\mathbb{N}\\}$ (or
$\\{\overline{p}_{a}^{(n,\infty)}:\ n\in\mathbb{N}\\}$) determines uniquely a
tail invariant probability measure $\mu_{a}$ on the path space of the Bratteli
diagram $B_{\infty}=(V,E)$.

###### Proof.

It suffices to show that

(8.12)
${F^{\prime}}^{T}(\overline{p}_{a}^{(n+1,\infty)})=\overline{p}_{a}^{(n,\infty)},\quad
n\in\mathbb{N}.$

We compute

$\displaystyle{F^{\prime}}^{T}\left\langle\frac{a^{i-1}}{H(n+1,a)(a+1)^{i}}:\
i\in\mathbb{N}\right\rangle=$ $\displaystyle\
\frac{1}{H(n+1,a)}\left\langle\sum_{i=j}^{\infty}\frac{a^{i}}{(a+1)^{i+1}}:\
j=0,1,2,\ldots\right\rangle$ $\displaystyle=$ $\displaystyle\
\frac{1}{H(n+1,a)}\left\langle\frac{a^{j}}{(a+1)^{j}}:\
j=0,1,2,\ldots\right\rangle$ $\displaystyle=$ $\displaystyle\
\frac{1}{H(n,a)}\left\langle\frac{a^{j}}{(a+1)^{j+1}}:\
j=0,1,2,\ldots\right\rangle$ $\displaystyle=$ $\displaystyle\
{\overline{p}_{a}^{(n,\infty)}}.$

Thus, we have

$\mu_{a}([\overline{e}])=\frac{a^{j-1}}{H(n,a)(a+1)^{j}}.$

and

$\mu_{a}(X^{(n)}_{j})=\frac{a^{j-1}}{H(n,a)(a+1)^{j}}H^{(n)}_{j}=\frac{a^{j-1}}{(a+1)^{n+j-1}}\left(\begin{array}[]{c}n+j-2\\\
n-1\end{array}\right).$

One can easily check that $\mu_{a}$ is a probability measure since, by
definition of $H(n,a)$, we have

$\sum_{j\geq 1}\mu_{a}(X^{(n)}_{j})=\frac{1}{H(n,a)}\sum_{j\geq
1}\frac{a^{j-1}}{(a+1)^{j}}H^{(n)}_{j}=1,\qquad n\in\mathbb{N}.$

∎

Our next goal is to show that every measure $\mu_{a}$ is ergodic. For this, we
need to use the notion of completely monotonic sequences and their relations
with the Hausdorff moment problem, see e.g. [Wid41], [Akh21].

Let $\overline{c}=\\{c_{i}:\ i\in\mathbb{N}\\}$ be a given sequence of
positive numbers. We define new sequences
$\Delta^{n}(\overline{c}),n\in\mathbb{N}_{0}$, where
$\Delta^{0}(\overline{c})=\overline{c}$ and the sequences
$\Delta^{1}(\overline{c}),\Delta^{2}(\overline{c}),\ldots$ (called the
sequences of differences of $\overline{c}$) are defined as follows:

$\Delta^{1}(\overline{c})=\left\\{c_{1}-c_{2},c_{2}-c_{3},c_{3}-c_{4},\ldots\right\\},$
$\Delta^{2}(\overline{c})=\Delta^{1}(\Delta^{1}(\overline{c})),$

and so on, $\Delta^{n}(\overline{c})=\Delta^{1}(\Delta^{n-1}(\overline{c}))$.

######  Remark 8.4.

In the literature, one can see another definition of the difference operator
$\delta$ acting on sequences $\overline{c}=\\{c_{n}\\}_{n}$: $\overline{c}$.
By definition, $\delta c_{n}=c_{n+1}-c_{n}$ and
$\delta^{k+1}(\overline{c})=\delta(\delta^{k}(\overline{c}))$. Then
$\Delta^{k}(\overline{c})=(-1)^{k}\delta^{k}(\overline{c})$.

The terms of the sequences $[\Delta^{k}(\overline{c})]_{i}$ can be found by
the following formulas:

$[\Delta^{2}\left(\overline{c}\right)]_{i}=c_{i}-2\cdot c_{i+1}+c_{i+2},$
$[\Delta^{3}(\overline{c})]_{i}=c_{i}-3c_{i+1}+3c_{i+2}-c_{i+3},$

and, in general,

$[\Delta^{k}(\overline{c})]_{i}=c_{i}-\left(\begin{array}[]{c}k\\\ 1\\\
\end{array}\right)c_{i+1}+\left(\begin{array}[]{c}k\\\ 2\\\
\end{array}\right)c_{i+2}-\ldots+(-1)^{k-1}\left(\begin{array}[]{c}k\\\ k-1\\\
\end{array}\right)c_{i+k-1}+(-1)^{k}c_{i+k},$

for all $i,k$.

It is said the sequence $\overline{c}$ is completely monotonic if
$[\Delta^{k}(\overline{c})]_{i}>0$ for each $k\in\mathbb{N}_{0}$ and
$i\in\mathbb{N}$.

Recall the following result proved by Hausdorff in [Hau21a], [Hau21b]

######  Theorem 8.5.

A sequence $\overline{c}=\\{c_{1},c_{2},c_{3},c_{4},\ldots\\}$ is completely
monotonic if and only if there exists a positive finite measure $\theta$ on
the interval $[0,1]$ such that, for every $i\in\mathbb{N}$,

$c_{i}=\int_{0}^{1}x^{i}\theta(dx).$

Moreover, there is a one-to-one correspondence between the set of all
completely monotonic sequences and the set of all finite positive measures on
the interval $[0,1]$.

Now, we can use a different characterization of the set $M_{1}(B_{\infty})$ of
tail invariant probability measures on the path space of the generalized
Bratteli diagram $B_{\infty}$.

Every measure $\mu\in M_{1}(B_{\infty})$ is uniquely determined by a sequence
of positive infinite vectors $\overline{p}^{(n,\infty)}=\langle
p^{(n,\infty)}_{i}:i\in\mathbb{N}\rangle$. Since $\mu$ is a probability
measure, the vector $\overline{p}^{(1,\infty)}$ is probability.

From (8.12) we get

(8.13) $p^{(n)}_{i}=p^{(n+1)}_{i}+p^{(n+1)}_{i+1}+\ldots.$

######  Lemma 8.6.

Every tail invariant probability measure $\mu\in M_{1}(B_{\infty})$ is
uniquely determined by a completely monotonic sequences
$\overline{p}=\left\langle p_{i}\ |\ i=1,2,\ldots\right\rangle$ in such a way
that, $p_{i}=p_{i}^{(1,\infty)},\ i=1,2,\ldots$. For each cylinder set
$[\overline{e}]$ such that $r(\overline{e})=i\in V_{n}$, we have
$\mu([\overline{e}])=[\Delta^{n-1}(\overline{p})]_{i}$, $n\in\mathbb{N}$.

###### Proof.

Fix some $n$, and show that $\overline{p}^{(n)}$ is a completely monotonic
sequence. Indeed, it follows from (8.13) that $[\Delta
p^{(n)}]_{i}=p^{(n+1)}_{i}>0$. For $\Delta^{2}$, we compute

$[\Delta^{2}p^{(n)}]_{i}=[\Delta p^{(n)}]_{i}-[\Delta
p^{(n)}]_{i+1}=p^{(n)}_{i}-2p^{(n)}_{i+1}+p^{(n)}_{i+2}=p^{(n+1)}_{i}-p^{(n+1)}_{i+1}=p^{(n+2)}_{i}>0.$

By induction, we deduce that
$\Delta^{k}(\overline{p}^{(n)})=\overline{p}^{(n+k)}$, or
$[\Delta^{k}p^{(n)}]_{i}=p^{(n+k)}_{i}>0$ for all natural numbers $n$ and $k$.
∎

It follows from Proposition 8.3 that the measure $\mu_{a}$ is determined by
the probability completely monotonic vectors (sequence)

${\overline{p}_{a}^{(1,\infty)}}=\left\langle\frac{1}{(a+1)},\frac{a}{{(a+1)}^{2}},\frac{a^{2}}{{(a+1)}^{3}},\ldots\right\rangle.$

######  Theorem 8.7.

The family of measures $\\{\mu_{a}:0<a<\infty\\}$ forms the set of all ergodic
probability tail invariant measures on the generalized Bratteli diagram
$B_{\infty}$.

###### Proof.

Assume that

(8.14) $\mu_{a}=\lambda\mu+(1-\lambda)\rho,$

where $0<\lambda<1$, and $\mu$ and $\rho$ are tail invariant probability
measures on $X_{B_{\infty}}$. It follows from Lemma 8.6 that $\mu$ and $\rho$
are determined by some completely monotonic positive vectors
$\overline{p}^{1}=\langle p^{(1)}_{1},p^{(1)}_{2}\ldots\rangle$ and
$\overline{w}^{1}=\langle w^{(1)}_{1},w^{(1)}_{2},\ldots\rangle$,
respectively. Then, we have

$\frac{a^{k-1}}{{(a+1)}^{k}}=\lambda p^{(1)}_{k}+(1-\lambda)w^{(1)}_{k}$

for each $k=1,2,\ldots$.

Use Theorem 8.5 and denote by $\theta_{a}$, $\theta_{\mu}$, and
$\theta_{\rho}$ the corresponding finite measures on the interval $[0,1]$
defined by the sequences ${\overline{y}}_{1}(a),\overline{p}^{(1)}$, and
$\overline{w}^{(1)}$. Then we have

$\frac{a^{k-1}}{{(a+1)}^{k}}=\int_{0}^{1}x^{k}\theta_{a}(dx),\ \
p^{(1)}_{k}=\int_{0}^{1}x^{k}\theta_{\mu}(dx),\ \
w^{(1)}_{k}=\int_{0}^{1}x^{k}\theta_{\rho}(dx),$

and for all $k\in\mathbb{N}$,

$\int_{0}^{1}x^{k}\theta_{a}(dx)=\lambda\int_{0}^{1}x^{k}\theta_{\mu}(dx)+(1-\lambda)\int_{0}^{1}x^{k}\theta_{\rho}(dx).$

This means that $\theta_{a}=\lambda\theta_{\mu}+(1-\lambda)\theta_{\rho}$. To
finish the proof, we observe that

$\theta_{a}=\frac{1}{a}\delta_{\frac{a}{a+1}}$

is a delta-measure concentrated at $\frac{a}{a+1}$. Indeed,

$\int_{0}^{1}x^{k}\theta_{a}(dx)=\frac{a^{k-1}}{{(a+1)}^{k}}=\frac{1}{a}\int_{0}^{1}x^{k}\delta_{\frac{a}{a+1}}(dx)$

for every $k$. Therefore, relation (8.14) is impossible, and this means that
$\mu_{a}$ is an ergodic measure. ∎

###  8.2. Subdiagrams of $B_{\infty}$

Now we present some natural subdiagrams of the generalized Bratteli diagram
$B_{\infty}$. Let $k$ be a fixed natural number; take the sequence of vertices
$W_{1}=\\{k\\},\ W_{2}=\\{k,k+1\\},\
\ldots,W_{n}=\\{k,k+1,\ldots,k+n-1\\},\ldots$. We will use this sequence to
define two subdiagrams of $B_{\infty}$. They are the vertex subdiagram
$\overline{B}(W,k)$ of $B_{\infty}$, supported by the sequence $\\{W_{n}\\}$
and an edge subdiagram $\overline{B}_{k}$ which is defined by the sequence of
incidence matrices $\overline{F}_{n}=(\overline{f}^{(n)}_{ij})$,
$n\in\mathbb{N}$, where

(8.15) $\overline{f}^{(n)}_{ij}=\begin{cases}1,&\textrm{if}\ j=i\ \textrm{or}\
j=i-1\ \textrm{and}\ i=k,\ldots,k+n-1\\\ 0,&\textrm{otherwise}\end{cases}$

We first consider the edge subdiagram $\overline{B}_{k}$. It is not hard to
see that, by definition, $\overline{B}_{k}$ is isomorphic to the classical
Pascal graph because for every $n$ the vertex $j\in W_{n}$ is the source for
exactly two edges connecting $j$ with the vertices $j$ and $j+1$ from
$W_{n+1}$.

It is well known (see, e.g. [MP05] or Section 5) that the path space
$X_{\overline{B}_{k}}$ of $\overline{B}_{k}$ supports uncountably many non-
atomic ergodic tail invariant measures $\nu_{p}$, $0<p<1$, such that for a
cylinder set $[\overline{e}]$ with the source in $k$ and range in $i\in
W_{n+1}$, we have

$\nu_{p}([\overline{e}])=p^{n+k-i}(1-p)^{i-k},$

where $i=k,k+1,\ldots,k+n$.

Our goal is to find out whether the measure extension $\widehat{\nu}_{p}$ on
$\widehat{X}_{\overline{B}_{k}}\subset X_{B_{\infty}}$ is finite or infinite.
For this, we use criterion (3.12) from Proposition 3.15. Denoting by
$F^{\prime}$ the incidence matrix of $B_{\infty}$, we observe that the entries
$\widetilde{f}^{(n)}_{ij}$ of the matrix
$\widetilde{F}_{n}=F^{\prime}-\overline{F}_{n}$ are identified with the edges
that were deleted from $B_{\infty}$ to produce the Pascal graph. It can be
seen that $\widetilde{f}^{(n)}_{ij}=1$ if and only if $i=k,\ldots k+n-1$ and
$1\leq j\leq i-2$. Hence, it follows from (3.12) that

(8.16) $\widehat{\nu}_{p}(\widehat{X}_{\overline{B}_{k}})<\infty\
\Longleftrightarrow\ I=\sum_{n\geq 1}\sum_{i\in V_{n+1}}\sum_{j\in
V_{n}}\widetilde{f}^{(n)}_{ij}H_{j}^{(n)}p^{n+k-i}(1-p)^{i-k}<\infty,$

where $H_{j}^{(n)}$ is the height of the $j$-th tower in $B_{\infty}$.

######  Proposition 8.8.

(1) The extension $\widehat{\nu}_{p}$ of the measure $\nu_{p}$ is infinite.

(2) The path space $\widehat{X}_{\overline{B}_{k}}$ has zero measure for any
ergodic tail invariant probability measure on $B_{\infty}$.

###### Proof.

(1) We will show that the series $I$ in (8.16) is divergent.

We fix some $n$ and, using (8.1), (8.4), and (8.2), obtain

$\displaystyle I_{n}:=$ $\displaystyle\ \sum\limits_{i\in
V_{n+1}}{\sum\limits_{j\in V_{n}}{\widetilde{f}}_{ij}^{(n)}}\
H_{j}^{(n)}{(1-p)}^{i-k}p^{n+k-i}$ $\displaystyle=$ $\displaystyle\
p^{n}H_{k}^{(n)}+\sum\limits_{i=k+1}^{k+n}p^{n+k-i}{(1-p)}^{i-k}\sum\limits_{j=1}^{i-2}H_{j}^{(n)}$
$\displaystyle=$ $\displaystyle\
p^{n}H_{k}^{(n)}+\sum\limits_{i=k+1}^{k+n}p^{n+k-i}(1-p)^{i-k}\cdot
H_{i-2}^{(n+1)}$ $\displaystyle=$ $\displaystyle\
p^{n}\left(\begin{array}[]{c}n+k-2\\\ n-1\\\
\end{array}\right)+\sum\limits_{i=k+1}^{k+n}p^{n+k-i}(1-p)^{i-k}\left(\begin{array}[]{c}n+i-3\\\
n\\\ \end{array}\right).$

Changing the index of summation, we get

$\displaystyle I_{n}=$ $\displaystyle\ p^{n}\left(\begin{array}[]{c}n+k-2\\\
n-1\\\
\end{array}\right)+\sum\limits_{i=0}^{n-1}p^{n-i-1}(1-p)^{i+1}\left(\begin{array}[]{c}n+i+k-2\\\
n\\\ \end{array}\right)$ $\displaystyle=$ $\displaystyle\
p^{n}\left(\begin{array}[]{c}n+k-2\\\ n-1\\\
\end{array}\right)+p^{n}\sum\limits_{i=0}^{n-1}\left(\frac{1-p}{p}\right)^{i+1}\left(\begin{array}[]{c}n+i+k-2\\\
n\\\ \end{array}\right).$

Then, the measure ${\widehat{\nu}}_{p}$ is infinite if the series

$I=\sum\limits_{n=1}^{\infty}p^{n}\cdot\left[\left(\begin{array}[]{c}n+k-2\\\
n-1\\\
\end{array}\right)+\sum\limits_{i=0}^{n-1}\left(\frac{1-p}{p}\right)^{i+1}\cdot\left(\begin{array}[]{c}n+i+k-2\\\
n\\\ \end{array}\right)\right]$

is divergent. For this, it suffices to show that

$J=\sum\limits_{n=1}^{\infty}p^{n}\cdot\left[\sum\limits_{i=0}^{n-1}\left(\frac{1-p}{p}\right)^{i+1}\left(\begin{array}[]{c}n+i+k-2\\\
n\\\ \end{array}\right)\right]=\infty$

The series $J$ can be represented in the following form and then estimated
from below:

$\displaystyle J=$ $\displaystyle\
\sum\limits_{i=0}^{\infty}\left(\frac{1-p}{p}\right)^{i+1}\sum_{n=i+1}^{\infty}p^{n}\left(\begin{array}[]{c}n+i+k-2\\\
n\\\ \end{array}\right)$ $\displaystyle=$ $\displaystyle\
\sum\limits_{i=0}^{\infty}(1-p)^{i+1}\cdot\sum\limits_{n=0}^{\infty}p^{n}\cdot\left(\begin{array}[]{c}n+2i+k-1\\\
k+i-2\\\ \end{array}\right)$ $\displaystyle\geq$ $\displaystyle\
\sum\limits_{i=0}^{\infty}(1-p)^{i+1}\cdot\sum\limits_{n=0}^{\infty}p^{n}\cdot\left(\begin{array}[]{c}n+i+k-2\\\
k+i-2\\\ \end{array}\right)$

Next, we use the equality

$\sum\limits_{n=0}^{\infty}\left(\begin{array}[]{c}n+s\\\ s\\\
\end{array}\right)\cdot p^{n}=\frac{1}{{(1-p)}^{s+1}}$

whenever $|p|<1$ and $s=0,1,2,\ldots$ Then,

$\sum\limits_{n=0}^{\infty}\left(\begin{array}[]{c}n+i+k-2\\\ i+k-2\\\
\end{array}\right)\cdot p^{n}=\frac{1}{{(1-p)}^{i+k-1}}.$

Finally, we conclude that

$J\geq\sum\limits_{i=0}^{\infty}(1-p)^{i+1}\frac{1}{(1-p)^{i+k-1}}=\sum\limits_{i=0}^{\infty}\frac{1}{{(1-p)}^{k-2}}=\infty.$

In this way we have proved that the extension ${\widehat{\nu}}_{p}$ of the
measure $\nu_{p}$ is infinite.

(2) Now we prove that $\mu_{a}(\widehat{X}_{\overline{B}_{k}})=0$ for every
$0<a<\infty$. For this, it is enough to show that
$\mu_{a}(X_{\overline{B}_{k}})=0$. Let $Y^{(n)}_{i}$ be the set of all paths
$x=(x_{n})$ from $X_{\overline{B}_{\infty}}$ such that the finite path
$(x_{1},...,x_{n})$ lies in $\overline{B}_{k}$ and $s(x_{n})=i\in W_{n}$. Then

(8.17) $X_{\overline{B}_{k}}=\bigcap_{n=1}^{\infty}\bigcup_{i\in
W_{n}}Y^{(n)}_{i}$

and

$\mu_{a}(\widehat{X}_{\overline{B}_{k}})=\lim\limits_{n\to\infty}\mu_{a}\left(\bigcup_{i\in
W_{n}}Y^{(n)}_{i}\right)$

where $\mu_{a}(Y^{(n)}_{i})$ can be found in Proposition 8.3. To find the
height of every tower $Y^{(n)}_{i}$, we use also the fact that
$\overline{B}_{k}$ is the Pascal graph. Therefore,

$\displaystyle\mu_{a}(\widehat{X}_{\overline{B}_{k}})=$ $\displaystyle\
\lim\limits_{n\to\infty}\left(\frac{a^{k}}{{(1+a)}^{n+k}}\cdot\left[\left(\begin{array}[]{c}n\\\
0\\\ \end{array}\right)+\left(\begin{array}[]{c}n\\\ 1\\\
\end{array}\right)\left(\frac{a}{a+1}\right)+\ldots+\left(\begin{array}[]{c}n\\\
n\\\ \end{array}\right)\left(\frac{a}{a+1}\right)^{n}\right]\right)$
$\displaystyle=$ $\displaystyle\
\lim\limits_{n\to\infty}\left[\frac{a^{k}}{{(1+a)}^{n+k}}\right]\cdot\left(1+\frac{a}{a+1}\right)^{n}$
$\displaystyle=$ $\displaystyle\
{\left[\frac{a}{a+1}\right]}^{k}\cdot\lim\limits_{n\to\infty}\frac{(1+2a)^{n}}{{(1+a)}^{2n}}$
$\displaystyle=$ $\displaystyle\ 0.$

∎

######  Remark 8.9.

Let a subdiagram $\overline{B}$ of $B_{\infty}$ be defined by the sequence
$\overline{W}_{n}=\\{1,\ldots,k\\}$. The incidence matrix $\overline{F}$ is
the low triangular $k\times k$ matrix such that $\overline{f}_{ij}=1$ if
$i\leq j$ and zero otherwise. Then the path space of $\overline{B}$ is
countable and the only tail invariant measure is the delta measure supported
by the infinite path going through vertex 1. Indeed, the ‘‘normalized’’
vectors ${\overline{y}}_{i}^{(n,m)}$ of the rows of the matrix
${\overline{G}^{\prime}}^{(n,m)}=\overline{F^{\prime}}^{m}$ have the form

${\overline{y}}_{i}^{(n,m)}=\frac{1}{S_{i}^{(m)}}\cdot\left\langle
S_{i}^{(m-1)},S_{i-1}^{(m-1)},\ldots,S_{1}^{(m-1)},0,\ldots,0\right\rangle,\
i=1,\ldots,k.$

We can easily show that
$\overline{y}_{i}^{(n,\infty)}=\lim_{m\to\infty}{\overline{y}}_{i}^{(n,m)}=\left\langle
1,0,\ldots,0\right\rangle$ because

$\lim_{m\to\infty}\frac{S_{i}^{(m-1)}}{S_{i}^{(m)}}=1,\quad\lim_{m\to\infty}\frac{S_{i}^{(m-1)}}{S_{i+1}^{(m)}}=0,\
i=1,\ldots,k.$

###  8.3. Values of the measures $\mu_{a}$ on subdiagrams $B(W,k)$

We consider now the vertex subdiagram $\overline{B}(W,k)$ of $B_{\infty}$ that
was defined at the beginning of this subsection. Recall that $k$ is fixed and
$W_{n}=\\{k,k+1,\ldots,k+n-1\\}$. For the reader’s convenience, we include
Figure 3:

Figure 3. The subdiagram $\overline{B}(W,k)$.

Let $h_{i}^{(n)}$ be the heights of the towers $\overline{X}_{i}^{(n)}$,
$i=k,k+1,\ldots,k+n-1$, inside $\overline{B}(W,k)$ where $n\geq 1$. It is
evident that $h_{k}^{(n)}=1$ for all $n$. Moreover, from the definition of the
subdiagram $\overline{B}(W,k)$, we get the relation

(8.18) $h_{i}^{(n+1)}=h_{k}^{(n)}+\ldots+h_{i}^{(n)},$

whenever $i=k,\ldots,k+n-1$, and $h_{k+n}^{(n+1)}=h_{k+n-1}^{(n+1)}$.

We recall that $H_{i}^{(n)}$ denotes the height of the tower $X_{i}^{(n)}$ in
the diagram $B_{\infty}$, $n,i=1,2,\ldots$. We know that
$H_{i}^{(n)}=\left(\begin{array}[]{c}i+n-2\\\ n-1\\\ \end{array}\right)$, see
(8.2), (8.4).

######  Lemma 8.10.

For $n=2,3,\ldots$ and $i=k+1,\ldots,k+n-1$, the heights of the towers
$\overline{X}_{i}^{(n)}$ and $X_{i}^{(n)}$ satisfy the relation

(8.19) $h_{i}^{(n)}=H_{i-k+1}^{(n)}-H_{i-k}^{(n+1)}.$

###### Proof.

It is obvious that (8.19) holds for $n=2$ and $i=k+1$. Assume that (8.19) is
true for some $n$. Then, using (8.18), we get that, for $i=k+1,\ldots,k+n-1$,

$\displaystyle h_{i}^{(n+1)}=$ $\displaystyle\
1+h_{k+1}^{(n)}+\ldots+h_{i}^{(n)}$ $\displaystyle=$ $\displaystyle\
\left[1+H_{2}^{(n)}+\ldots+H_{i-k+1}^{(n)}\right]-\left[H_{1}^{(n+1)}+\ldots+H_{i-k}^{(n+1)}\right]$
$\displaystyle=$ $\displaystyle\ H_{i-k+1}^{(n+1)}-H_{i-k}^{(n+2)}.$

It remains to show that (8.19) holds for $i=k+n$:

$\displaystyle h_{k+n}^{(n+1)}=h_{k+n-1}^{(n+1)}=$ $\displaystyle\
H_{n}^{(n+1)}-H_{n-1}^{(n+2)}$ $\displaystyle=$ $\displaystyle\
\left(\begin{array}[]{c}2n-1\\\ n\\\
\end{array}\right)-\left(\begin{array}[]{c}2n-1\\\ n+1\\\ \end{array}\right)$
$\displaystyle=$ $\displaystyle\ \frac{(2n-1)!\cdot
2}{(n+1)!\cdot(n-1)!}=\frac{(2n)!}{(n+1)!\cdot n!}$ $\displaystyle=$
$\displaystyle\ \left(\begin{array}[]{c}2n\\\ n\\\
\end{array}\right)-\left(\begin{array}[]{c}2n\\\ n+1\\\ \end{array}\right)$
$\displaystyle=$ $\displaystyle\ H_{(n+1)}^{(n+1)}-H_{(n)}^{(n+2)}.$

∎

For $0<a<\infty$, let $\mu_{a}$ be the ergodic probability measure defined in
Theorem 8.7. We use (8.17) to describe the path space
$\overline{X}_{\overline{B}_{k}}$ of the subdiagram $\overline{B}(W,k)$ (we
write $X_{\overline{B}_{k}}$ instead of $\overline{X}_{\overline{B}(W,k)}$ for
convenience).

######  Proposition 8.11.

Let $Z_{n}:=\bigcup_{i\in W_{n}}Y^{(n)}_{i}$ and let $Y_{i}^{(n)}$ denote the
inner tower of the subdiagram $\overline{B}_{k}$ corresponding to the vertex
$i\in W_{n}$. Set $\mu_{n,a}=\mu_{a}(Z_{n})$. Then

(8.20) $\mu_{a}(\overline{X}_{\overline{B}_{k}})=\lim_{n\to\infty}\mu_{n,a},$

and $\mu_{a}(\overline{X}_{\overline{B}_{k}})>0$ if and only if $0<a<1$.

###### Proof.

Since the sequence of sets $Z_{n}:=\bigcup_{i\in W_{n}}Y^{(n)}_{i}$ is
decreasing, we conclude that (8.20) holds.

Further, we note that $\mu_{1,a}=\dfrac{a^{k-1}}{(a+1)^{k}}$. Using the
definition of $Z_{n}$, we can write

$\mu_{n,a}=\frac{1}{{(a+1)}^{n}}\left[h_{k}^{(n)}\left(\frac{a}{a+1}\right)^{k-1}+h_{k+1}^{(n)}\left(\frac{a}{a+1}\right)^{k}+\ldots+h_{k+n-1}^{(n)}\left(\frac{a}{a+1}\right)^{k+n-2}\right].$

Denoting $z=\frac{a}{a+1}$ and
$I_{n}=h_{k}^{(n)}+h_{k+1}^{(n)}z+\ldots+h_{k+n-1}^{(n)}z^{n-1}$, we can write

$\mu_{n,a}=\frac{a^{k-1}}{{(a+1)}^{n+k-1}}I_{n}.$

We want to represent $I_{n+1}$ in terms of $I_{n}$. Note that the following
relation holds:

$\displaystyle I_{n+1}=$ $\displaystyle\ \sum_{j=0}^{n}h_{k+j}^{(n+1)}z^{j}$
$\displaystyle=$ $\displaystyle\
\sum_{j=0}^{n-1}\left(\sum_{l=0}^{j}h^{(n)}_{k+l}\right)z^{j}+\left(h_{k}^{(n)}+\ldots+h_{k+n-1}^{(n)}\right)z^{n}$
$\displaystyle=$ $\displaystyle\
\sum_{l=0}^{n-1}h^{(n)}_{k+l}z^{l}(1+z+\ldots+z^{n-l})$ $\displaystyle=$
$\displaystyle\ \frac{1}{1-z}\sum_{l=0}^{n-1}h^{(n)}_{k+l}z^{l}(1-z^{n-l+1})$
$\displaystyle=$ $\displaystyle\
\frac{1}{1-z}\left[\sum_{l=0}^{n-1}h^{(n)}_{k+l}z^{l}-z^{n+1}\sum_{l=0}^{n-1}h^{(n)}_{k+l}\right]$
$\displaystyle=$ $\displaystyle\
\frac{1}{1-z}\left[I_{n}-z^{n+1}h^{(n+1)}_{k+n-1}\right].$

Because $\frac{1}{1-z}=a+1$, we obtain that

$\displaystyle\mu_{n+1,a}=$ $\displaystyle\ \frac{a^{k-1}}{{(a+1)}^{n+k}}\cdot
I_{n+1}$ $\displaystyle=$ $\displaystyle\
\frac{a^{k-1}}{{(a+1)}^{n+k-1}}\cdot\left[I_{n}-{\left(\frac{a}{a+1}\right)}^{n+1}h_{k+n-1}^{(n+1)}\right]$
$\displaystyle=$ $\displaystyle\
\mu_{n,a}-\frac{a^{k+n}}{{(a+1)}^{2n+k}}h_{k+n-1}^{(n+1)}$

It follows from (8.19) (see also (8.2) and (8.4)) that

$\displaystyle h_{k+n-1}^{(n+1)}=$ $\displaystyle\
H_{n}^{(n+1)}-H_{n-1}^{(n+2)}$ $\displaystyle=$ $\displaystyle\
\left(\begin{array}[]{c}2n-1\\\ n\\\
\end{array}\right)-\left(\begin{array}[]{c}2n-1\\\ n+1\\\ \end{array}\right)$
$\displaystyle=$ $\displaystyle\ \frac{(2n-1)!\cdot 2}{(n+1)!\cdot(n-1)!}$
$\displaystyle=$ $\displaystyle\ \frac{(2n)!}{(n+1)!\cdot n!}$
$\displaystyle=$ $\displaystyle\ \frac{1}{n+1}\left(\begin{array}[]{c}2n\\\
n\\\ \end{array}\right)$

Thus, we obtain that, for all $n\in\mathbb{N}$,

$\displaystyle\mu_{n,a}-\mu_{n+1,a}=$ $\displaystyle\
\frac{a^{k+n}}{{(a+1)}^{2n+k}}\frac{1}{(n+1)}\left(\begin{array}[]{c}2n\\\
n\\\ \end{array}\right)$ $\displaystyle=$ $\displaystyle\
\left(\frac{a}{a+1}\right)^{k}\frac{1}{n+1}\left(\begin{array}[]{c}2n\\\ n\\\
\end{array}\right)\left(\frac{a}{{(a+1)}^{2}}\right)^{n}$

It follows from this relation that

$\mu_{1,a}-\mu_{m+1,a}=\left(\frac{a}{a+1}\right)^{k}\sum_{n=1}^{m}\frac{1}{n+1}\left(\begin{array}[]{c}2n\\\
n\\\ \end{array}\right)\left(\frac{a}{{(a+1)}^{2}}\right)^{n}$

Recall that $C_{n}=\dfrac{1}{n+1}\left(\begin{array}[]{c}2n\\\ n\\\
\end{array}\right)$ is called the Catalan number, and the generating function
for the corresponding series is well known, see e.g. [Sta15]

(8.21) $C(x)=\sum_{n=0}^{\infty}\frac{1}{n+1}\left(\begin{array}[]{c}2n\\\
n\\\ \end{array}\right)x^{n}=\frac{1-\sqrt{1-4x}}{2x},\quad|x|<\frac{1}{4}.$

Using this fact, we can substitute $x=\dfrac{a}{{(a+1)}^{2}}$ in (8.21) and
find that

$\displaystyle\mu_{a}(\widehat{X}_{{\overline{B}}_{k}})=$ $\displaystyle\
\lim\limits_{m\to\infty}\mu_{m+1,a}$ $\displaystyle=$ $\displaystyle\
\mu_{1,a}-\left(\frac{a}{a+1}\right)^{k}\sum_{n=1}^{\infty}\frac{1}{n+1}\left(\begin{array}[]{c}2n\\\
n\\\ \end{array}\right)\left(\frac{a}{{(a+1)}^{2}}\right)^{n}$
$\displaystyle=$ $\displaystyle\ \left(\frac{a}{a+1}\right)^{k}\left[\
\frac{1}{a}-\sum_{n=1}^{\infty}\frac{1}{n+1}\left(\begin{array}[]{c}2n\\\ n\\\
\end{array}\right)\ \left(\frac{a}{{(a+1)}^{2}}\right)^{n}\right]$
$\displaystyle=$ $\displaystyle\ \left(\frac{a}{a+1}\right)^{k}\left[\
\frac{1}{a}-C\left(\frac{a}{{(a+1)}^{2}}\right)+1\right].$

By a direct computation, we obtain that

$C\left(\frac{a}{{(a+1)}^{2}}\right)=\frac{(a+1)[(a+1)-|a-1|]}{2a}=\begin{cases}\dfrac{a+1}{a}\
&\ \mbox{if}\ a\geq 1\\\ a+1\ &\ \mbox{if}\ 0<a<1.\end{cases}$

Finally,

$\mu_{a}(\widehat{X}_{{\overline{B}}_{k}})=\begin{cases}0,\ &\ \mbox{if}\
a\geq 1\\\ \left(\dfrac{a}{a+1}\right)^{k-1}(1-a),\ &\ \mbox{if}\
0<a<1.\end{cases}$

∎

######  Remark 8.12.

We consider the generalized Bratteli diagram $B_{\infty}$ and the tail
invariant measure $\mu_{a}$ on the path space $X_{B_{\infty}}$. Recall that
the measure $\mu_{a}$ defines the probability distribution on the vertices
$i\in V_{1}$ of the first level:
$\mu_{a}(\\{i\\})=\dfrac{a^{i-1}}{(1+a)^{i}},i=1,2,\ldots$. Define a Markov
kernel $P=(p_{ij})$ by setting

$p_{ij}=\frac{a^{j-i}}{(1+a)^{j-i+1}},\qquad j\geq i,$

and $p_{ij}=0$ if $j<i$.

Hence, $\mathrm{Prob}(i\to i)=\frac{1}{1+a}$ and

$\mathrm{Prob}(i\to\\{j|j>i\\})=\sum_{j>i}p_{ij}=\frac{a}{1+a}.$

It follows from Proposition 8.11 that the following result holds.

######  Corollary 8.13.

The following are equivalent:

(i) $\mathrm{Prob}(i\to i)>\mathrm{Prob}(i\to\\{j\ |\ j>i\\})$,

(ii) $\mu_{a}(\widehat{X}_{{\overline{B}(W,i)}})>0$,

(iii) $0<a<1$.

###  8.4. Tail invariant probability measures on the subdiagram $B(W,k)$

We consider again the vertex subdiagrams $B(W,k)$, $k\in\mathbb{N},$ defined
in subsection 8.3. We recall that the heights of the towers in this subdiagram
have been found in Lemma 8.10 by the formula:

(8.22) $h_{i}^{(n)}=H_{i-k+1}^{(n)}-H_{i-k}^{(n+1)},\ i=k+1,\ldots,k+n-1,$

where

$H_{i}^{(n)}=\left(\begin{array}[]{c}i+n-2\\\ n-1\\\ \end{array}\right),\quad
n,i=1,2,\ldots.$

Let the matrices ${G^{\prime}}^{(n,m)}=\\{{g^{\prime}}_{ij}^{(n,m)}\\}$
($m\geq 1$) be defined as above, that is they are the product of the incidence
matrices between the levels $V_{n}$ and $V_{n+m}$, and let
$G^{(n,m)}=\\{g_{ij}^{(n,m)}\\}$ denote the corresponding stochastic matrices,
where $i=k,\ldots,k+n+m-1,\ j=k,\ldots,k+n-1$ and $n,m=1,2,\ldots$.

######  Lemma 8.14.

The following formula holds:

(8.23)
${g^{\prime}}_{ij}^{(n,m)}=\begin{dcases}S_{i-j+1}^{(m-1)},&i=k,\ldots,k+n-1,\quad
j=k,\ldots,i,\\\ S_{i-j+1}^{(m-1)},&i=k+n,\quad j=k,\ldots,k+n-1,\\\
S_{i-j+1}^{(m-1)}-S_{m+n+k-j+1}^{(i-k-n-1)},&i=k+n+1,\ldots,k+n+m-1,\quad
j=k,\ldots,k+n-1.\end{dcases}$

For $i=n+m+k$ and $j=k,k+1,\ldots,k+n-1$, we have the equality
${g^{\prime}}_{n+m+k,j}^{(n,m)}={g^{\prime}}_{n+m+k-1,j}^{(n,m)}$.

###### Proof.

For $m=1$, ${G^{\prime}}^{(n,1)}=F_{n}$ is the matrix of $(k+n)\times(k+n-1)$
size, and therefore we can write

${G^{\prime}}^{(n,1)}=\left[\begin{array}[]{rrrrrrrrrrrrr}1&0&0&0&\ldots&\ldots&\ldots&0\\\
1&1&0&0&\ldots&\ldots&\ldots&0\\\ 1&1&1&0&\ldots&\ldots&\ldots&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\\
1&1&1&1&\ldots&\ldots&\ldots&1\\\ 1&1&1&1&\ldots&\ldots&\ldots&1\\\
\end{array}\right],$

where the entries of the last two rows equal 1. The rows are indexed by
$i=k,k+1,\ldots,k+n$, and the columns are indexed by $j=k,\ldots,k+n-1$.

Writing the rows of the matrices ${G^{\prime}}^{(n,m)}$ and
${G^{\prime}}^{(n,m+1)}$ as vectors, we obtain the following relations:

${\overline{g}^{\prime}}_{s}^{(n,m+1)}=\sum_{i=1}^{s}{\overline{g}^{\prime}}_{i}^{(n,m)},\quad
s=k,\ldots,k+n+m-1,$

and

(8.24)
${\overline{g}^{\prime}}_{k+n+m}^{(n,m+1)}={\overline{g}^{\prime}}_{k+n+m-1}^{(n,m+1)},\quad
n,m=1,2,\ldots.$

Using these relations and the induction assumption, we get that for
$s=k,\ldots,k+n+m-1,\ j=k,\ldots,k+n-1$

${g^{\prime}}_{sj}^{(n,m+1)}=\sum_{i=j}^{s}{g^{\prime}}_{ij}^{(n,m)}=\sum_{i=j}^{s}S_{i-j+1}^{(m-1)}$

if $s=k,\ldots,k+n$, and

${g^{\prime}}_{sj}^{(n,m+1)}=\sum_{i=j}^{s}S_{i-j+1}^{(m-1)}-\sum_{i=k+n+1}^{s}S_{m+n+k-j+1}^{(i-k-n-1)},$

if $s=k+n+1,\ldots,k+n+m-1$. Further, we use

$S_{l}^{(0)}+S_{l}^{(1)}+\ldots+S_{l}^{(u)}=S_{l+1}^{(u)}.$

and get that

$\sum_{i=j}^{s}S_{i-j+1}^{(m-1)}=\sum_{i=1}^{s-j+1}S_{i}^{(m-1)}=S_{s-j+1}^{(m)},\quad
s=k,\ldots,k+n+m-1$

and

$\sum_{i=k+n+1}^{s}S_{m+n+k-j+1}^{(i-k-n-1)}=\sum_{u=0}^{(s-k-n-1)}S_{m+n+k-j+1}^{(u)}=S_{m+n+k-j+2}^{(s-k-n-1)}.$

Therefore, we proved that

${g^{\prime}}_{sj}^{(n,m+1)}=S_{s-j+1}^{(m)},\quad s=k,\ldots,k+n,$

and

${g^{\prime}}_{sj}^{(n,m+1)}=S_{s-j+1}^{(m)}-S_{m+n+k-j+2}^{(s-k-n-1)},\quad
s=k+n+1,\ldots,k+n+m-1.$

relation (8.23) holds. ∎

We will now transform formulas (8.22) and (8.23) and write them in a more
convenient form. For this,

(8.25) $\displaystyle h_{i}^{(n)}=$ $\displaystyle\
\left(\begin{array}[]{c}n+i-k-1\\\ n-1\\\
\end{array}\right)-\left(\begin{array}[]{c}n+i-k-1\\\ n\\\ \end{array}\right)$
$\displaystyle=$ $\displaystyle\ \left(\begin{array}[]{c}n+i-k-1\\\ n-1\\\
\end{array}\right)\cdot\left[1-\frac{(n+i-k-1)!}{n!\cdot(i-k-1)!}\cdot\frac{(n-1)!\cdot(i-k)!}{(n+i-k-1)!}\right]$
$\displaystyle=$ $\displaystyle\ \left(\begin{array}[]{c}n+i-k-1\\\ n-1\\\
\end{array}\right)\cdot\left[1-\frac{i-k}{n}\right]$ $\displaystyle=$
$\displaystyle\ H_{i-k+1}^{(n)}\cdot\left[1-\frac{i-k}{n}\right].$

Similarly, we calculate for $i=k+n+1,\ldots,k+n+m-1$, $j=k,\ldots,k+n-1$

${g^{\prime}}_{ij}^{(n,m)}=S_{i-j+1}^{(m-1)}-S_{(m+n+k-j+1)}^{(i-k-n-1)}=S_{i-j+1}^{(m-1)}\cdot\left[1-\frac{(i-k-n)\cdot\ldots\cdot(i-j)}{m\cdot\ldots\cdot(m+n+k-j)}\right].$

Setting

$r_{ij}=\frac{(i-k-n)\cdot\ldots\cdot(i-j)}{m\cdot\ldots\cdot(m+n+k-j)},\quad
i=k+n+1,\ldots,k+n+m-1,\ j=k,\ldots,k+n-1,$

and $r_{ij}=0$ for $i=k,\ldots,k+n-1$, $j=k,\ldots,i$, or for $i=k+n$,
$j=k,\ldots,k+n-1$, we obtain that

(8.26)
${g^{\prime}}_{ij}^{(n,m)}=S_{i-j+1}^{(m-1)}\cdot\left[1-r_{ij}\right],$

######  Proposition 8.15.

The standard Bratteli diagram $\overline{B}(W,k)$ admits uncountable many tail
invariant probability measures $\nu_{a}$, $0\leq a\leq 1$.

###### Proof.

We first calculate the entries $g_{ij}^{(n,m)}$ of the stochastic matrices
$G^{(n,m)}$. Then we find the probability vectors
${\overline{q}}^{(n,\infty,k)}=\left\langle q_{j}^{(n,\infty,k)}\ |\
j=k,\ldots,k+n-1\right\rangle$ as the limit points of vectors
${\overline{g}}_{i}^{(n,m)}=\left\langle g_{ij}^{(n,m)}\ |\
j=k,\ldots,k+n-1\right\rangle$ as $m\to\infty$.

Using (4.10), we have

$\begin{aligned} g_{ij}^{(n,m)}=&\
{g^{\prime}}_{ij}^{(n,m)}\cdot\frac{h_{j}^{(n)}}{h_{i}^{(n+m)}}\\\ =&\
S_{i-j+1}^{(m-1)}\cdot\left[1-r_{ij}\right]\cdot\frac{h_{j}^{(n)}}{H_{i-k+1}^{(n+m)}\cdot\left[1-\frac{i-k}{n+m}\right]}=\\\
=&\
\left[S_{i-j+1}^{(m-1)}\cdot\frac{1}{H_{i}^{(n+m)}}\right]\cdot\left(1-r_{ij}\right)\cdot\frac{H_{i}^{(n+m)}}{H_{i-k+1}^{(n+m)}}\cdot
h_{j}^{(n)}\cdot\left[\frac{n+m}{n+m+k-i}\right]\end{aligned}.$

Further, we calculate

$\frac{H_{i}^{(n+m)}}{H_{i-k+1}^{(n+m)}}=\left(\begin{array}[]{c}n+m+i-2\\\
n+m-1\\\ \end{array}\right)\cdot\left(\begin{array}[]{c}n+m+i-k-1\\\ n+m-1\\\
\end{array}\right)^{-1}=$
$\frac{(n+m+i-2)!}{(n+m-1)!\cdot(i-1)!}\cdot\frac{(n+m-1)!\cdot(i-k)!}{(n+m+i-k-1)!}=\frac{(n+m+i-k)\cdot\ldots\cdot(n+m+i-2)}{(i-k+1)\cdot\ldots\cdot(i-1)}$

Now assume that $m\to\infty$ and $\dfrac{i_{m}}{m}\to a$. We remark that
$0\leq a\leq 1$ because $k\leq{i}_{m}\leq m+n+k$. In what follows, we assume
that $0<a<1$.

We compute successively (see the proofs of Lemma 8.2 and Proposition 8.3):

$p_{j}^{(n,\infty)}={\lim\limits_{m\to\infty}\left[S_{i-j+1}^{(m-1)}\cdot\frac{1}{H_{i}^{(n+m)}}\right]}=\frac{a^{j-1}}{{(1+a)}^{n+j-1}},$
$\lim\limits_{m\to\infty}{\frac{H_{i}^{(n+m)}}{H_{i-k+1}^{(n+m)}}}=\lim\limits_{m\to\infty}\frac{(n+m-i-k)\cdot\ldots\cdot(n+m+i-2)}{(i-k+1)\cdot\ldots\cdot(i-1)}=\left(\frac{1+a}{a}\right)^{k-1},$
${\lim\limits_{m\to\infty}}{(1-r_{ij}})=\lim\limits_{m\to\infty}\left[1-\frac{(i-k-n)\cdot\ldots\cdot(i-j)}{m\cdot\ldots\cdot(m+n+k-j)}\right]=\left[1-a^{n+k-j+1}\right],$

and

$\lim\limits_{m\to\infty}\frac{n+m}{n+m+k-i}=\frac{1}{1-a}.$

As a consequence of the above formulas, we have

(8.27)
$p_{j}^{(n,\infty,k)}=\frac{a^{j-k}}{{(1+a)}^{n+j-k}}\cdot\frac{\left[1-a^{n+k-j+1}\right]}{1-a},$

and

$q_{j}^{(n,\infty,k)}=h_{j}^{(n)}p_{j}^{(n,\infty,k)}$

for $j=k,k+1,\ldots,k+n-1$.

It remains to check that the sequence of vectors

${\overline{p}}^{(n,\infty,k)}=\left\langle p_{j}^{(n,\infty,k)}\ |\
j=k,\ldots,k+n-1\right\rangle$

satisfies relation (3.6) of Theorem 3.6. Recall that
${F^{\prime}}_{n}={G^{\prime}}^{(n,1)}$.

We should verify that, for every $l=k,\ldots,k+n-1$, the relation

$\sum\limits_{j=l}^{k+n}p_{j}^{(n+1,\infty,k)}=p_{l}^{(n,\infty,k)}$

holds. For this,

$\sum\limits_{j=l}^{k+n}p_{j}^{(n+1,\infty,k)}=\frac{1}{{(1+a)}^{n+1}(1-a)}\sum\limits_{j=l}^{k+n}\frac{a^{j-k}}{{(1+a)}^{j-k}}(1-a^{n+k-j+2})$
$=\frac{1}{{(1+a)}^{n+1}(1-a)}\left[\left(\frac{a}{1+a}\right)^{l-k}\sum\limits_{j=0}^{n+k-l}\left(\frac{a}{1+a}\right)^{j}-a^{n+2}\left(\frac{1}{1+a}\right)^{l-k}\sum\limits_{j=0}^{n+k-l}\left(\frac{1}{1+a}\right)^{j}\right]$
$=\frac{1}{(1+a)^{n+1}\cdot(1-a)}\times$
$\times\left\\{(1+a)\left(\frac{a}{1+a}\right)^{l-k}\left[1-\left(\frac{a}{1+a}\right)^{n+k+1-l}\right]-\left(\frac{1+a}{a}\right)\cdot\frac{a^{n+2}}{{(1+a)}^{l-k}}\cdot\left[1-\left(\frac{1}{1+a}\right)^{n+k+1-l}\right]\right\\}$
$=\frac{1}{{(1+a)}^{n+k-l}(1-a)}\cdot\left[a^{l-k}-\frac{a^{n+1}}{{(1+a)}^{n+k+1-l}}-{a}^{n+1}+\frac{a^{n+1}}{{(1+a)}^{n+k+1-l}}\right]$
$=\frac{a^{l-k}}{(1+a)^{n+k-l}(1-a)}\cdot\left[1-a^{n+k-l+1}\right]=p_{l}^{(n,\infty,k)}$

as follows from (8.27).

We have proved that the sequence of vectors
$\\{{\overline{p}}^{(n,\infty,k)},\ n=1,2,\ldots\\}$ determines an invariant
probability measure $\nu_{a}$ on the subdiagram ${\overline{B}}(W,k)$ for
every $0\leq a\leq 1$. ∎

We discuss the ergodicity of $\nu_{a}$ in the next proposition.

######  Lemma 8.16.

Let $\nu$ be a tail invariant probability measure on the subdiagram
$\overline{B}(W,k)$ and let

${\overline{p}}^{(n,\infty,k)}=\left\langle p_{j}^{(n,\infty,k)}\ |\ k\leq
j\leq k+n-1\right\rangle$

be the sequence of vectors defining the measure $\nu$ as in Theorem 3.6. Then
the measure $\nu$ is completely determined by a sequence of numbers
$\\{p_{n},\ n=1,2,\ldots\\}$, where $p_{n}=p_{k+n-1}^{(n,\infty,k)}$.

###### Proof.

(Sketch) We know that, for every $n=1,2,\ldots$ and for
$l=k,k+1,\ldots,k+n-1$, the following relation holds

(8.28) $\sum\limits_{j=l}^{k+n}p_{j}^{(n+1,\infty,k)}=p_{l}^{(n,\infty,k)}.$

It follows directly from (8.28) that
$p_{k+n-1}^{(n+1,\infty,k)}=p_{k+n-1}^{(n,\infty,k)}-p_{k+n}^{(n+1,\infty,k)}=p_{n}-p_{n+1}$
where $n\in\mathbb{N}$. Using (8.28) and beginning with the first level, we
consequently find that $p_{k}^{(k,\infty,1)}=p_{1}$,
$p_{k}^{(k,\infty,2)}=p_{1}-p_{2}$, $p_{k+1}^{(k+1,\infty,2)}=p_{2}$, and so
on. In particular, the vector $\overline{p}^{(k+1,\infty,4)}$ has the entries
$\langle p_{1}-3p_{2}+p_{3},p_{2}-3p_{3},p_{3}-p_{4},p_{4}\rangle$.

This computation shows that, for every vector $\overline{p}^{(n,\infty,k)}$,
all coordinates $p_{j}^{(n,\infty,k)}$ are represented as a linear combination
of the terms of the sequence $\\{p_{n}\\}$. Moreover, to write
$p_{j}^{(n,\infty,k)}$ as a linear combination, we use only numbers
$p_{1},\ldots,p_{n}$. ∎

######  Lemma 8.17.

For $\nu$ as in Lemma 8.16, the sequence of numbers $\\{p_{n}\\}$ where
$p_{n}=p_{k+n-1}^{(n,\infty,k)}$ is completely monotonic.

###### Proof.

Now we prove that for every tail invariant probability measure $\nu$ on the
subdiagram ${\overline{B}}_{k}$, the sequence $\\{p_{n}\ |\ n=1,2,\ldots\\}$
defined above is completely monotonic. Let $\nu=\nu_{a}$.

By (8.27), we have

$p_{n}(a)=p_{n}=p^{(n,\infty,k)}_{k+n-1}=\frac{a^{n-1}}{{(1+a)}^{2n-1}}\frac{(1-a^{2})}{(1-a)}=\frac{a^{n-1}}{{(1+a)}^{2n-2}}.$

Therefore,

$(\Delta
p)_{n}=p_{n}-p_{n+1}=\frac{a^{n-1}}{{(1+a)}^{2n-2}}-\frac{a^{n}}{{(1+a)}^{2n}}=\frac{a^{n-1}(1+a+a^{2})}{{(1+a)}^{2n}},$

Similarly, we find

$(\Delta^{2}p)_{n}=\frac{a^{n-1}\cdot\left(1+a+a^{2}\right)^{2}}{{(1+a)}^{2n+2}},$

and, in general,

$(\Delta^{l}p)_{n}=\frac{a^{n-1}{(1+a+a^{2})}^{l}}{{(1+a)}^{2n+2l-2}},\quad
n=1,2,\ldots,\ l=1,2,\ldots.$

We have proved that $p_{n}(a)$ is a completely monotonic sequence. Now take
any invariant probability measure $\nu$ on $\overline{B}(W,k)$. Because the
set $M_{1}(X_{\overline{B}(W,k)})$ of all ergodic measures on
$\overline{B}(W,k)$ is contained in $\\{\nu_{a}:0<a<1\\}$, then $\nu$ is an
integral over the measures $\nu_{a}$, i.e., for every continuous function
$f\colon X_{\overline{B}(W,k)}\to\mathbb{R}$,

$\int\limits_{X_{\overline{B}(W,k)}}f(x)\;d\nu(x)=\int\limits_{0}^{1}\left[\int\limits_{X_{\overline{B}(W,k)}}f(x)\;d\nu_{a}(x)\right]d\rho(a),$

where $\rho$ is a probability measure on $(0,1)$.

In particular, $p_{n}(\nu)=\int\limits_{0}^{1}p_{n}(a)\;d\rho(a)$,
$n\in\mathbb{N}$. This equality implies that the following relations hold

$(\Delta^{l}p_{n}(\nu))_{n}=\int\limits_{0}^{1}(\Delta^{l}p(a))_{n}\rho(da)$

for $n,l=1,2,\ldots$. This means that the sequence $\\{p_{n}(\nu)\\}$ is
completely monotonic. ∎

######  Proposition 8.18.

Every measure $\nu_{a}$ defined in Proposition 8.15 is ergodic.

###### Proof.

We will show that every measure $\nu_{a}$ described in Proposition 8.15 is
ergodic. The proof is similar to the proof of Theorem 8.7. Each invariant
probability measure $\nu$ on ${\overline{B}}(W,k)$ determines a completely
monotonic sequence $\\{p_{n}(\nu)|\ n=1,2,\ldots\\}$ as shown in Lemmas 8.16
and 8.17. By Hausdorff theorem (Theorem 8.5) there exists a unique probability
measure $m$ on the interval $[0,1]$ such that
$\int\limits_{0}^{1}x^{n}m(dx)=p_{n}(\nu)$, $n\in\mathbb{N}$. For
$\nu=\nu_{a}$, we have $m=m_{a}$ and

$\int\limits_{0}^{1}x^{n}m_{a}(dx)=\frac{a^{(n-1)}}{(1+a)^{2n-2}}=\frac{{(1+a)}^{2}}{a}\cdot\int\limits_{0}^{1}x^{n}\delta_{\frac{a}{{(1+a)}^{2}}}(dx)$

where $n\in\mathbb{N}$. This implies that

$m_{a}=\frac{{(1+a)}^{2}}{a}\cdot\delta_{\frac{a}{{(1+a)}^{2}}}.$

Now, we use the same arguments as in Theorem 8.7 to prove that $\nu_{a}$
cannot be represented as a linear convex combination
$\nu_{a}=\lambda\nu^{(1)}+(1-\lambda)\nu^{(2)}$, where $0<\lambda<1$ and
$\nu^{(1)}$ and $\nu^{(2)}$ are different invariant finite measures on
${\overline{B}}_{k}$. Therefore each measure $\nu_{a}$ is an ergodic measure.
∎

######  Remark 8.19.

We observe that for $a=1$ the measure $\nu_{1}$ is infinite. Indeed, it
follows from (8.27) that, for $a=1$,

$\overline{p}^{(n,\infty,k)}=\left\langle\frac{n+1}{2^{n}},\frac{n}{2^{n-1}},\ldots,1\right\rangle$

and therefore $\overline{q}^{(n,\infty,k)}$ has the following form:

$\overline{q}^{(n,\infty,k)}=\left\langle h_{k}^{(n)}\frac{n+1}{2^{n}},\
h_{k+1}^{(n)}\frac{n}{2^{n-1}},\ldots,h_{k+n-1}^{(n)}\right\rangle.$

Since $h^{(n)}_{j}\to\infty$ as $n\to\infty$, we conclude that $\nu_{1}$ is
infinite.

######  Proposition 8.20.

The extension $\widehat{\nu}_{a}$ of each measure $\nu_{a}$ onto the set
$\mathcal{R}(X_{\overline{B}(W,k)})$ is finite, $0<a<1$. Moreover, the
measures $\widehat{\nu}_{a}$ are pairwise singular, $0<a<1$.

###### Proof.

The proof follows from condition (ii) of Theorem 3.13. Indeed, for $k\geq 2$,
we have

$\sum_{n=1}^{\infty}\left[\sum\limits_{i=k}^{k+n}\sum\limits_{j=1}^{k-1}{f^{\prime}}_{ij}^{(n)}H_{j}^{(n)}p_{i}^{(n+1,\infty,k)}+H_{k+n}^{(n)}\
p_{k+n}^{(n+1,\infty,k)}\right]$
$=\sum\limits_{n=1}^{\infty}\left[\sum\limits_{i=k}^{k+n}p_{i}^{(n+1,\infty,k)}\cdot\left(\sum\limits_{j=1}^{k-1}H_{j}^{(n)}\right)+H_{k+n}^{(n)}p_{k+n}^{(n+1,\infty,k)}\right]$
$=\sum\limits_{n=1}^{\infty}\left[\frac{1}{{(1+a)}^{n}}\sum\limits_{i=k}^{k+n}\frac{a^{i-k}}{{(1+a)}^{i-k}}\cdot\frac{\left(1-a^{n+k-i+2}\right)}{(1-a)}\sum\limits_{j=1}^{k-1}H_{j}^{(n)}+H_{k+n}^{(n)}\frac{a^{n}}{(1+a)^{2n}}\right]$
$=\sum\limits_{n=1}^{\infty}\frac{1}{{(1+a)}^{n}}\cdot\sum\limits_{i=0}^{n}\left(\frac{a}{1+a}\right)^{i}\cdot\frac{\left(1-a^{n-i+2}\right)}{(1-a)}\cdot\sum\limits_{j=1}^{k-1}H_{j}^{(n)}+\sum\limits_{n=1}^{\infty}H_{k+n}^{(n)}\frac{a^{n}}{(1+a)^{2n}}$
$<\frac{k-1}{1-a}\cdot\sum\limits_{n=1}^{\infty}{H_{k-1}^{(n)}}\frac{1}{(1+a)^{n}}\
\sum\limits_{i=0}^{n}\left(\frac{a}{1+a}\right)^{i}+\sum\limits_{n=1}^{\infty}{H_{k+n}^{(n)}\
\frac{a^{n}}{{(1+a)}^{2n}}}$
$<\frac{k-1}{1-a}\cdot\sum\limits_{n=1}^{\infty}\left(\begin{array}[]{c}n+k-3\\\
n-1\\\
\end{array}\right)\frac{2}{{(1+a)}^{n}}+\sum\limits_{n=1}^{\infty}\left(\begin{array}[]{c}2n+k-2\\\
n-1\\\ \end{array}\right)\cdot\left(\frac{a}{(1+a)^{2}}\right)^{n}$
$<\frac{k-1}{1-a}\cdot\sum\limits_{n=1}^{\infty}{(n+k-3)}^{k-2}\frac{2}{{(1+a)}^{n}}+\sum\limits_{n=1}^{\infty}\left(\begin{array}[]{c}2n+k-2\\\
n-1\\\ \end{array}\right)\cdot\left(\frac{a}{(1+a)^{2}}\right)^{n}.$

Note that both power series are convergent. The series

$\sum\limits_{n=1}^{\infty}{(n+k-3)}^{(k-2)}\frac{1}{{(1+a)}^{n}}<\infty$

because his radius of convergence is $1$ and $\frac{1}{a+1}<1$. Similarly, the
series

$\sum\limits_{n=1}^{\infty}\left(\begin{array}[]{c}2n+k-2\\\ n-1\\\
\end{array}\right)\cdot\left(\frac{a}{(1+a)^{2}}\right)^{n}<\infty$

because his radius of convergence is $\frac{1}{4}$ and
$\frac{a}{(1+a)^{2}}<\frac{1}{4}$ for $0\leq a<1$.

∎

Question. The generalized Bratteli diagram $B_{\infty}$ supports two
collections of tail invariant measures: $\\{\mu_{a}\ |\ 0<a<\infty\\}$ and
$\\{\widehat{\nu}_{a}\ |\ 0<a<1\\}$. How are the measures from the two
collections related?

######  Remark 8.21.

We note that the extended measures $\widehat{\nu}_{a}$ are pairwise singular.
Indeed, we know that the measures $\nu_{a}$ are pairwise singular, hence they
are supported by pairwise disjoint sets $Y_{a}$. The set $\widehat{Y}_{a}$
consists of all infinite paths tail equivalent to a path from $Y_{a}$. Then
the sets $\widehat{Y}_{a}$ are pairwise disjoint and this means that
$\widehat{\overline{\nu}}_{a}$ are pairwise singular.

##  9\. Generalized Bratteli-Vershik systems

In this section, we discuss orders and the corresponding Vershik maps for
generalized Bratteli diagrams. In particular, we consider the cases of
$\mathbb{Z}$-infinite and $\mathbb{N}$-infinite Pascal-Bratteli diagrams (see
Section 5) and the diagram $B_{\infty}$ studied in Section 8.

###  9.1. Vershik maps on infinite Pascal-Bratteli diagrams

In this subsection, we consider infinite Pascal-Bratteli-Vershik dynamical
systems. First, we define a so-called natural order on $\mathbb{Z}$-infinite
and $\mathbb{N}$-infinite Pascal-Bratteli diagrams.

First, let $B$ be a $\mathbb{Z}$-infinite Pascal-Bratteli diagram. Let
$\overline{t}=(\ldots t_{-1},t_{0},t_{1},\ldots)$ be a vertex in $V_{n}$ for
$n\geq 2$. Recall that we denote $I_{\overline{t}}=\\{i_{1},\ldots,i_{l}\\}$,
where $i_{1}<i_{2}<\ldots<i_{l}$, the set of indexes such that $\overline{t}$
has nonzero entries exactly at positions $\\{i_{j}\\}_{j=1}^{l}$. Then we have
$\overline{t}=\sum_{j=1}^{l}t_{i_{j}}\overline{e}^{(i_{j})}$ and
$\sum_{j=1}^{l}t_{i_{j}}=n$. Then $s(r^{-1}(\overline{t}))=\\{\overline{s}\in
V_{n-1}:\overline{s}=\overline{s}(\overline{t},i)=\overline{t}-\overline{e}^{(i)},\;i\in
I_{\overline{t}}\\}$ and the set $r^{-1}(\overline{t})$ is in a one-to-one
correspondence with the set of vertices $s(r^{-1}(\overline{t}))$. Define an
order on $r^{-1}(\overline{t})$ as follows: for any two edges $e,f\in
r^{-1}(\overline{t})$ with $s(e)=\overline{s}(\overline{t},i)$ and
$s(f)=\overline{s}(\overline{t},j)$, where $i,j\in I_{\overline{t}}$, we have
$e<f$ if $i<j$. The order on $\mathbb{Z}$-IPB diagram defined by this rule is
called the natural order. In the same manner, we define the natural order on
$\mathbb{N}$-IPB diagram.

It is convenient to present any $\overline{t}\in V_{n}$ as a pair
$(I_{\overline{t}},c(\overline{t}))$, where $c(\overline{t})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$. Let
$x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)$ be a path of $X_{B}$,
where ${\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots$ are the vertices,
${\overline{t}}^{(n)}\in V_{n}$, $n=1,2,\ldots$ Then
$I_{\overline{t}^{(1)}}\subset I_{\overline{t}^{(2)}}\subset\ldots$ and
$c({\overline{t}}^{(n+1)})$ is obtained from $c({\overline{t}}^{(n)})$ by
adding the number "1" either to a component of $c({\overline{t}}^{(n)})$ (then
$I_{\overline{t}^{(n)}}=I_{\overline{t}^{(n+1)}}$) or to an additional
position.

Now we are able to determine the sets $X_{\max}$ and $X_{\min}$ for both
versions of IPB diagrams. The sets $X_{\max}$ and $X_{\min}$ are described as
follows.

######  Remark 9.1.

(1) The set $X_{\max}$ consists of all infinite paths
$x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)$ such that for every
$n=1,2,\ldots$ either $c({\overline{t}}^{(n+1)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}+1\right\rangle$ or
$c({\overline{t}}^{(n+1)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}},1\right\rangle$, where the number "1" is at
position $i_{l+1}>i_{l}$ and $c({\overline{t}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$.

(2) Similarly, the set $X_{\min}$ consists of all infinite paths
$x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)$ such that for every
$n=1,2,\ldots$ either $c({\overline{t}}^{(n+1)})=\left\langle
t_{i_{1}}+1,\ldots,t_{i_{l}}\right\rangle$ or
$c({\overline{t}}^{(n+1)})=\left\langle
1,t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$, where the number "1" is at
position $i_{0}<i_{1}$ and $c({\overline{t}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$.

Moreover, we can represent the set ${X}_{\max}$ as a disjoint union of two
subsets $X^{u}_{\max}$ and $X^{c}_{\max}$, where

  1. (1)

$X^{u}_{\max}=\\{x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)\in
X_{\max}$ such that for infinitely many $n$, if
$c({\overline{t}}^{(n)})=\left\langle t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$
then $c({\overline{t}}^{(n+1)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}},1\right\rangle\\}$

  2. (2)

$X^{c}_{\max}=\\{x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)\in
X_{\max}$ for which there exists $n$ with
$c\left({\overline{t}}^{(n)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$ such that for $m=1,2,\ldots$ we have
$c\left({\overline{t}}^{(n+m)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}+m\right\rangle\\}$.

In other words, the set $X_{\max}^{c}$ consists of all infinite maximal paths
such that, starting from some level, we obtain the next vertex of the path by
adding ‘‘1’’ at the same coordinate $i_{l}$. The set $X_{\max}^{u}$ is the
complement of the set $X_{\max}^{c}$. In a similar way, we divide $X_{\min}$
into the union of the sets $X^{u}_{\min}$ and $X^{c}_{\min}$.

######  Lemma 9.2.

For both versions of infinite Pascal-Bratteli diagrams, the set
${X}^{u}_{\max}$ is uncountable and the sets ${X}^{c}_{\max}$ and
${X}^{c}_{\min}$ are infinite countable. For $\mathbb{Z}$-infinite Pascal-
Bratteli diagram the set ${X}^{u}_{\min}$ is uncountable, and for
$\mathbb{N}$-infinite Pascal-Bratteli diagram the set ${X}^{u}_{\min}$ is
empty.

###### Proof.

First we prove that the set ${X}^{u}_{\max}$ is uncountable for both versions
of IPB. Indeed, for an infinite path $x=(\overline{t}^{(n)})\in X_{\max}$, we
can obtain the next vertex $\overline{t}^{(n+1)}$ from the vertex
$\overline{t}^{(n)}$ by adding ‘‘1’’ either to the last non-zero coordinate
$i_{l}$ of the vector $\overline{t}^{(n)}$ or any of the zero coordinates
after the coordinate $i_{l}$. Recall that each path in $X^{u}_{\max}$ is
obtained by adding ‘‘1’’ to the zero coordinates infinitely many times. Thus,
each path $x\in X^{u}_{\max}$ is completely determined by two infinite
sequences $\overline{i}=\\{i_{1}<i_{2}<i_{3}<\ldots\\}$ and
$c(x)=\\{t_{i_{1}},t_{i_{2}},t_{i_{3}},\ldots\\}$, where
$i_{1},i_{2},i_{3},\ldots$ are positions in $\mathbb{Z}$ (or in $\mathbb{N}$),
such that all but finitely many vertices of $x$ have non-zero coordinates at
those positions, and each natural number $t_{i_{l}}$ is the maximal possible
value of the non-zero coordinate $i_{l}$ for a vertex of $x$. To find a vertex
${\overline{t}}^{(n)}$ of the path $x$, we choose a number $l\geq 1$ such that
$t_{i_{1}}+\ldots+t_{i_{l-1}}<n\leq t_{i_{1}}+\ldots+t_{i_{l}}$ and then set
$I_{{\overline{t}}^{(n)}}=\\{i_{1},\ldots,i_{l}\\}$ and
$c\left({\overline{t}}^{(n)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{l-1}},n-(t_{i_{1}}+\ldots+t_{i_{l-1}})\right\rangle$.
Therefore, the set ${X}^{u}_{\max}$ is uncountable. Similarly, for
$\mathbb{Z}$-infinite Pascal-Bratteli diagram, the set $X_{\min}^{u}$ is
uncountable. It is easy to see that for $\mathbb{N}$-infinite Pascal-Bratteli
diagram, the set $X_{\min}^{u}$ is empty.

Now we prove that the set $X^{c}_{\max}$ is countable. Each $x\in
X^{c}_{\max}$ is completely determined by two finite sequences
$\overline{i}=\\{i_{1}<\ldots<i_{s}\\}$ and
$c(x)=\\{t_{i_{1}},\ldots,t_{i_{(s-1)}},t_{i_{s}}=\infty\\}$, where
$i_{1},\ldots,i_{s}$ are positions in $\mathbb{Z}$ (or in $\mathbb{N}$) such
that all but finitely many vertices of $x$ have non-zero coordinates at those
positions, each number $t_{i_{l}}$, for $1\leq l\leq s-1$ is a natural number
which is the maximal possible value of the non-zero coordinate $i_{l}$ for a
vertex of $x$, and $t_{i_{s}}=\infty$. To find a vertex ${\overline{t}}^{(n)}$
of the path $x$, we repeat the procedure above whenever $n\leq
t_{i_{1}}+\ldots+t_{i_{s-1}}$. For $n>t_{i_{1}}+\ldots+t_{i_{s-1}}$ we have
$I_{{\overline{t}}^{(n)}}=\\{i_{1},\ldots,i_{s}\\}$ and
$c({\overline{t}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{s-1}},n-(t_{i_{1}}+\ldots+t_{i_{s-1}})\right\rangle$.
Thus, the set $X^{c}_{\max}$ is countable. Similarly, the set $X^{c}_{\min}$
is countable.

∎

######  Remark 9.3.

(1) For both IPB, the sets $X^{c}_{\max}$ and $X^{c}_{\min}$ contain a special
subset $X^{s}$ consisting of those paths $x=x(i)$ such that
$\overline{i}=\\{i\\}$ and $c(x)=\\{\infty\\}$ for $i\in\mathbb{Z}$ or
$i\in\mathbb{N}$. Then for $x=(\overline{t}^{(n)})_{n\in\mathbb{N}}\in X^{s}$,
we have $I_{{\overline{t}}^{(n)}}=\\{i\\}$, and
$c({\overline{t}}^{(n)})=\left\\{n\right\\}$. The paths of $X^{s}$ are maximal
and minimal simultaneously. Moreover, $X^{s}=X_{\max}\cap X_{\min}$.

(2) For $\mathbb{N}$-IPB diagram, we have $X^{u}_{\min}=\emptyset$ which
implies $X_{\min}=X^{c}_{\min}$.

(3) For any ${\overline{t}}^{(n)}\in V_{n}$, it is easy to find a unique
finite minimal path and a unique finite maximal path that join
${\overline{t}}^{(n)}$ with $V_{1}$. Namely, let ${\overline{t}}^{(n)}$ be
defined by a pair
$\left({I}_{\overline{t}^{(n)}},c({\overline{t}}^{(n)})\right)$, where

$I_{{\overline{t}}^{(n)}}=\left\\{i_{1}<i_{2}<\ldots<i_{l}\right\\}\mbox{ and
}c({\overline{t}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle.$

Let $x_{\min}=({\overline{t}}^{(1)},\ldots,{\overline{t}}^{(n)})$ and
$x_{\max}=({\overline{s}}^{(1)},\ldots,{\overline{s}}^{(n)})$, where
${\overline{s}}^{(n)}={\overline{t}}^{(n)}$, be the minimal and the maximal
paths between $V_{1}$ and ${\overline{t}}^{(n)}$. Then

$c({\overline{t}}^{(n-1)})=\left\langle
t_{i_{1}}-1,\ldots,t_{i_{l}}\right\rangle$

and

$I_{{\overline{t}}^{(n-1)}}=\begin{cases}\left\\{i_{1}<\ldots<i_{l}\right\\}&\text{if
}\left(t_{i_{1}}-1\right)>0\\\ \left\\{i_{2}<\ldots<i_{l}\right\\}&\text{if
}\left(t_{i_{1}}-1\right)=0.\\\ \end{cases}$

To determine the next vertices
${\overline{t}}^{(n-2)},\ldots,{\overline{t}}^{(1)}$ one must subtract in turn
the number "1" from the first component of the vectors
${\overline{t}}^{(n-1)},\ldots,{\overline{t}}^{(2)}$ and then determine the
sets $I_{{\overline{t}}^{(n-2)}},\ldots,I_{{\overline{t}}^{(1)}}$ as above. In
the end, we will have $I_{{\overline{t}}^{(1)}}=\\{i_{l}\\}$,
$c({\overline{t}}^{(1)})=\left\langle 1\right\rangle$.

Similarly, we determine $x_{\max}$. We have
$c({\overline{s}}^{(n-1)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}-1\right\rangle$ and

$I_{{\overline{s}}^{(n-1)}}=\begin{cases}\left\\{i_{1}<\ldots<i_{l}\right\\}&\text{if
}\left(t_{i_{l}}-1\right)>0\\\ \left\\{i_{1}<\ldots<i_{l-1}\right\\}&\text{if
}\left(t_{i_{l}}-1\right)=0.\\\ \end{cases}$

To determine the next vertices
${\overline{s}}^{(n-2)},\ldots,{\overline{s}}^{(1)}$ one must subtract in turn
the number ‘‘1’’ from the last component of the vectors
${\overline{s}}^{(n-1)},\ldots,{\overline{s}}^{(2)}$ and then determine the
sets $I_{{\overline{s}}^{(n-2)}},\ldots,I_{{\overline{s}}^{(1)}}$ as above. We
will have $I_{{\overline{s}}^{(1)}}=\\{i_{1}\\}$ and
$c({\overline{t}}^{(1)})=\left\langle 1\right\rangle.$

In Theorem 9.4, we describe the sets $Succ(x)$ and $Pred(y)$ for
$\mathbb{Z}$-IPB and $\mathbb{N}$-IPB diagrams (see Subsection 3.3 for
definitions).

######  Theorem 9.4.

Let $B$ be a $\mathbb{Z}$\- or $\mathbb{N}$-infinite Pascal-Bratteli diagram
and the sets $X^{c}_{\min}$, $X^{u}_{\min}$, $X^{c}_{\max}$, $X^{u}_{\max}$ be
as above. Then

(i) if $x\in X^{u}_{\max}$ then $Succ(x)=\emptyset$;

(ii) if $x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)\in X^{c}_{\max}$
such that

$I_{{\overline{t}}^{(n)}}=\\{i_{1}<\ldots<i_{l}\\}\mbox{ and
}c({\overline{t}}^{(n)})=\\{t_{i_{1}},\ldots,t_{i_{(l-1)}},t_{i_{l}}=\infty\\}$

then $Succ(x)=\\{x(i_{l})\\}\subset X^{s}$ (see Part (1) of Remark 9.3);

(iii) if $y\in X^{u}_{\min}$ then $Pred(y)=\emptyset$.

(iv) if $y=({\overline{s}}^{(1)},{\overline{s}}^{(2)},\ldots)\in X^{c}_{\min}$
such that

$I_{{\overline{s}}^{(n)}}=\\{i_{1}>\ldots>i_{l}\\}\mbox{ and
}c({\overline{s}}^{(n)})=\\{t_{i_{1}},\ldots,t_{i_{(l-1)}},t_{i_{l}}=\infty\\}$

then $Pred(y)=\\{x(i_{l})\\}\subset X^{s}$.

###### Proof.

(i) Let $x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)\in X^{u}_{\max}$
with $I_{{\overline{t}}^{(n)}}=\left\\{i_{1}<\ldots<i_{l}\right\\}$ and
$c\left({\overline{t}}^{(n)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$, where
$\left\\{i_{1}<\ldots<i_{l}\right\\}$ and $\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$ depend on $n$ and $i_{l}\to\infty$ as
$n\to\infty$. Assume that
$y=({\overline{s}}^{(1)},{\overline{s}}^{(2)},\ldots)\in Succ(x)$, $y\in
X_{\min}$. Then for infinitely many $n$, there are $\overline{z}\in V_{n+1}$
and edges
$e_{\overline{z}{\overline{t}}^{(n)}}^{(n)},e_{\overline{z}{\overline{s}}^{(n)}}^{(n)}\in{r}^{-1}(\overline{z})$
with $s(e_{\overline{z}{\overline{t}}^{(n)}}^{(n)})=\overline{t}^{(n)}$ and
$s(e_{\overline{z}{\overline{s}}^{(n)}}^{(n)})=\overline{s}^{(n)}$ such that
$e_{\overline{z}{\overline{s}}^{(n)}}^{(n)}$ is the successor of
$e_{\overline{z}{\overline{t}}^{(n)}}^{(n)}$. Then

$\overline{z}={\overline{t}}^{(n)}+\overline{e}^{(i)}={\overline{s}}^{(n)}+\overline{e}^{(j)}$

for some $i,j\in\mathbb{Z}$ (or $i,j\in\mathbb{N}$).

Note that $i<i_{l}$ because otherwise
$e_{\overline{z}{\overline{t}}^{(n)}}^{(n)}$ would be the maximal edge of
${r}^{(-1)}(\overline{z})$ and would have no successor. Thus, $i_{k-1}\leq
i<i_{k}$ for some $k\in\\{1,\ldots,l\\}$. If $i=i_{k-1}$, since
$e_{\overline{z}{\overline{s}}^{(n)}}^{(n)}$ is the successor of
$e_{\overline{z}{\overline{t}}^{(n)}}^{(n)}$, we have

$I_{{\overline{s}}^{(n)}}=\left\\{i_{1}<\ldots<i_{l}\right\\}\mbox{ and
}c\left({\overline{s}}^{(n)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{k-1}}+1,t_{i_{k}}-1,\ldots,t_{i_{l}}\right\rangle$

If $i>i_{k-1}$, we have

$I_{{\overline{s}}^{(n)}}=\left\\{i_{1}<\ldots<i_{k-1}<i<\ldots<i_{l}\right\\}\mbox{
and }c\left({\overline{s}}^{(n)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{k-1}},1,t_{i_{k}}-1,\ldots,t_{i_{l}}\right\rangle.$

Let $y^{(n)}_{\min}$ be the minimal path between ${\overline{s}}^{(n)}$ and
$V_{1}$. According to the Part (3) of Remark 9.3, the path $y^{(n)}_{\min}$
starts from the position $i_{l}$ or $i\geq i_{l-1}$ of $V_{1}$. Since
${y^{(n)}}_{\min}$ is a part of $y$, the minimal path $y$ starts from the same
position of $V_{1}$. Thus, there is no a such $y$ because both
$i_{l}\to\infty$ and $i_{l-1}\to\infty$ as $n\to\infty$.

(ii) Let $x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)\in
X_{\max}^{c}$ and let $N$ be such that
$c\left({\overline{t}}^{(N)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$ and for $m=1,2,\ldots$ we have
$c\left({\overline{t}}^{(N+m)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}+m\right\rangle$. Assume that
$y=({\overline{s}}^{1},{\overline{s}}^{2},\ldots)\in Succ(x)$, $y\in
X_{\min}$. Choose infinitely many $n>N$, such that there are $\overline{z}\in
V_{n+1}$ and
$e_{\overline{z}{\overline{t}}^{(n)}}^{(n)},e_{\overline{z}{\overline{s}}^{(n)}}^{(n)}\in{r}^{-1}(\overline{z})$
such that $e_{\overline{z}{\overline{s}}^{(n)}}^{(n)}$ is the successor of
$e_{\overline{z}{\overline{t}}^{(n)}}^{(n)}$. Then
$\overline{z}={\overline{t}}^{(n)}+\overline{e}^{(i)}$ for some
$i\in\mathbb{Z}$ (or $i\in\mathbb{N}$). Let us observe that $i<i_{l}$ and by
the same arguments as before we get: if $i=i_{k-1}$,

$I_{{\overline{s}}^{(n)}}=\left\\{i_{1}<\ldots<i_{l}\right\\}\mbox{ and
}c\left({\overline{s}}^{(n)}\right)=\left\langle
t_{i_{1}},\ldots,t_{i_{k-1}}+1,t_{i_{k}}-1,\ldots,t_{i_{l}}+(n-N)\right\rangle.$

If $i_{k-1}<i<i_{k}$ then

$I_{{\overline{s}}^{(n)}}=\left\\{i_{1}<\ldots<i_{k-1}<i<\ldots<i_{l}\right\\}\mbox{
and }c\left({\overline{s}}^{(n)}\right)=\left\langle t_{i_{1}},\ldots
t_{i_{k-1}},1,t_{i_{k}}-1,\ldots,t_{i_{l}}+(n-N)\right\rangle.$

Form the minimal path $y^{(n)}_{\min}$ between ${\overline{s}}^{(n)}$ and
$V_{1}$ according to the procedure from Part (3) of Remark 9.3. Then the
vertices of this path between the levels $1$ and
$n-(t_{i_{1}}+\ldots+t_{i_{l}})$, coincide with the vertices between the same
levels of the minimal path $x(i_{l})\in X^{s}$. Since
$(n-(t_{i_{1}}+\ldots+t_{i_{l}}))\to\infty$ as $n\rightarrow\infty$, we have
$Succ(x)=\\{x(i_{l})\\}$.

(iii) and (iv) In a similar way, we determine the sets $Pred(y)$ for $y\in
X_{\min}$. ∎

Theorem 9.5 concerns Vershik maps on $\mathbb{Z}$\- and $\mathbb{N}$-infinite
Pascal Bratteli diagrams and properties of the sets $X_{\min}$, $X_{\max}$.

######  Theorem 9.5.

(1) The Vershik map $\varphi_{B}\colon X\setminus X_{\max}\rightarrow
X\setminus X_{\min}$ can be extended to a continuous surjection
$\varphi\colon[(X\setminus X_{\max})\cup X^{c}_{\max}]\rightarrow[(X\setminus
X_{\min}\ )\cup X^{s}]$ such that $\varphi=\varphi_{B}$ on $X\setminus
X_{\max}$ and $\varphi(x)=x(i_{l})$, where $x$ and $x(i_{l})$ are described in
the proof of Theorem (9.4). Similarly, the inverse map $\varphi_{B}^{-1}\colon
X\setminus X_{\min}\rightarrow X\setminus X_{\max}$ can be extended to a
surjection $\psi\colon[(X\setminus X_{\min})\cup
X^{c}_{\min}]\rightarrow[(X\setminus X_{\max})\cup X^{s}]$ such that
$\psi=\varphi_{B}^{-1}$ on $X\setminus X_{\min}$ and $\psi(y)=x{(i_{l})}$,
where $y$ and $x(i_{l})$ are as above.

(2) There exists a continuous one-to-one map $f\colon X_{\max}\rightarrow
X_{\min}$ for $\mathbb{Z}$-IPB diagram and a continuous surjection $f\colon
X_{\max}\rightarrow X_{\min}$ for $\mathbb{N}$-IPB diagram.

(3) The sets $X_{\max}$ and $X_{\min}$ are nowhere dense in $X_{B}$ in both
cases of $\mathbb{Z}$-IPB and $\mathbb{N}$-IPB.

(4) Let $\overline{d}=\langle d_{1},d_{2},d_{3},\ldots\rangle$ be a
probability vector and $\mu_{d\overline{}}$ be an invariant measure on
$\mathbb{Z}$-IPB or $\mathbb{N}$-IPB described in Section 5. Then
$\mu_{\overline{d}}(X_{\max})=\mu_{\overline{d}}(X_{\min})=0$ whenever
$d_{i}<1$ for each $i=1,2,\ldots$. If $\overline{d}$ is a such vector that
$d_{j}=1$ and $d_{i}=0$ for $i\neq j$, $j\in\mathbb{Z}$ or $j\in\mathbb{N}$
then $\mu_{\overline{d}}$ is a $\delta$-measure concentrated on the path
${x\overline{}}{(j)}=({\overline{s}}^{(1,j)},{\overline{s}}^{(2,j)},\ldots)$,
where $s_{i}^{(n,j)}=n$ for $i=j$ and $s_{i}^{(n,j)}=0$ for $i\neq j$,
$i,n=1,2,\ldots$. Moreover, ${x\overline{}}{(j)}$ is both a maximal and a
minimal path.

###### Proof.

Part (1) is a consequence of Theorem 9.4.

(2) First consider a $\mathbb{Z}$-IPB diagram. Let
$x=({\overline{t}}^{(1)},{\overline{t}}^{(2)},\ldots)\in X_{\max}$ with
$I_{{\overline{t}}^{(n)}}=\\{i_{1}<\ldots<i_{l}\\}$ and
$c({\overline{t}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$, where $l$ depends on $n$. Define a
path $y=({\overline{s}}^{(1)},{\overline{}\overline{s}}^{(2)},\ldots)$, such
that $c({\overline{s}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$ and
$I_{{\overline{s}}^{(n)}}=\\{i^{{}^{\prime}}_{1}>i^{{}^{\prime}}_{2}>\ldots>i^{{}^{\prime}}_{l}\\}$,
where

$i^{{}^{\prime}}_{1}=i_{1},\;i^{{}^{\prime}}_{2}=2i_{1}-i_{2},\ldots,i^{{}^{\prime}}_{l}=2i_{1}-i_{l},$

for $n=1,2,\ldots$. It is obvious that $y\in X_{\min}$. We set $f(x)=y$. Then
it is not hard to see that $f\colon X_{\max}\rightarrow X_{\min}$ is a
continuous one-to-one mapping. Moreover, $f(X_{\max}^{u})=X_{\min}^{u}$,
$f(X_{\max}^{c})=X_{\min}^{c}$ and $f$ is equal to the identity on $X^{s}$.

Similarly we define a continuous surjection $f\colon X_{\max}\rightarrow
X_{\min}=X_{\min}^{c}$ for $\mathbb{N}$-IPB diagram. We define vertices
${\overline{s}}^{(n)}$, $n\geq 1$ in the same manner if
$i^{{}^{\prime}}_{l}\geq 1$. If $i^{{}^{\prime}}_{l}<1<i^{{}^{\prime}}_{l-1}$
then we determine ${\overline{s}}^{(n)}$ by the sets
$I_{{\overline{s}}^{(n)}}=\\{i^{{}^{\prime}}_{1}>i^{{}^{\prime}}_{2}>\ldots>i^{{}^{\prime}}_{l-1}>1\\}$
and $c({\overline{s}}^{(n)})=\left\langle
t_{i_{1}},\ldots,t_{i_{l}}\right\rangle$.

(3) It is easy to prove the property (3) if we take into consideration the
structure of the sets $X_{\max}$ and $X_{\min}$ .

(4) Assume that $B=(V,E)$ is a $\mathbb{Z}$-IPB diagram. Denote

$X_{\max}^{(n)}=\bigcup_{\overline{s}\in
V_{n}}[E_{\max}(V_{1},\overline{s})],$

where $[E_{\max}(V_{1},\overline{s})]$ is a cylinder set corresponding to the
maximal path $E_{\max}(V_{1},\overline{s})$ of $E(V_{1},\overline{s})$ for
$\overline{s}\in V_{n}$. Then $X_{\max}^{(n)}$ is an open set for every
$n\in\mathbb{N}$ and

$X_{\max}=\bigcap_{n\in\mathbb{N}}X_{\max}^{(n)}.$

Let $\overline{d}=\langle d_{i},\ i\in\mathbb{Z}\rangle$ be a probability
vector such that $d^{*}=\sup_{i\geq 1}d_{i}<1$. We have

$\mu_{\overline{d}}(X_{\max}^{(n)})=\mu_{\overline{d}}\left(\bigcup_{\overline{s}\in
V_{n}}E_{\max}({\overline{s}}^{(0)},\overline{s})\right)=\sum_{\overline{s}\in
V_{n}}\prod_{i\in\mathbb{Z}}d_{i}^{s_{i}}.$

We also have

$\mu_{\overline{d}}\left(X_{\max}^{(n+1)}\right)=\sum_{\overline{s}\in
V_{n+1}}\prod_{i\in\mathbb{Z}}d_{i}^{s_{i}}\leq
d^{\ast}\cdot\sum_{\overline{s}\in
V_{n}}\prod_{i\in\mathbb{Z}}d_{i}^{s_{i}}=d^{\ast}\cdot\mu_{\overline{d}}\left(X_{\max}^{(n)}\right).$

Hence

$\mu_{\overline{d}}(X_{\max})=\mu_{\overline{d}}\left(\bigcap_{n=1}^{\infty}X_{\max}^{(n)}\right)=0.$

Similarly, $\mu_{\overline{d}}(X_{\min})=0$. The same properties are true for
$\mathbb{N}$-IPB diagram and can be proved similarly. The proof of the last
part of (4) is obvious.

∎

######  Remark 9.6.

The map $g\colon X_{B}\rightarrow X_{B}$ such that $g=\varphi_{B}$ on
$X\setminus X_{\max}$ and $g=f$ on $X_{\max}$ is not continuous on the entire
space $X_{B}$. It is an one-to-one Borel mapping if $({X}_{B},E)$ is a
$\mathbb{Z}$-IPB diagram and it is a Borel surjection if $({X}_{B},E)$ is a
$\mathbb{N}$-IPB diagram.

###  9.2. The Vershik map on the generalized Bratteli diagram $B_{\infty}$

In this subsection, we consider Vershik maps defined on the path spaces of the
generalized Bratteli diagram $B_{\infty}$ and its subdiagram
$\overline{B}(W,k)$ where $k\geq 1$ is fixed and
$W=(W_{n}),W_{n}=\\{k,\ldots,k+n-1\\}$. We denote by $X_{B_{\infty}}$ and
$X_{k}:=X_{\overline{B}(W,k)}$ the corresponding path spaces. For both
diagrams, $B_{\infty}$ and $\overline{B}(W,k)$, the set of all orders is
uncountable and we do not consider all possible orders. We present here a few
examples of orders such that the corresponding Vershik maps have different
properties.

Let $\omega$ be the left-to-right order for every vertex $v\in V_{n}$, $n>1$.
For the standard Bratteli diagram $\overline{B}(W,k)$, we can easily find the
sets $X_{\max}(\omega)$ and $X_{min(\omega)}$ of infinite maximal and minimal
paths: $X_{min(\omega)}$ is a singleton containing of the vertical path
$X_{\min}$ through vertices $k$ for every level, $X_{\max}(\omega)$ is a
countable set $\\{y_{1},y_{2},\ldots,y_{\infty}\\}$ where $y_{i}$ is the path
going through the vertices $k,k+1,\ldots k+i,k+i,\ldots$ for every $i\geq 1$
and $y_{\infty}$ is the rightmost infinite path. In other words, $y_{i}$ is a
slanting path connecting $k$ and $k+i$ and then going vertically through the
vertices $k+i$.

Let $C$ be the countable set of eventually vertical paths in $X_{k}$. The set
$C$ contains $X_{\max}\setminus\\{y_{\infty}\\}$.

######  Lemma 9.7.

(1) The Vershik map $\varphi_{B}:X_{k}\setminus X_{\max}(\omega)\to
X_{k}\setminus X_{\min}(\omega)$ admits a continuous surjective extension
$\varphi$ by setting $\varphi(y_{i})=X_{\min}$ for all $i=1,2,\ldots,\infty$.

(2) Let $\mathcal{R}$ denote the tail equivalence relation. Then
$C=\bigcup_{i\geq 1}\mathcal{R}(y_{i})$ where $y_{i}\in X_{\max}(\omega)$.

The proof is obvious. One only checks that this map $\varphi$ is continuous.

The set $D=X_{k}\setminus C$ is characterized by the property:

$x=(x_{n})\in D\ \Longleftrightarrow\
|\\{n\in\mathbb{N}:r(x_{n})>s(x_{n})\\}|=\infty.$

We can show that the tail equivalence relation is minimal up to a countable
set.

######  Proposition 9.8.

Let $x\in D$. Then $\mathcal{R}(x)$ is dense in $X_{k}$.

###### Proof.

We show that for every fixed $x\in D$ and every cylinder set $[\overline{e}]$,
one has $\mathcal{R}(x)\cap[\overline{e}]\neq\emptyset$. Let
$r(\overline{e})=j\in V_{n}$. Take $m>n$ such that $r(x_{m})=l>j$. By
definition of the subdiagram $\overline{B}(W,k)$, there exists a finite path
connecting the vertices $j$ and $l$. This means that that the
$\mathcal{R}$-orbit of $x$ will visit $[\overline{e}]$. We also note that
$\mathcal{R}(x_{\infty})$ is a dense orbit. ∎

######  Remark 9.9.

(1) A similar result holds for the diagram $B_{\infty}$. As in the case of
$\overline{B}(W,k)$, we have countably many maximal infinite paths and the
unique minimal path through the leftmost vertex. The extension of the Vershik
map is continuous and surjective.

(2) Proposition 9.8 is also true for the diagram $B_{\infty}$.

######  Lemma 9.10.

(1) For the generalized Bratteli diagram $B_{\infty}$ and its subdiagram
$\overline{B}(W,k)$ there exist orders $\tau$ and $\tau(k)$, respectfully,
such that the Vershik map is an essentially minimal homeomorphism of the path
space with a unique fixed point.

(2) There exists an order $\omega$ of $B_{\infty}$ such that both sets
$X_{\max}$ and $X_{\min}$ are countable and $Succ(x)=\emptyset$ for every
$x\in X_{\max}$ and $Pred(x)=\emptyset$ for every $x\in X_{\min}$.

###### Proof.

(1) We define an order $\tau$ on $B_{\infty}$ by the following rule: $\tau$ is
the left-to-right order for every vertex $v\in V_{2n+1}$ and $\tau$ is the
right-to-left order for every vertex $v\in V_{2n}$, $n=0,1,\ldots$. Similarly,
we consider the case of the subdiagram $\overline{B}(W,k)$. Then, it is
obvious that $X_{\max}=X_{\min}=\\{z\\}$, where $z$ is the vertical path that
goes through the first vertex of $B_{\infty}$ (or the vertex $k$ for
$\overline{B}(W,k)$.

(2) Now we define an order $\omega$ on the diagram $B_{\infty}$ that satisfies
statement (2). For each $i>2$, let $\omega_{i}$ be defined as follows:

$\omega_{i}=\\{(i-1)<1<2<\ldots<(i-2)<i\\}$

and $\omega_{2}=\\{1<2\\}$ (note that $|r^{-1}(1)|=1$). Then, for every level
$V_{n}$, $n\geq 1$ and for every vertex $i$, the cycle $\omega_{i}$ defines a
stationary linear order on $r^{-1}(i)$. In such a way the diagram $B_{\infty}$
is supplied with the stationary order $\omega$. It follows from this
definition that the set $X_{\max}(\omega)$ of all infinite maximal paths
consists of all vertical paths $x_{i}$, that is $x_{i}$ passes through the
vertex $i$ for all levels $V_{n}$, $i\in\mathbb{N}$. The set
$X_{\min}(\omega)$ is formed by the infinite path $x_{1}$ and two sequences of
infinite paths, $\\{y_{i}=(y_{i}(n):i\geq 1\\}$ and $\\{z_{m}=(z_{m}(n)):m\geq
2\\}$, where $s(y_{i}(n))=i+n-1,r(y_{i}(n))=i+n$ and
$s(z_{m}(n))=r(z_{m}(n))=1$ for $1\leq n<m$ and
$s(z_{m}(n))=n-m+1,r(z_{m}(n))=n-m+2$ for $n\geq m$. In other words, $y_{i}$
is the right-slanting pass that begins at $i$ and goes through the vertices
$i+1,i+2,...$, and $z_{m}$ is the path that goes vertically through the vertex
1 up to the level $V_{m}$ and then is parallel to $y_{i}$. It can be easily
checked that for this order $\omega$ we have $Succ(x)=\emptyset$ and
$Pred(y)=\emptyset$ for $x\in X_{\max}$ or for $y\in X_{\min}$ respectively. ∎

######  Corollary 9.11.

For the the ordered Bratteli diagram $({B}_{\infty},\omega)$, the Vershik map
$\varphi\colon X\setminus X_{\max}\to X\setminus X_{\min}$ cannot be extended
to a continuous map of the entire space $X_{B_{\infty}}$. However, there
exists a one-to-one Borel extension of $\varphi$ acting on the space
$X_{B_{\infty}}$.

Acknowledgments. We are very grateful to our colleagues, especially, P.
Jorgensen, T. Raszeja, and S. Sanadhya for the numerous valuable discussions.
S.B. and O.K. are also grateful to the Nicolas Copernicus University in Toruń
for its hospitality and support. S.B. acknowledges the hospitality of AGH
University of Krakow during his visit. O.K. is supported by the NCN (National
Science Center, Poland) Grant 2019/35/D/ST1/01375 and by the program
‘‘Excellence Initiative - Research University’’ for the AGH University of
Krakow.

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