By (5.27) and (3.9), (3.11), for $\theta_{i}\in\varTheta^{+}_{\psi}$, we have,
see (4.15),

$\langle\theta_{i}^{l}\rangle\leq(\alpha+1)^{l}\langle\theta_{i}\rangle$

since

$\displaystyle\langle\theta_{i}^{l}\rangle=\int_{X}\int_{X}a_{i}(x-y)\theta_{i}^{l-1}(y)dxdy+\langle\theta_{i}^{l-1}\rangle\leq(\alpha+1)\langle\theta_{i}^{l-1}\rangle.$

Then, see (5.28),

$\langle\theta_{i}\rangle^{c_{0}}\langle\theta^{1}_{i}\rangle^{c_{1}}\cdots\langle\theta_{i}^{l}\rangle^{c_{l}}\cdots\leq\langle\theta_{i}\rangle^{c_{0}+c_{1}+\cdots}(\alpha+1)^{c_{1}+2c_{2}+\cdots}=\langle\theta_{i}^{l}\rangle^{m_{i}}(\alpha+1)^{n}.$

We use this in (5.3) and obtain, see also (5.38),

$\displaystyle\varOmega^{m_{i}}_{\tau_{i},n}(\theta_{i})\leq\exp\left(m_{i}\vartheta_{0}+\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)\bigg{[}\langle\theta_{i}\rangle^{m_{i}}[m_{i}(\alpha+1)]^{n}$
$\displaystyle+c_{a}^{n}\langle\psi\rangle^{m_{i}}\sum_{k=1}^{n}w_{k}(m_{i},n)\left(\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)^{k}\bigg{]}.$

Now we use this in (5.44), and finally arrive at the following estimate

$\displaystyle\widehat{\Omega}^{m_{i}}_{\tau_{i}}(\theta_{i}|t)$
$\displaystyle\leq$
$\displaystyle\exp\left(m_{i}\vartheta_{0}+\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)\bigg{[}e^{(\alpha+1)m_{i}t}\langle\theta_{i}\rangle^{m_{i}}$
$\displaystyle+$
$\displaystyle\langle\psi\rangle^{m_{i}}\sum_{q=0}^{\infty}\frac{(c_{a}t)^{q}}{q!}\sum_{k=1}^{\infty}w_{k}(m_{i},q)\left(\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)^{k}\bigg{]},$

where we took into account that $w_{k}(m_{i},q)=0$ for $k>q$, see (5.30). By
(5.29) we have

$\displaystyle\sum_{q=0}^{\infty}\frac{(c_{a}t)^{q}}{q!}\sum_{k=1}^{\infty}w_{k}(m_{i},q)\left(\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)^{k}$ (5.47)
$\displaystyle\quad=\sum_{k=1}^{\infty}\frac{1}{k!}\left(\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)^{k}\sum_{l=0}^{k}(-1)^{k-l}{k\choose
l}\sum_{q=0}^{\infty}\frac{(c_{a}t)^{q}}{q!}(m_{i}+l)^{q}$
$\displaystyle\quad=e^{m_{i}c_{a}t}\sum_{k=1}^{\infty}\frac{1}{k!}\left(\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)^{k}\sum_{l=0}^{k}(-1)^{k-l}{k\choose l}e^{lc_{a}t}$
$\displaystyle\quad=e^{m_{i}c_{a}t}\bigg{[}\exp\left((e^{c_{a}t}-1)\tau_{i}\langle\psi\rangle
e^{\vartheta_{0}}\right)-1\bigg{]}$
$\displaystyle\quad\leq\tau_{i}\varepsilon(1+\varepsilon)^{m_{i}}\exp\left(\langle\psi\rangle
e^{\vartheta_{0}}\right),$

where $t$ is as in (5.43) and $\tau_{i}\leq 1$, $\varepsilon<1$. We use now
(5.47) in (5.3) and then turn (5.43) into the following estimate

$\displaystyle\mu_{t}(\varPhi^{m}_{\tau}(\theta|\cdot))$ $\displaystyle\leq$
$\displaystyle\exp\left((m_{0}+m_{1})\vartheta_{0}+(\tau_{0}+\tau_{1})\langle\psi\rangle
e^{\vartheta_{0}}\right)$ $\displaystyle\times$
$\displaystyle\left[e^{(\alpha+1)m_{0}t}\langle\theta_{0}\rangle^{m_{0}}+\tau_{0}\varepsilon(1+\varepsilon)^{m_{0}}\exp\left(\langle\psi\rangle
e^{\vartheta_{0}}\right)\right]$ $\displaystyle\times$
$\displaystyle\left[e^{(\alpha+1)m_{1}t}\langle\theta_{1}\rangle^{m_{1}}+\tau_{1}\varepsilon(1+\varepsilon)^{m_{1}}\exp\left(\langle\psi\rangle
e^{\vartheta_{0}}\right)\right],\quad t<\log(1+\varepsilon)/c_{a}.$

For each $\gamma\in\Gamma_{*}^{2}$ and a decreasing sequence of positive
$\tau_{k}\to 0$, the sequence $\varPhi^{m}_{\tau}(\theta|\gamma)$,
$\tau_{i}=\tau^{k}$, $i=0,1$, is nondecreasing, see (5.32) and (2.55). By
(5.3) and the Beppo Levi monotone convergence lemma we conclude that the
pointwise limit, see (5.24), (5.22) and (2.8), (2.14),

$\displaystyle\varPhi^{m}_{0}(\theta|\gamma)=\lim_{k\to+\infty}\varPhi^{m}_{\tau^{k}}(\theta|\gamma)=H^{m}_{\theta}(\gamma)=\sum_{(\mathbf{x}^{m_{0}},\mathbf{y}^{m_{1}})\in\gamma}\theta^{\otimes
m}(\mathbf{x}^{m_{0}},\mathbf{y}^{m_{1}}),$ (5.49)

is $\mu_{t}$-integrable. Moreover, by the same lemma and (5.3) it follows that

$\mu_{t}(\varPhi^{m}_{0}(\theta|\cdot))=\chi_{\mu_{t}}^{(m)}(\theta^{\otimes
m})\leq\varkappa_{t}^{|m|}\|\theta_{0}\|^{m_{0}}_{L^{1}(X)}\|\theta_{1}\|^{m_{1}}_{L^{1}(X)},$
(5.50)

with

$\varkappa_{t}=e^{\vartheta_{0}+(\alpha+1)t},$ (5.51)

which by (2.9) yields the property in question for
$t<\log(1+\varepsilon)/c_{a}$. Since this length of the validity interval is
independent of $\mu_{0}$, the further continuation can be done by the literal
repetition of the procedure above. $\square$

######  Remark 5.5.

According to Proposition 4.3, see also [1, Theorem 3.5], the type of the
solution $\mu_{t}$ obtained in Lemma 5.3 does not exceed
$\exp(\vartheta_{0}+\alpha t)$, which is a more precise estimate than that
given in (5.50), (5.51).

##  6\. Existence: Approximating Models

The aim of this section is introducing approximating models, for which the
corresponding processes can be constructed by employing explicitly derived
Markov transition functions. By this result the process in question will be
obtained as the limit of such approximating processes. Similarly as in [15],
the Kolmogorov operators for the approximating models are obtained from that
in (1).

###  6.1. The models

We begin by introducing the basic function, cf. (2.29)

$\psi_{\sigma}(x)=\frac{1}{1+\sigma|x|^{d+1}},\qquad\sigma\in(0,1],\ \ x\in
X=\mathds{R}^{d},$ (6.1)

and then define

$a_{i}^{\sigma}(x,y)=a_{i}(x-y)\psi_{\sigma}(x),\qquad i=0,1,$ (6.2)

where $a_{i}$ are the jump kernels that appear in (1). It is clear that these
$a^{\sigma}_{i}$ satisfy, cf. (3.1), (3.3),

$\displaystyle\max_{i=0,1}\sup_{(x,y)\in X^{2}}a^{\sigma}_{i}(x,y)\leq\|a\|,$
(6.3) $\displaystyle\max\left\\{\sup_{y\in
X}\int_{X}|x|^{l}a_{i}^{\sigma}(x,y)dx;\ \sup_{y\in
X}\int_{X}|x|^{l}a_{i}^{\sigma}(y,x)dx\right\\}\leq\bar{a}^{(l)}_{i},\quad
l=0,\dots,d+1,\quad i=0,1.$

The Kolmogorov operator corresponding to the approximation model is obtained
by replacing in (1) $a_{i}(x-y)$ with $a_{i}^{\sigma}(x,y)$, $i=0,1$; that is,
it has the form

$\displaystyle({L}^{\sigma}F)({\gamma})$ $\displaystyle=$
$\displaystyle\sum_{x\in{\gamma}_{0}}\int_{X}a_{0}^{\sigma}(x,y)\exp\left(-\sum_{z\in{\gamma}_{1}}\phi_{0}(z-y)\right)\left[F({\gamma}\setminus
x\cup_{0}y)-F({\gamma})\right]dy\qquad$ $\displaystyle+$
$\displaystyle\sum_{x\in{\gamma}_{1}}\int_{X}a_{1}^{\sigma}(x,y)\exp\left(-\sum_{z\in{\gamma}_{0}}\phi_{1}(z-y)\right)\left[F({\gamma}\setminus
x\cup_{1}y)-F({\gamma})\right]dy.$

Noteworthy, in the approximating model the kernels corresponding to the jumps
from $x$ to $y$ get smaller if $x$ goes away from the origin. Now we introduce
$L^{\Delta,\sigma}$ by replacing in (4.1) $a_{i}(x-y)$ with
$a_{i}^{\sigma}(x,y)$, $i=0,1$, where one should take into account that
$a^{\sigma}_{i}(x,y)\neq a^{\sigma}_{i}(y,x)$. Then, cf. (4.5),

$\mu(L^{\sigma}KG)=\langle\\!\langle
L^{\Delta,\sigma}k_{\mu},G\rangle\\!\rangle,\qquad\sigma\in[0,1],$ (6.5)

holding for each $\mu\in\mathcal{P}_{\rm exp}$ and $G\in\mathcal{G}_{\infty}$.
For $\sigma=0$, $L^{\Delta,\sigma}$ coincides with the operator given in
(4.1). Clearly, for all $\sigma\in[0,1]$, $L^{\Delta,\sigma}$ satisfies (4.16)
and similar estimates, which allows one to define bounded operators
$(L^{\Delta,\sigma})^{n}_{\vartheta^{\prime}\vartheta}$, $n\in\mathds{N}$, and
thus construct, cf. (4.30),

$\varXi^{\sigma}_{\vartheta^{\prime}\vartheta}(t)=1+\sum_{n=1}^{\infty}\frac{t^{n}}{n!}(L^{\Delta,\sigma})^{n}_{\vartheta^{\prime}\vartheta},\qquad
t<T(\vartheta^{\prime},\vartheta),$ (6.6)

where the latter is the same as in (4.21). Similarly, one obtains
$\widehat{L}^{\sigma}$ by making the aforementioned replacements in (4.23),
and then defines, cf. (4.28),

$\varSigma^{\sigma}_{\vartheta\vartheta^{\prime}}(t)=1+\sum_{n=1}^{\infty}\frac{t^{n}}{n!}(\widehat{L}^{\sigma})^{n}_{\vartheta\vartheta^{\prime}},\qquad
t<T(\vartheta^{\prime},\vartheta).$ (6.7)

Thereafter, one sets

$k^{\sigma}_{t}=\varXi^{\sigma}_{\vartheta^{\prime}\vartheta}(t)k_{0},\quad
G^{\sigma}_{t}=\varSigma^{\sigma}_{\vartheta\vartheta^{\prime}}(t)G_{0},\qquad
t<T(\vartheta^{\prime},\vartheta),$ (6.8)

holding for each $k_{0}\in\mathcal{K}_{\vartheta}$,
$G_{0}\in\mathcal{G}_{\vartheta^{\prime}}$, $\vartheta^{\prime}>\vartheta$.
For $\sigma=0$, both $k^{\sigma}_{t}$ and $G^{\sigma}_{t}$ coincide with those
that appear in (4.29), (4.32), etc. Let us now consider the Fokker-Planck
equation

$\mu_{t_{2}}(F)=\mu_{t_{1}}(F)+\int_{t_{1}}^{t_{2}}\mu_{s}(L^{\sigma}F)ds,\qquad
F\in\mathcal{D}(L),$ (6.9)

where the latter is as in Definition 3.1. Since $L^{\sigma}$ satisfies all the
estimates used in the proof of Lemma 5.3, see, e.g., (6.3), we have the
following.

######  Proposition 6.1.

For each $\mu\in\mathcal{P}_{\rm exp}$ and $\sigma\in[0,1]$, the Fokker-Planck
equation (6.9) with $\mu_{0}=\mu$ has exactly one solution
$t\mapsto\mu^{\sigma}_{t}\in\mathcal{P}_{\rm exp}$ defined by the map
$t\mapsto k_{t}^{\sigma}\in\mathcal{K}^{\star}$ constructed with the help of
(6.6), (6.8), similarly as in the case $\sigma=0$, see Remark 4.4. Let also
$\vartheta$ be such that $k_{0}\in\mathcal{K}_{\vartheta}$. Then for each
$\sigma\in[0,1]$, $\vartheta^{\prime}>\vartheta$ and
$G\in\mathcal{G}_{\vartheta^{\prime}}$, the following holds, see (4.33),

$\langle\\!\langle k^{\sigma}_{t},G\rangle\\!\rangle=\langle\\!\langle
k_{0},G^{\sigma}_{t}\rangle\\!\rangle,\qquad
t<T(\vartheta^{\prime},\vartheta),$ (6.10)

where $G^{\sigma}_{t}=\varSigma^{\sigma}_{\vartheta\vartheta^{\prime}}(t)G$,
see (6.7), (6.8).

###  6.2. The weak convergence

Our aim is to prove that $\mu_{t}^{\sigma}\Rightarrow\mu_{t}$ as $\sigma\to
0$. We begin by proving the following statement.

######  Proposition 6.2.

Let $\\{\mu_{n}\\}_{n\in\mathds{N}}\subset\mathcal{P}_{\rm exp}$ be such that
the type of each $\mu_{n}$ does not exceed $e^{\vartheta}$,
$\vartheta\in\mathds{R}$, and $\mu_{n}\Rightarrow\mu$ for some
$\mu\in\mathcal{P}(\Gamma^{2}_{*})$. Then $\mu\in\mathcal{P}_{\rm exp}$ and
its type $\leq e^{\vartheta}$. Moreover, for each
$G\in\mathcal{G}_{\vartheta}$, it follows that

$\langle\\!\langle k_{\mu_{n}},G\rangle\\!\rangle\to\langle\\!\langle
k_{\mu},G\rangle\\!\rangle,\qquad n\to+\infty.$ (6.11)

###### Proof.

Since $\widehat{F}^{m}_{\tau}(\mathbf{v}|\cdot)\in C_{\rm b}(\Gamma^{2}_{*})$,
see Proposition 2.9 and (2.55), the assumed convergence yields
$\mu_{n}(F)\to\mu(F)$, holding for all $F\in\widehat{\mathcal{F}}$, including
$F=\varPhi^{m}_{\tau}$, see (5.24), (5.22). Therefore, by (5.49), (5.50) we
have

$\mu(\varPhi^{m}_{\tau})\leq\sup_{n}\mu_{n}(\varPhi^{m}_{\tau})\leq
e^{|m|\vartheta}\|\theta_{0}\|^{m_{0}}_{L^{1}(X)}\|\theta_{1}\|^{m_{1}}_{L^{1}(X)},$

holding for all $\theta_{0},\theta_{1}\in\varTheta_{\psi}^{+}$. As in the
proof of Lemma 5.1, this yields $\mu\in\mathcal{P}_{\rm exp}$ and its type
does not exceed $e^{\vartheta}$. The validity of (6.11) follows by the fact
just mentioned. ∎

Now we prove that the solutions of the Pokker-Planck equations (1.2) and (6.9)
have the property $\mu_{t}^{\sigma}\Rightarrow\mu_{t}$ as $\sigma\to 0$,
holding for each $t>0$. We obtain this result by proving a bit more general
statement, which will be used in the subsequent part of this paper.

######  Lemma 6.3.

Let $\\{\mu^{\sigma}\\}_{\sigma\in(0,1]}\subset\mathcal{P}_{\rm exp}$ be such
that the type of each $\mu^{\sigma}$ does not exceed $e^{\vartheta_{0}}$ for
some $\vartheta_{0}\in\mathds{R}$, and $\mu^{\sigma}\Rightarrow\mu$ as
$\sigma\to 0$. Let also $t\mapsto\mu_{t}^{\sigma}$, $\sigma\in(0,1]$,
$\mu_{t}^{\sigma}|_{t=0}=\mu^{\sigma}$, be the solution of the Fokker-Planck
equation (6.9) mentioned in Proposition 6.1. Then for each $t>0$, it follows
that $\mu_{t}^{\sigma}\Rightarrow\mu_{t}$ as $\sigma\to 0$, where $\mu_{t}$ is
the solution of (1.2) with $\mu_{t}|_{t=0}=\mu$.

Noteworthy, by Proposition 6.2 it follows that the limiting measure $\mu$ in
Lemma 6.3 is sub-Poissonian and its type does not exceed $e^{\vartheta_{0}}$.
The proof of Lemma 6.3 is based on the following statement.

######  Lemma 6.4.

For a given $t>0$, let $k_{t}^{\sigma}$ and $k_{t}$ be the correlation
functions of the measures $\mu_{t}^{\sigma}$ and $\mu_{t}$ mentioned in Lemma
6.3. Then there exists $\tilde{\vartheta}(t)\in\mathds{R}$ such that

$\forall G\in\mathcal{G}_{\tilde{\vartheta}(t)}\qquad\langle\\!\langle
k^{\sigma}_{t},G\rangle\\!\rangle\to\langle\\!\langle
k_{t},G\rangle\\!\rangle,\qquad{\rm as}\ \ \sigma\to 0.$ (6.12)

###### Proof.

As the type of each $\mu^{\sigma}$ does not exceed $e^{\vartheta_{0}}$, both
$k_{t}^{\sigma}$ and $k_{t}$ lie in $\mathcal{K}_{\vartheta(t)}$ with
$\vartheta(t)=\vartheta_{0}+\alpha t$, see Remark 4.4 and the proof of
Proposition 6.1. Moreover, $k_{t}^{\sigma}$ and $G$ satisfy (6.10) with
appropriate $\vartheta,\vartheta^{\prime}$. Recall that the map
$\vartheta^{\prime}\mapsto T(\vartheta^{\prime},\vartheta)$ attains its
maximum $T_{*}(\vartheta)$ given in (5.11).

Let now the convergence stated in (6.12) hold for a given $t\geq 0$. By the
assumed convergence $\mu^{\sigma}\Rightarrow\mu$ and Proposition 6.2 this
certainly holds for $t=0$. Our aim is to prove that it holds also for all
$t+s$, $s\leq s_{0}$, with a possibly $t$-dependent $s_{0}>0$. Set
$\bar{\vartheta}_{t}=\vartheta(t)+\delta(\vartheta(t))$, see (5.11). For
$s<T_{*}(\vartheta(t))$, the norm of
$\varXi_{\bar{\vartheta}_{t}\vartheta(t)}(s)$ satisfies

$\|\varXi_{\bar{\vartheta}_{t}\vartheta(t)}(s)\|\leq\frac{T_{*}(\vartheta(t))}{T_{*}(\vartheta(t))-s},$

see (4.32). In the same way, one estimates also the norm of
$\varXi^{\sigma}_{\bar{\vartheta}_{t}\vartheta(t)}(s)$, $\sigma\in(0,1]$ since
the norms of the corresponding
$(L^{\Delta,\sigma})^{n}_{\bar{\vartheta}_{t}\vartheta(t)}$ have the same
bounds as for $\sigma=0$. For $\sigma\in(0,1]$, we write

$q^{\sigma}_{s}=k_{t+s}-k^{\sigma}_{t+s}=\varXi_{\bar{\vartheta}_{t}\vartheta(t)}(s)k_{t}-\varXi^{\sigma}_{\bar{\vartheta}_{t}\vartheta(t)}(s)k^{\sigma}_{t},\qquad
s<T_{*}(\vartheta(t)).$ (6.13)

Note that

$\forall G\in\mathcal{G}_{\vartheta(t)}\qquad\langle\\!\langle
q^{\sigma}_{0},G\rangle\\!\rangle\to 0\qquad{\rm as}\ \ \sigma\to 0.$ (6.14)

At the same time, (6.13) can be written in the form

$\displaystyle q^{\sigma}_{s}$ $\displaystyle=$
$\displaystyle\varXi_{\bar{\vartheta}_{t}\vartheta(t)}(s)q^{\sigma}_{0}-\varPi^{\sigma}_{\bar{\vartheta}_{t}\vartheta(t)}(s)k^{\sigma}_{t},$
(6.15) $\displaystyle\varPi^{\sigma}_{\bar{\vartheta}_{t}\vartheta(t)}(s)$
$\displaystyle:=$
$\displaystyle\int_{0}^{s}\frac{d}{du}\left[\varXi_{\bar{\vartheta}_{t}\vartheta}(s-u)\varXi^{\sigma}_{{\vartheta}\vartheta(t)}(u)\right]du,$

where $s$ and $\vartheta\in(\vartheta(t),\bar{\vartheta}_{t})$ are chosen in
such a way that

$s<\min\\{T(\bar{\vartheta}_{t},\vartheta);T({\vartheta},\vartheta(t))\\},$
(6.16)

and hence the Bochner integral in the second line of (6.15) makes sense, see
(6.6). Since the map $(\vartheta,\vartheta^{\prime})\mapsto
T(\vartheta^{\prime},\vartheta)$ is continuous, see (4.21), one can pick
$\vartheta_{1}<\vartheta$ and $\vartheta_{2}>\vartheta$, $\vartheta$ being as
in (6.16), such that

$s<\min\\{T(\bar{\vartheta}_{t},\vartheta_{2});T({\vartheta}_{1},\vartheta(t))\\}.$
(6.17)

Keeping this in mind, we use an evident identical extension of (4.31) to all
$\sigma\leq 1$ and obtain

$\displaystyle\varPi^{\sigma}_{\bar{\vartheta}_{t}\vartheta(t)}(s)=-\int_{0}^{s}\varXi_{\bar{\vartheta}_{t}\vartheta_{2}}(s-u)\widetilde{L}^{\Delta,\sigma}_{\vartheta_{2}\vartheta_{1}}\varXi^{\sigma}_{{\vartheta}_{1}\vartheta(t)}(u)du,$
(6.18)
$\displaystyle\widetilde{L}^{\Delta,\sigma}_{\vartheta_{2}\vartheta_{1}}:={L}^{\Delta}_{\vartheta_{2}\vartheta_{1}}-{L}^{\Delta,\sigma}_{\vartheta_{2}\vartheta_{1}}.$

We apply this in (6.15) and get

$q^{\sigma}_{s}=\varXi_{\bar{\vartheta}_{t}\vartheta(t)}(s)q^{\sigma}_{0}+\int_{0}^{s}\varXi_{\bar{\vartheta}_{t}\vartheta_{2}}(s-u)\widetilde{L}^{\Delta,\sigma}_{\vartheta_{2}\vartheta_{1}}k_{t+u}^{\sigma}du.$

Note that $\widetilde{L}^{\Delta,\sigma}$ can be written in the same form as
$L^{\Delta}$, see (4.1), in which $a_{i}(x-y)$, $i=0,1$, ought to be replaced
by $\tilde{a}^{\sigma}_{i}(x,y):=a_{i}(x-y)(1-\psi_{\sigma}(x))$. Now let us
turn to picking $s_{0}$ and $\vartheta_{j}$, $j=1,2$, such that (6.17) holds
for $s\leq s_{0}$. First we set
$\vartheta_{1}=\vartheta(t)+\delta(\vartheta(t))/2$, see (5.11). By (4.21) and
(5.11) we then get

$T(\bar{\vartheta}_{t},\vartheta_{1})=T_{*}(\vartheta(t))/2<T(\vartheta_{1},\vartheta(t)).$

Now we fix some $\epsilon\in(0,1)$ and set

$s_{0}=\epsilon T_{*}(\vartheta(t))/2=\epsilon
T(\bar{\vartheta}_{t},\vartheta_{1}).$ (6.19)

Since the map $\vartheta\mapsto T(\vartheta^{\prime},\vartheta)$ is
continuous, one can pick $\vartheta_{2}\in(\vartheta_{1},\bar{\vartheta}_{t})$
such that $s_{0}<T(\bar{\vartheta}_{t},\vartheta_{2})$, see (6.19). Then
(6.17) holds for these $\vartheta_{j}$, $j=1,2$, and $s\leq s_{0}$. Now we
take $G\in\mathcal{G}_{\bar{\vartheta}_{t}}$ and set

$G_{s}=\varSigma_{\vartheta_{2}\bar{\vartheta}_{t}}(s)G.$ (6.20)

Note that
$G_{s}\in\mathcal{G}_{\vartheta_{2}}\subset\mathcal{G}_{\vartheta(t)}$; that
is, $G_{s}$ can be considered as an element of $\mathcal{G}_{\vartheta(t)}$
since

$G_{s}=I_{\vartheta(t)\vartheta_{2}}\varSigma_{\vartheta_{2}\bar{\vartheta}_{t}}(s)G,$

where $I_{\vartheta(t)\vartheta_{2}}=\varSigma_{\vartheta(t)\vartheta_{2}}(0)$
is the embedding operator. For these $G$ and $G_{s}$, by (6.17) and (6.10) we
then have

$\displaystyle\langle\\!\langle
q_{s}^{\sigma},G\rangle\\!\rangle=\langle\\!\langle
q_{0}^{\sigma},G_{s}\rangle\\!\rangle+R^{\sigma}(s),$ (6.21) $\displaystyle
R^{\sigma}(s):=\int_{0}^{s}\langle\\!\langle\widetilde{L}^{\Delta,\sigma}_{\vartheta_{2}\vartheta_{1}}k^{\sigma}_{t+u},G_{s-u}\rangle\\!\rangle
du.$

In view of (6.14), it remains to prove that $R^{\sigma}(s)\to 0$ as $\sigma\to
0$. To this end, we split $R^{\sigma}(s)$ into four terms in accord with the
structure of $\widetilde{L}^{\Delta,\sigma}$, see (4.1). Thus, we write

$\displaystyle R^{\sigma}(s)=\sum_{j=1}^{4}R^{\sigma}_{j}(s),$ (6.22)

with

$\displaystyle R^{\sigma}_{1}(s)$ $\displaystyle=$
$\displaystyle\int_{0}^{s}\bigg{(}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\sum_{y\in\eta_{0}}\int_{X}\tilde{a}^{\sigma}_{0}(x,y)e(\tau^{0}_{y};\eta_{1})(\Upsilon^{0}_{y}k_{t+u}^{\sigma})(\eta_{0}\setminus
y\cup x,\eta_{1})dx\bigg{)}$ $\displaystyle\times$ $\displaystyle
G_{s-u}(\eta_{0},\eta_{1})\lambda(d\eta_{0})\lambda(d\eta_{1})\bigg{)}du,$
$\displaystyle=$
$\displaystyle\int_{0}^{s}\bigg{(}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}\tilde{a}^{\sigma}_{0}(x,y)e(\tau^{0}_{y};\eta_{1})(\Upsilon^{0}_{y}k_{t+u}^{\sigma})(\eta_{0}\cup
x,\eta_{1})$ $\displaystyle\times$ $\displaystyle G_{s-u}(\eta_{0}\cup
y,\eta_{1})dxdy\bigg{)}\lambda(d\eta_{0})\lambda(d\eta_{1})\bigg{)}du$
$\displaystyle R^{\sigma}_{2}(s)$ $\displaystyle=-$
$\displaystyle\int_{0}^{s}\bigg{(}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}\tilde{a}^{\sigma}_{0}(x,y)e(\tau^{0}_{y};\eta_{1})(\Upsilon^{0}_{y}k_{t+u}^{\sigma})(\eta_{0}\cup
x,\eta_{1})$ $\displaystyle\times$ $\displaystyle G_{s-u}(\eta_{0}\cup
x,\eta_{1})dxdy\bigg{)}\lambda(d\eta_{0})\lambda(d\eta_{1})\bigg{)}du,$
$\displaystyle R^{\sigma}_{3}(s)$ $\displaystyle=$
$\displaystyle\int_{0}^{s}\bigg{(}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}\tilde{a}^{\sigma}_{1}(x,y)e(\tau^{1}_{y};\eta_{0})(\Upsilon^{1}_{y}k_{t+u}^{\sigma})(\eta_{0},\eta_{1}\cup
x)$ $\displaystyle\times$ $\displaystyle G_{s-u}(\eta_{0},\eta_{1}\cup
y)dxdy\bigg{)}\lambda(d\eta_{0})\lambda(d\eta_{1})\bigg{)}du,$ $\displaystyle
R^{\sigma}_{4}(s)$ $\displaystyle=-$
$\displaystyle\int_{0}^{s}\bigg{(}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}\tilde{a}^{\sigma}_{1}(x,y)e(\tau^{1}_{y};\eta_{1})(\Upsilon^{1}_{y}k_{t+u}^{\sigma})(\eta_{0},\eta_{1})$
$\displaystyle\times$ $\displaystyle G_{s-u}(\eta_{0},\eta_{1}\cup
x)dxdy\bigg{)}\lambda(d\eta_{0})\lambda(d\eta_{1})\bigg{)}du.$

By (4.1), (4.6), (4.7) and (3.2) for each $\vartheta\in\mathds{R}$, $i=0,1$,
$s\geq 0$ and $(\eta_{0},\eta_{1})\in\Gamma_{0}^{2}$, we have

$\displaystyle\left|(\Upsilon^{i}_{y}k^{\sigma}_{s})(\eta_{0},\eta_{1})\right|\leq\|k_{s}^{\sigma}\|_{\vartheta}\exp\bigg{(}\vartheta(|\eta_{0}|+|\eta_{1}|)\bigg{)}\int_{\Gamma_{0}}e^{\vartheta|\xi|}e\left(|t^{i}_{y}|;\xi\right)\lambda(d\xi)$
(6.24)
$\displaystyle=\|k_{s}^{\sigma}\|_{\vartheta}\exp\bigg{(}\vartheta(|\eta_{0}|+|\eta_{1}|)\bigg{)}\sum_{n=0}^{\infty}\frac{1}{n!}e^{n\vartheta}\left(\int_{X}\left[1-e^{-\phi_{i}(x-y)}\right]dx\right)^{n}$
$\displaystyle\leq\|k_{s}^{\sigma}\|_{\vartheta}\exp\bigg{(}\vartheta(|\eta_{0}|+|\eta_{1}|)+\varphi
e^{\vartheta}\bigg{)}.$

By (6.18) we know that
$k^{\sigma}_{t+s}\in\mathcal{K}^{\star}_{\vartheta(t+u)}\subset\mathcal{K}^{\star}_{\vartheta_{1}}$,
which by (2.12) implies $\|k^{\sigma}_{t+s}\|_{\vartheta_{1}}\leq 1$. We take
this into account in (6.2), and also that $\tau_{y}^{i}(x)\leq 1$, see (4.6),
and then estimate the summands in (6.22) as follows

$\left|R^{\sigma}_{j}(s)\right|\leq\int_{X}r^{\sigma}_{j}(y)g_{j}(y)dy,$
(6.25)

with

$\displaystyle
r^{\sigma}_{1}(y)=\int_{X}(1-\psi_{\sigma}(x))a_{0}(x-y)dx,\quad
r^{\sigma}_{3}(y)=\int_{X}(1-\psi_{\sigma}(x))a_{1}(x-y)dx,$ (6.26)
$\displaystyle r^{\sigma}_{2}(y)=(1-\psi_{\sigma}(y))\bar{a}^{0}_{0},\qquad
r^{\sigma}_{4}(y)=(1-\psi_{\sigma}(y))\bar{a}^{0}_{1},$

see (3.3). It is clear that $r^{\sigma}_{j}(y)\leq r^{1}_{j}(y)$,
$j=1,\dots,4$, and

$\forall y\in X\qquad r^{\sigma}_{j}(y)\to 0\quad\sigma\to 0,\ \ =1,\dots,4.$
(6.27)

Furthermore,

$\displaystyle
g_{1}(y)=g_{2}(y)=c(\vartheta_{1})\int_{0}^{s}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\left|G_{u}(\eta_{0}\cup
y,\eta_{1})\right|e^{\vartheta_{1}(|\eta_{0}|+|\eta_{1}|)}\lambda(d\eta_{0})\lambda(d\eta_{1})du,$
(6.28) $\displaystyle
g_{3}(y)=g_{4}(y)=c(\vartheta_{1})\int_{0}^{s}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\left|G_{u}(\eta_{0},\eta_{1}\cup
y)\right|e^{\vartheta_{1}(|\eta_{0}|+|\eta_{1}|)}\lambda(d\eta_{0})\lambda(d\eta_{1})du,$

where $c(\vartheta_{1})=\exp(\vartheta_{1}+\varphi e^{\vartheta_{1}})$. Let us
show that each $g_{j}$, $j=1,\dots,4$, is integrable for all $s\leq s_{0}$.
Since $G_{u}\in\mathcal{G}_{\vartheta_{2}}$, by (4.32) and (6.20) for $u\leq
s\leq s_{0}$, see (6.19), we have

$|G_{u}|_{\vartheta_{2}}\leq\frac{T(\bar{\vartheta}_{t},\vartheta_{2})}{T(\bar{\vartheta}_{t},\vartheta_{2})-s_{0}}|G|_{\bar{\vartheta}_{t}}=:C_{G}.$
(6.29)

By (6.28) we then have

$\displaystyle\int_{X}g_{1}(y)dy$ $\displaystyle\leq$ $\displaystyle
c(\vartheta_{1})\int_{0}^{s}\left(\int_{\Gamma_{0}}\int_{\Gamma_{0}}\int_{X}\left|G_{u}(\eta_{0}\cup
y,\eta_{1})\right|e^{\vartheta_{1}(|\eta_{0}|+|\eta_{1}|)}\lambda(d\eta_{0})\lambda(d\eta_{1})dy\right)du\qquad$
$\displaystyle=$ $\displaystyle
c(\vartheta_{1})e^{-\vartheta_{1}}\int_{0}^{s}\left(\int_{\Gamma_{0}}\int_{\Gamma_{0}}|\eta_{0}|\left|G_{u}(\eta_{0},\eta_{1})\right|e^{\vartheta_{1}(|\eta_{0}|+|\eta_{1}|)}\lambda(d\eta_{0})\lambda(d\eta_{1})\right)du$
$\displaystyle\leq$
$\displaystyle\frac{sc(\vartheta_{1})}{(\vartheta_{2}-\vartheta_{1})e^{1+\vartheta_{1}}}C_{G}.$

Clearly, the same estimate holds for the remaining $g_{j}$. Then by the
dominated convergence theorem and (6.27), (6.25) it follows that
$R^{\sigma}(s)\to 0$ as $\sigma\to 0$, holding for all $s\leq s_{0}$, see
(6.19). By (6.14) and (6.21) this yields $\langle\\!\langle
q^{\sigma}_{s},G\rangle\\!\rangle\to 0$ as $\sigma\to 0$.

To complete the proof of this statement, let us consider the following
sequences, cf. (6.19),

$t_{l}=t_{l-1}+s_{0l},\quad s_{0l}=\epsilon T_{*}(\vartheta_{t_{l-1}})/2,\quad
t_{0}=0,\quad l\in\mathds{N}.$ (6.31)

Now we may use the construction just made and the induction in $l$, which
yields (6.12) holding for all $t\leq t_{l}$. Thus, the proof will follow if we
show $t_{l}\to+\infty$ as $l\to+\infty$. Set $\sup_{l}t_{l}=:t_{*}$. By (6.31)
we have $t_{l}=s_{01}+\cdots+s_{0l}$. Hence, $t_{*}<\infty$ yields $s_{0l}\to
0$, $l\to+\infty$. By passing to the limit in the second formula in (6.31) we
then get $T_{*}(\vartheta_{t_{*}})=0$, which is impossible, see (5.11). ∎

Proof of Lemma 6.3. By Lemma 6.4 it follows that
$\mu_{t}^{\sigma}(F)\to\mu_{t}(F)$, $\sigma\to 0$, holding for all
$F\in\mathcal{F}_{\infty}$, which by Proposition 5.2 yields that
$\mu_{t}^{\sigma}(F)\to\mu_{t}(F)$, $\sigma\to 0$, for all
$F\in\widetilde{\mathcal{F}}$. Then the property in question follows by claim
(ii) of Proposition 2.8. $\square$
We end up this subsection with the following complement to Lemma 6.3. For
$F\in\widetilde{\mathcal{F}}$, see (2.49), and a sequence
$\\{\mu_{n}\\}_{n\in\mathds{N}}\subset\mathcal{P}_{\rm exp}$ as in Proposition
6.2, consider

$\tilde{\mu}_{n}(d\gamma)=C^{-1}_{n}F(\gamma)\mu_{n}(d\gamma),\qquad
n\in\mathds{N}.$ (6.32)

where

$C_{n}=\mu_{n}(F)>0,$ (6.33)

since each $F\in\widetilde{\mathcal{F}}$ is strictly positive.

######  Proposition 6.5.

Let $\tilde{\mu}_{n}$ and $\mu_{n}$ be as in (6.32) and assume that
$\mu_{n}\Rightarrow\mu$ as $n\to+\infty$. Then
$\tilde{\mu}_{n}\Rightarrow\tilde{\mu}$, where

$\tilde{\mu}(d\gamma)=C^{-1}F(\gamma)\mu(d\gamma),\qquad C=\mu(F).$

###### Proof.

By assumption, $C_{n}\to C$. Take any $F^{\prime}\in\widetilde{\mathcal{F}}$
and set $F^{\prime\prime}=F^{\prime}F$, which is an element of
$\widetilde{\mathcal{F}}$ since the latter is closed under multiplication, see
Proposition 2.8. Then
$\tilde{\mu}_{n}(F^{\prime})=C_{n}^{-1}\mu_{n}(F^{\prime\prime})\to
C^{-1}\mu(F^{\prime\prime})$ as $n\to+\infty$. Since $\widetilde{\mathcal{F}}$
is convergence determining, see claim (ii) of Proposition 2.8, the sequence
$\\{\tilde{\mu}_{n}\\}_{n\in\mathds{N}}$ converges to some
$\tilde{\mu}\in\mathcal{P}_{\rm exp}$ (by Proposition 6.2), such that
$\tilde{\mu}(F^{\prime})=C^{-1}\mu(F^{\prime\prime})$. This implies that
$\tilde{\mu}$ is as stated since $\widetilde{\mathcal{F}}$ is separating. ∎

##  7\. Existence: Approximating Processes

The aim of this section is proving Theorem 3.5 by constructing path measures
for the model described by $L^{\sigma}$ introduced in the preceding section.
This will be done in a direct way by means of the corresponding Markov
transition functions.

###  7.1. The transition function

We start by introducing the real linear space of signed measures on
$\Gamma^{2}_{*}$, see [8, Chapter 4], which we denote by $\mathcal{M}$. That
is, each $\mu\in\mathcal{M}$ is a $\sigma$-additive map
$\mu:\mathcal{B}(\Gamma^{2}_{*})\to\mathds{R}$ taking finite values only. Let
$\mathcal{M}^{+}$ be the set of $\mu\in\mathcal{M}$ such that
$\mu(\mathbb{A})\geq 0$ for all $\mathbb{A}\in\mathcal{B}(\Gamma^{2}_{*})$.
Then the Jordan decomposition of a given $\mu\in\mathcal{M}$ is the unique
representation $\mu=\mu^{+}-\mu^{-}$, $\mu^{\pm}\in\mathcal{M}^{+}$, in view
of which the cone $\mathcal{M}^{+}$ is generating. Set
$|\mu|=\mu^{+}+\mu^{-}$. Then

$\|\mu\|=|\mu|(\Gamma^{2}_{*})$ (7.1)

is a norm, additive on the cone $\mathcal{M}^{+}$. According to [8,
Proposition 4.1.8, page 119], $\mathcal{M}$ is a Banach space with this norm.
Set $\Psi_{1}=1+\Psi$, where the latter was defined in (2.30), and then define

$\mathcal{M}_{n}=\\{\mu\in\mathcal{M}:\|\mu\|_{n}:=|\mu|(\Psi_{1}^{n})<\infty\\},\qquad
n\in\mathds{N}.$ (7.2)

By the same [8, Proposition 4.1.8, page 119] $\mathcal{M}_{n}$ with the norm
$\|\mu\|_{n}$ is also a real Banach space. In the sequel, we extend (7.2) to
$n=0$ by setting $\mathcal{M}_{0}=\mathcal{M}$ and $\|\mu\|_{0}=\|\mu\|$.
Additionally, for $n\in\mathds{N}_{0}$, we set

$\varphi_{n}(\mu)=\mu(\Psi_{1}^{n}).$ (7.3)

Now for $\beta>0$, define

$\|\mu\|_{\beta}=\int_{\Gamma^{2}_{*}}\exp\left(\beta\Psi(\gamma)\right)|\mu|(d\gamma),\quad\mathcal{M}_{\beta}=\\{\mu\in\mathcal{M}:\|\mu\|_{\beta}<\infty\\},$
(7.4)

and also

$\varphi_{\beta}(\mu)=\int_{\Gamma^{2}_{*}}\exp\left(\beta\Psi(\gamma)\right)\mu(d\gamma).$
(7.5)

It is clear that

$\forall\mu\in\mathcal{M}_{+}\qquad\qquad\|\mu\|_{n}=\varphi_{n}(\mu),\qquad\|\mu\|_{\beta}=\varphi^{\beta}(\mu),$
(7.6)

holding for all $n\in\mathds{N}_{0}$ and $\beta>0$. In our construction, we
essentially use the cones of positive elements

$\mathcal{M}_{n}^{+}=\mathcal{M}_{n}\cap\mathcal{M}^{+},\qquad\mathcal{M}_{\beta}^{+}=\mathcal{M}_{\beta}\cap\mathcal{M}^{+},\qquad\beta>0,\
n\in\mathds{N}.$ (7.7)

For a given $\mathcal{N}\subset\mathcal{M}$, by $\overline{\mathcal{N}}$ we
denote the closure of $\mathcal{N}$ in $\|\cdot\|$ defined in (7.1). The proof
of the next statement is completely analogous to that of [15, Lemma 7.4 and
Corollary 7.5 pages 39, 40], and thus is omitted here.

######  Proposition 7.1.

For each $n\in\mathds{N}$ and $\beta>0$, it follows that
$\overline{\mathcal{M}_{\beta}}=\overline{\mathcal{M}_{n}}=\mathcal{M}$ and
also
$\overline{\mathcal{M}^{+}_{\beta}}=\overline{\mathcal{M}^{+}_{n}}=\mathcal{M}^{+}$.

Finally, we denote $\mathcal{M}^{+,1}=\mathcal{P}(\Gamma^{2}_{*})$ and also

$\mathcal{M}^{+,1}_{n}=\mathcal{M}^{+,1}\cap\mathcal{M}_{\beta},\qquad\mathcal{M}^{+,1}_{n}=\mathcal{M}^{+,1}\cap\mathcal{M}_{n}.$
(7.8)

By (2.37) it follows that

$\forall\beta>0\ \ \forall n\in\mathds{N}\qquad\qquad\mathcal{P}_{\rm
exp}\subset\mathcal{M}_{\beta}^{+,1}\subset\mathcal{M}_{n}^{+,1}.$ (7.9)

Now for $\sigma\in(0,1]$, we set, cf. (6.1),

$\displaystyle\varPsi^{\sigma}(\gamma)$ $\displaystyle=$
$\displaystyle\sum_{x\in\gamma_{0}}\int_{X}a^{\sigma}_{0}(x,y)\exp\left(-\sum_{z\in\gamma_{1}}\phi_{0}(z-y)\right)dy$
$\displaystyle+$
$\displaystyle\sum_{x\in\gamma_{1}}\int_{X}a^{\sigma}_{1}(x,y)\exp\left(-\sum_{z\in\gamma_{0}}\phi_{1}(z-y)\right)dy,\qquad\gamma\in\Gamma^{2}_{*}.$

By (2.29) and (6.1) it follows that
$\psi(x)\leq\psi_{\sigma}(x)\leq\psi(x)/\sigma$, $\sigma\in(0,1]$. For these
values of $\sigma$, by (2.30) and (3.3), (4.15) we then have

$\varPsi^{\sigma}(\gamma)\leq(\alpha/\sigma)\varPsi(\gamma),\qquad\gamma\in\Gamma^{2}_{*}.$
(7.10)

As mentioned above, the transition function in question will be constructed
directly, i.e., by the formula

$p^{\sigma}_{t}(\gamma,\cdot)=S^{\sigma}(t)\delta_{\gamma}(\cdot),\qquad t>0,\
\ \gamma\in\Gamma^{2}_{*},$ (7.11)

where $\delta_{\gamma}$ is the Dirac measure centered at $\gamma$ and
$S^{\sigma}=\\{S^{\sigma}(t)\\}_{t\geq 0}$ is the stochastic semigroup of
bounded linear operators acting in $\mathcal{M}$, generated by the dual
$L^{\dagger,\sigma}$ of $L^{\sigma}$ defined in (6.1). The mentioned duality
means that

$\mu(L^{\sigma}F)=(L^{\dagger,\sigma}\mu)(F),\qquad F\in\mathcal{D}(L).$
(7.12)

Recall that the domains of all $L^{\sigma}$, $\sigma\in[0,1]$ are the same,
i.e., are as in Definition 3.1. By (6.1) and (6.2) we then get

$\displaystyle(L^{\dagger,\sigma}\mu)(\mathbb{A})$ $\displaystyle=$
$\displaystyle-\int_{\Gamma_{*}^{2}}\mathds{1}_{\mathbb{A}}(\gamma)\varPsi^{\sigma}(\gamma)\mu(d\gamma)+\int_{\Gamma^{2}_{*}}\varOmega^{\sigma}(\mathbb{A}|\gamma)\mu(d\gamma)$
$\displaystyle=:$
$\displaystyle(A\mu)(\mathbb{A})+(B\mu)(\mathbb{A}),\quad\mathbb{A}\in\mathcal{B}(\Gamma^{2}_{*}).$

Here

$\displaystyle\varOmega^{\sigma}(\mathbb{A}|\gamma)$ $\displaystyle=$
$\displaystyle\sum_{x\in\gamma_{0}}\int_{X}a^{\sigma}_{0}(x,y)\exp\left(-\sum_{z\in\gamma_{1}}\psi_{0}(z-y)\right)\mathds{1}_{\mathbb{A}}(\gamma_{0}\setminus
x\cup_{0}y,\gamma_{1})dy$ $\displaystyle+$
$\displaystyle\sum_{x\in\gamma_{1}}\int_{X}a^{\sigma}_{1}(x,y)\exp\left(-\sum_{z\in\gamma_{0}}\psi_{1}(z-y)\right)\mathds{1}_{\mathbb{A}}(\gamma_{0},\gamma_{1}\setminus
x\cup_{1}y)dy.$

Note that $A$ in (7.1) is just the multiplication operator by
$\varPsi^{\sigma}$, and the following holds

$\varOmega^{\sigma}(\Gamma^{2}_{*}|\gamma)=\varPsi^{\sigma}(\gamma).$ (7.15)

Now we set

$\mathcal{D}(L^{\dagger,\sigma})=\\{\mu\in\mathcal{M}:|\mu|(\varPsi^{\sigma})<\infty\\},$
(7.16)

which might have sense if we show that $B$ can act on
$\mu\in\mathcal{D}(L^{\dagger,\sigma})$. By writing $\mu=\mu^{+}-\mu^{-}$ we
conclude that it is enough to show $B\mu\in\mathcal{M}$ for positive
$\mu\in\mathcal{D}(L^{\dagger,\sigma})$ only. Since $B$ itself is positive, by
(7.6) and (7.15) we have that

$\displaystyle\|B\mu\|=(B\mu)(\Gamma^{2}_{*})=\int_{\Gamma^{2}_{*}}\varPsi^{\sigma}(\gamma)\mu(d\gamma)=\|A\mu\|,$
(7.17)

which yields
$L^{\dagger,\sigma}:\mathcal{D}(L^{\dagger,\sigma})\to\mathcal{M}$. Clearly,
$(A,\mathcal{D}(L^{\dagger,\sigma}))$ is closed and the following holds

$\mathcal{M}_{1}\subset\mathcal{D}(L^{\dagger,\sigma}),$ (7.18)

see (7.10) and (7.2).

######  Remark 7.2.

Note that $\delta_{\gamma}\in\mathcal{D}(L^{\dagger,\sigma})$, since
$\delta_{\gamma}(\varPsi^{\sigma})=\varPsi^{\sigma}(\gamma)<\infty$, holding
for all $\gamma\in\Gamma^{2}_{*}$, see (7.10) and (2.32). At the same time,
$\delta_{\gamma}$ is evidently not sub-Poissonian.

Along with constructing the semigroup $S^{\sigma}$, see (7.11), in Lemma 7.3
below we obtain a number of complementary results, which we then exploit for
proving Theorem 3.5. To this end, for $n\in\mathds{N}$ and a positive $\mu$,
let us consider, cf. (7.3),

$\displaystyle\varphi_{n}(B\mu)$ $\displaystyle=$
$\displaystyle\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)(B\mu)(d\gamma)=\int_{\Gamma^{2}_{*}}\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)\varOmega^{\sigma}(d\gamma|\gamma^{\prime})\mu(d\gamma^{\prime})$
$\displaystyle=$
$\displaystyle\int_{\Gamma^{2}_{*}}\bigg{(}\sum_{x\in\gamma_{0}}\int_{X}a_{0}^{\sigma}(x,y)\exp\left(-\sum_{z\in\gamma_{1}}\phi_{0}(z-y)\right)\varPsi_{1}^{n}(\gamma\setminus
x\cup_{0}y)dy\bigg{)}\mu(d\gamma)$ $\displaystyle+$
$\displaystyle\int_{\Gamma^{2}_{*}}\bigg{(}\sum_{x\in\gamma_{1}}\int_{X}a_{1}^{\sigma}(x,y)\exp\left(-\sum_{z\in\gamma_{0}}\phi_{1}(z-y)\right)\varPsi_{1}^{n}(\gamma\setminus
x\cup_{1}y)dy\bigg{)}\mu(d\gamma).$

Since $\varPsi_{1}(\gamma\setminus x\cup_{i}y)=1+\varPsi(\gamma\setminus
x\cup_{i}y)=\varPsi_{1}(\gamma)+\psi(y)-\psi(x)$, $i=0,1$, see (2.30), then

$\varPsi^{n}_{1}(\gamma\setminus x\cup_{i}y)\leq
2^{n}\varPsi^{n}_{1}(\gamma),$ (7.20)

which by (7.10) yields in (7.1) the following estimate

$\forall\mu\in\mathcal{M}_{n+1}^{+}\qquad\|B\mu\|_{n}=\varphi_{n}(B\mu)\leq
2^{n}\alpha\sigma^{-1}\|\mu\|_{n+1},$ (7.21)

and hence

$\forall n\in\mathds{N}_{0}\qquad
B:\mathcal{M}_{n+1}^{+}\to\mathcal{M}_{n}^{+}.$ (7.22)

In a similar way, one shows that
$\|A\mu\|_{n}\leq(\alpha/\sigma)\|\mu\|_{n+1}$ and

$-A:\mathcal{M}_{n+1}^{+}\to\mathcal{M}_{n}^{+},$ (7.23)

which finally yields that
$L^{\dagger,\sigma}:\mathcal{M}_{n+1}^{+}\to\mathcal{M}_{n}^{+}$, holding for
all $n\in\mathds{N}_{0}$. By means of (7.21) and the corresponding estimate
for $A$ we then define bounded linear operators

$(L^{\dagger,\sigma})^{l}_{n,n+l}:\mathcal{M}_{n}\to\mathcal{M}_{n+l},\qquad
l\in\mathds{N},$ (7.24)

the norms of which satisfy

$\|(L^{\dagger,\sigma})^{l}_{n,n+l}\|\leq\left(\frac{\alpha}{\sigma}\right)^{l}(2^{n}+1)(2^{n+1}+1)\cdots(2^{n+l-1}+1).$
(7.25)

Next, similarly as in (4.27) we also define bounded operators
$(L^{\dagger,\sigma}_{\beta^{\prime},\beta})^{l}:\mathcal{M}_{\beta}\to\mathcal{M}_{\beta^{\prime}}$,
$\beta>\beta^{\prime}>0$, see (7.4). To estimate their norms, for a given
$\mu=\mu^{+}-\mu^{-}\in\mathcal{D}(L^{\dagger,\sigma})$, we write

$\displaystyle
L^{\dagger,\sigma}\mu=(A+B)(\mu^{+}-\mu^{-})=(B\mu^{+}-A\mu^{-})-(B\mu^{-}-A\mu^{+})=:\mu_{1}^{+}-\mu_{1}^{-}.$

It is clear that $\mu_{1}^{\pm}\in\mathcal{M}^{+}$. Then

$\|L^{\dagger,\sigma}\mu\|_{\beta^{\prime}}\leq\|\mu_{1}^{+}\|_{\beta^{\prime}}+\|\mu_{1}^{-}\|_{\beta^{\prime}}=\|A\mu^{+}\|_{\beta^{\prime}}+\|A\mu^{-}\|_{\beta^{\prime}}+\|B\mu^{+}\|_{\beta^{\prime}}+\|B\mu^{-}\|_{\beta^{\prime}}.$
(7.26)

Here we used the additivity of the norms on the positive cone as well as the
positivity of $B$ and $-A$, see (7.22), (7.23). Now by (7.10) we have

$\varPsi^{\sigma}(\gamma)\exp\left(\beta^{\prime}\varPsi(\gamma)\right)\leq\frac{\alpha}{e\sigma(\beta-\beta^{\prime})}\exp\left(\beta\varPsi(\gamma)\right),$

which for $\mu\in\mathcal{M}^{+}_{\beta}$ yields

$\|A\mu\|_{\beta^{\prime}}\leq\frac{\alpha}{e\sigma(\beta-\beta^{\prime})}\|\mu\|_{\beta}.$
(7.27)

Next, similarly as in (7.1), (7.21) by (7.10) we have

$\displaystyle\int_{\Gamma^{2}_{*}}\exp\left(\beta^{\prime}\varPsi(\gamma)\right)(B\mu)(d\gamma)=\int_{\Gamma^{2}_{*}}\int_{\Gamma^{2}_{*}}\exp\left(\beta^{\prime}\varPsi(\gamma^{\prime})\right)\varOmega^{\beta}(d\gamma^{\prime}|\gamma)\mu(d\gamma)$
$\displaystyle\quad=\int_{\Gamma^{2}_{*}}\exp\left(\beta^{\prime}\varPsi(\gamma)\right)\bigg{(}\sum_{x\in\gamma_{0}}\int_{X}a_{0}^{\sigma}(x,y)\exp\left(-\sum_{z\in\gamma_{1}}\phi_{0}(z-y)+\beta^{\prime}(\psi(y)-\psi(x))\right)dy$
$\displaystyle\quad+\sum_{x\in\gamma_{1}}\int_{X}a_{1}^{\sigma}(x,y)\exp\left(-\sum_{z\in\gamma_{0}}\phi_{1}(z-y)+\beta^{\prime}(\psi(y)-\psi(x))\right)dy\bigg{)}\mu(d\gamma)$
$\displaystyle\quad\leq
e^{\beta^{\prime}}\int_{\Gamma^{2}_{*}}\varPsi^{\sigma}(\gamma)\exp\left(\beta^{\prime}\varPsi(\gamma)\right)\mu(d\gamma)\leq\frac{e^{\beta^{\prime}}\alpha}{\sigma
e(\beta-\beta^{\prime})}\|\mu\|_{\beta},\qquad\mu\in\mathcal{M}_{\beta}^{+}.$

We combine now this estimate with (7.27) and obtain in (7.26) the following,
cf. (4.27)

$\|(L^{\dagger,\sigma}_{\beta^{\prime},\beta})^{l}\|\leq\left(\frac{l}{eT_{\sigma}(\beta,\beta^{\prime})}\right)^{l},\qquad
l\in\mathds{N},$ (7.28)

with

$T_{\sigma}(\beta,\beta^{\prime})=\frac{\sigma(\beta-\beta^{\prime})}{\alpha
e^{\beta}}.$ (7.29)

By (7.24)), for each $l\in\mathds{N}$ and $\mu\in\mathcal{M}_{\beta}$, we have
that $(L^{\dagger,\sigma})^{l}\mu\in\mathcal{M}_{\beta^{\prime}}$,
$\beta^{\prime}<\beta$, and the following holds

$(L^{\dagger,\sigma}_{\beta^{\prime},\beta})^{l}\mu=(L^{\dagger,\sigma})^{l}\mu,\qquad
l\in\mathds{N}.$ (7.30)

In the next statement, we employ a perturbation technique for constructing
stochastic semigroups of bounded linear operators in ordered Banach spaces
with norms additive on the cones of positive elements. Note that the spaces
defined in (7.2), (7.4) have this property, see (7.6), (7.7). The details of
this technique can be found in our previous work [15, subsect. 7.1.1]. Here we
just recall that a semigroup $S=\\{S(t)\\}_{t\geq 0}$ of such operators is
called stochastic (resp. substochastic) if each $S(t)$ is positive and
$\|S(t)u\|=\|u\|$ (resp. $\|S(t)u\|\leq\|u\|$) for each positive $u$ and
$t>0$. Also, for a given $n\in\mathds{N}$ and

$\mathcal{D}^{\sigma}_{n}:=\\{\mu\in\mathcal{M}_{n}:\
|\mu|(\varPsi^{\sigma}\varPsi_{1}^{n})<\infty\\},$ (7.31)

cf. (7.2), (7.16), by the trace of $A$ in $\mathcal{M}_{n}$ we mean the
operator $(A,\mathcal{D}^{\sigma}_{n})$ acting therein.

######  Lemma 7.3.

For each $\sigma\in(0,1]$, the closure of
$(L^{\dagger,\sigma},\mathcal{D}(L^{\dagger,\sigma}))$, see (7.16), is the
generator of a stochastic semigroup, $S^{\sigma}=\\{S^{\sigma}(t)\\}_{t\geq
0}$, in $\mathcal{M}$ such that
$S^{\sigma}(t):\mathcal{M}_{n}\to\mathcal{M}_{n}$, holding for each $t>0$ and
$n\in\mathds{N}$. For each $n\in\mathds{N}$, the restrictions
$S^{\sigma}(t)|_{\mathcal{M}_{n}}$ constitute a $C_{0}$-semigroup on
$\mathcal{M}_{n}$. Additionally, for each $\beta>0$ and
$\beta^{\prime}\in(0,\beta)$,
$S^{\sigma}(t):\mathcal{M}^{+}_{\beta}\to\mathcal{M}^{+}_{\beta^{\prime}}$ for
$t<T_{\sigma}(\beta,\beta^{\prime})$, see (7.29).

###### Proof.

We basically follow the way of proving [15, Lemma 7.6], based on the Thieme-
Voigt theorem [22] in the form adapted to the context of the present work, see
[15, Assumption 7.1 and Proposition 7.2]. Thus, we begin by mentioning that
all the items of Assumption 7.1 _ibid_ are satisfied. That is: (i) each
$\mathcal{M}_{n}$ is dense in $\mathcal{M}$, see Proposition 7.1; (ii) each
$\mathcal{M}_{n}$ is a Banach space (by the aforementioned [8, Proposition
4.1.8, page 119]); (iii) each cone $\mathcal{M}_{n}^{+}$, $n\in\mathds{N}$, is
$\mathcal{M}^{+}\cap\mathcal{M}_{n}$ and $\|\cdot\|_{n}$ is additive on this
cone, see (7.6); (iv) each $\mathcal{M}_{n}^{+}$ is dense in
$\mathcal{M}^{+}$, see Proposition 7.1. Now we can apply [15, Proposition
7.2], which amounts to checking that:

  * (i)

$-A$ and $B$ map $\mathcal{D}(L^{\dagger,\sigma})\cap\mathcal{M}^{+}$ to
$\mathcal{M}^{+}$, which follows by the very definition of $A$ and (7.17);

  * (ii)

$(A,\mathcal{D}(L^{\dagger,\sigma}))$ generates a substochastic semigroup,
$S_{0}^{\sigma}=\\{S_{0}^{\sigma}(t)\\}_{t\geq 0}$, such that (a)
$S^{\sigma}_{0}(t):\mathcal{M}_{n}\to\mathcal{M}_{n}$, (b) the restrictions
$S_{0}^{\sigma}(t)|_{\mathcal{M}_{n}}$ constitute a $C_{0}$-semigroup on
$\mathcal{M}_{n}$ generated by the trace of $A$ in $\mathcal{M}_{n}$;

  * (iii)

$B:\mathcal{D}^{\sigma}_{n}\to\mathcal{M}_{n}$ and $\varphi((A+B)\mu)=0$,
holding for all $\mu\in\mathcal{D}(L^{\dagger,\sigma})\cap\mathcal{M}^{+}$;

  * (iv)

there exist positive $c_{n}$ and $\varepsilon_{n}$ such that the following
holds

$\forall\mu\in\mathcal{D}^{\sigma}_{n}\cap\mathcal{M}^{+}\qquad\varphi_{n}((A+B)\mu)\leq
c_{n}\varphi_{n}(\mu)-\varepsilon_{n}\|A\mu\|.$ (7.32)

The semigroup $S_{0}^{\sigma}$ mentioned in item (ii) consists of the
multiplication operators

$(S_{0}^{\sigma}(t)\mu)(d\gamma)=\exp\left(-t\varPsi^{\sigma}(\gamma)\right)\mu(d\gamma),$
(7.33)

which is certainly such that (a) holds for each $n\in\mathds{N}$. To check the
strong continuity of $S_{0}^{\sigma}$, we take $\mu\in\mathcal{M}^{+}$ and
$\varepsilon>0$, and then show that $\|\mu-S_{0}^{\sigma}(t)\mu\|<\varepsilon$
whenever $t$ is smaller than an $\varepsilon$-specific $\delta>0$. The
validity of such estimates for an arbitrary $\mu\in\mathcal{M}$ then simply
follows by the Jordan decomposition. Since $\mathcal{D}(L^{\dagger,\sigma})$
is dense in $\mathcal{M}$, see (7.18) and Proposition 7.1, one finds
$\mu^{\prime}\in\mathcal{D}(L^{\dagger,\sigma})\cap\mathcal{M}^{+}$ such that
$\|\mu-\mu^{\prime}\|<\varepsilon/3$. By (7.33) and (7.10) we then have

$\displaystyle\|\mu-S^{\sigma}_{0}(t)\mu\|\leq\|\mu-\mu^{\prime}\|+\|S^{\sigma}_{0}(t)(\mu-\mu^{\prime})\|+\|\mu^{\prime}-S^{\sigma}_{0}(t)\mu^{\prime}\|$
(7.34) $\displaystyle\leq
2\|\mu-\mu^{\prime}\|+t\int_{\Gamma^{2}_{*}}\varPsi^{\sigma}(\gamma)\mu^{\prime}(d\gamma)\leq\frac{2}{3}\varepsilon+\frac{t\alpha}{\sigma}\|\mu^{\prime}\|_{1}<\varepsilon,$

for $t<\sigma\varepsilon/3\alpha\|\mu^{\prime}\|_{1}$. Moreover, (7.33) can be
considered as the definition of bounded linear operators acting in a given
$\mathcal{M}_{n}$. These operators constitute a $C_{0}$ semigroup, which can
be proved similarly as in (7.34). Its generator is then obviously the trace of
$A$ in $\mathcal{M}_{n}$, see (7.31). Thus, it remains to prove the validity
of (7.32), that is, the validity of

$\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)(L^{\dagger,\sigma}\mu)(d\gamma)+\varepsilon_{n}\int_{\Gamma^{2}_{*}}\varPsi^{\sigma}(\gamma)\mu(d\gamma)\leq
c_{n}\int_{\Gamma^{2}+*}\varPsi_{1}^{n}(\gamma)\mu(d\gamma),$ (7.35)

holding for all $\mu\in\mathcal{D}^{\sigma}_{n}\cap\mathcal{M}^{+}$ and
certain positive $c_{n}$ and $\varepsilon_{n}$. This clearly amounts to
proving that each of the summands in the left-hand side of (7.35) is
$\leq(c_{n}/2)\mu(\varPsi_{1}^{n})$ with a properly chosen $c_{n}$. We begin
by proving this for the first summand. By (7.12) we have that

$\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)(L^{\dagger,\sigma}\mu)(d\gamma)=\int_{\Gamma^{2}_{*}}(L^{\sigma}\varPsi_{1}^{n})(\gamma)\mu(d\gamma)$
(7.36)

Similarly as in obtaining (7.20) we have

$\left|\varPsi_{1}^{n}(\gamma\setminus
x\cup_{i}y)-\varPsi_{1}^{n}(\gamma)\right|\leq(2^{n}-1)\left|\psi(y)-\psi(x)\right|\varPsi_{1}^{n-1}(\gamma).$
(7.37)

Set

$b_{i}(x)=\int_{X}a_{i}(x-y)\left|\psi(y)-\psi(x)\right|dy,\qquad i=0,1.$
(7.38)

Assume first that $|x|\geq|y|$. Then

$\displaystyle\left|\psi(y)-\psi(x)\right|=\psi(y)-\psi(x)=(|x|^{d+1}-|y|^{d+1}|\psi(x)\psi(y)$
(7.39)
$\displaystyle\leq\psi(x)\left[\left(|x-y|+|y|\right)^{d+1}-|y|^{d+1}\right]\psi(y)$
$\displaystyle=\psi(x)\sum_{l=1}^{d+1}{d+1\choose
l}|x-y|^{l}|y|^{d+1-l}\psi(y)$
$\displaystyle\leq\psi(x)\sum_{l=1}^{d+1}{d+1\choose
l}|x-y|^{l}=:\psi(x)h(x-y),$

where we have used the fact that $|y|^{d+1-l}\psi(y)\leq 1$ for all $l\geq 1$
and $y\in X$. For $|x|<|y|$, we have

$\displaystyle\left|\psi(y)-\psi(x)\right|=(|y|^{d+1}-|x|^{d+1})\psi(x)\psi(y)$
$\displaystyle\leq\psi(x)\left[\left(|y-x|+|x|\right)^{d+1}-|x|^{d+1}\right]\psi(y)$
$\displaystyle\leq\psi(x)\sum_{l=1}^{d+1}{d+1\choose
l}|x-y|^{l}|y|^{d+1-l}\psi(y)$ $\displaystyle\leq\psi(x)h(x-y).$

Now we use these two estimates in (7.38) and obtain

$b_{i}(x)\leq\psi(x)\bar{\alpha},\qquad\bar{\alpha}:=\max_{i=0,1}\int_{X}a_{i}(x)h(x)dx,$
(7.40)

cf. (4.15) and (3.3). By (7.37) and the latter estimate we obtain

$\displaystyle\left|(L^{\sigma}\varPsi_{1}^{n})(\gamma)\right|\leq\sum_{x\in\gamma_{0}}\int_{X}a_{0}(x-y)\left|\varPsi_{1}^{n}(\gamma\setminus
x\cup_{0}y)-\varPsi_{1}^{n}(\gamma)\right|dy$ (7.41)
$\displaystyle+\sum_{x\in\gamma_{1}}\int_{X}a_{1}(x-y)\left|\varPsi_{1}^{n}(\gamma\setminus
x\cup_{1}y)-\varPsi_{1}^{n}(\gamma)\right|dy$
$\displaystyle\leq(2^{n}-1)\bar{\alpha}\varPsi_{1}^{n-1}(\gamma)\left(\sum_{x\in\gamma_{0}}\psi(x)+\sum_{x\in\gamma_{1}}\psi(x)\right)\leq(2^{n}-1)\bar{\alpha}\varPsi_{1}^{n}(\gamma),$

holding for all $\sigma\in[0,1]$, including $\sigma=0$. By (7.41) and (7.36)
we then have

$\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)(L^{\dagger,\sigma}\mu)(d\gamma)\leq(2^{n}-1)\bar{\alpha}\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)\mu(d\gamma).$
(7.42)

By (7.10) we have
$\varPsi^{\sigma}(\gamma)\leq(\alpha/\sigma)\varPsi_{1}^{n}(\gamma)$ holding
for all $n\in\mathds{N}$ and $\gamma\in\Gamma_{*}^{2}$, which then yields

$\int_{\Gamma^{2}_{*}}\varPsi^{\sigma}(\gamma)\mu(d\gamma)\leq\frac{\alpha}{\sigma}\int_{\Gamma^{2}_{*}}\varPsi_{1}^{n}(\gamma)\mu(d\gamma).$

The latter estimate together with (7.42) yields the validity of (7.35) with
$\varepsilon_{n}=1$ and $c_{n}=(2^{n}-1)\bar{\alpha}+\alpha/\sigma$.

It remains now to prove the concluding statement of the lemma. We proceed by
defining the following bounded operators

$S^{\sigma}_{\beta^{\prime},\beta}(t)=I_{\beta^{\prime},\beta}+\sum_{l=1}^{\infty}\frac{t^{l}}{l!}(L^{\dagger,\sigma}_{\beta^{\prime},\beta})^{l},\qquad
t<T_{\sigma}(\beta,\beta^{\prime}),$ (7.43)

acting from $\mathcal{M}_{\beta}$ to $\mathcal{M}_{\beta^{\prime}}$, see
(7.29). Here the powers $(L^{\dagger,\sigma}_{\beta^{\prime},\beta})^{l}$
satisfy (7.28) and $I_{\beta^{\prime},\beta}$ is the embedding operator. By
(7.43) and (7.30), for each $\mu\in\mathcal{M}_{\beta}$, one has

$S^{\sigma}_{\beta^{\prime},\beta}(t)\mu=S^{\sigma}(t)\mu,$ (7.44)

where $S^{\sigma}(t)$ is the same as in the first part of the lemma. Then the
positivity of $S^{\sigma}_{\beta^{\prime},\beta}(t)$ follows by the positivity
of the latter. This completes the whole proof. ∎

Thus, the lemma just proved yields the existence of the semigroup $S^{\sigma}$
which we use in (7.11) to obtain the Markov transition function
$p_{t}^{\sigma}$. The fact that $t\mapsto p^{\sigma}_{t}$ satisfies the
corresponding conditions, see [11, eqs. (1.3) – (1.6), page 156], follows
directly by (7.11). We will use this function to construct the finite
dimensional marginals of the stochastic process corresponding to the
approximating model described by $L^{\sigma}$. This will be done in the next
subsection.

###  7.2. Constructing path measures

By means of the semigroup $S^{\sigma}$ constructed in Lemma 7.3 we may have

$\hat{\mu}^{\sigma}_{t}(\cdot)=(S^{\sigma}(t)\mu)(\cdot)=\int_{\Gamma^{2}_{*}}p^{\sigma}_{t}(\gamma,\cdot)\mu(d\gamma),\qquad\mu\in\mathcal{M}.$
(7.45)

Recall that here $\sigma\in(0,1]$ and $S^{\sigma}$ is stochastic. The latter
means that $\hat{\mu}^{\sigma}_{t}$ is in $\mathcal{P}(\Gamma^{2}_{*})$
whenever $\mu$ has this property. Moreover, $\hat{\mu}^{\sigma}_{t}$ may be in
$\mathcal{M}_{n}\cap\mathcal{P}(\Gamma^{2}_{*})$ under the corresponding
condition. However, so far we do not know whether $S^{\sigma}$ preserves
$\mathcal{P}_{\rm exp}$.

######  Lemma 7.4.

For given $\mu\in\mathcal{P}_{\rm exp}$, let
$t\mapsto\mu_{t}^{\sigma}\in\mathcal{P}_{\rm exp}$, $t>0$, be the solution of
(6.9), see Proposition 6.1. Let also $\hat{\mu}^{\sigma}_{t}$ be as in (7.45)
with the same $\mu$. Then, for all $t>0$, it follows that
${\mu}^{\sigma}_{t}=\hat{\mu}^{\sigma}_{t}$.

###### Proof.

By (7.9) and (7.18) it follows that $\mathcal{P}_{\rm
exp}\subset\mathcal{D}(L^{\dagger,\sigma})$, which means that
$t\mapsto\hat{\mu}^{\sigma}_{t}$ is the classical solution of the initial
value problem

$\frac{d}{dt}\hat{\mu}_{t}^{\sigma}=L^{\dagger,\sigma}\hat{\mu}^{\sigma}_{t},\qquad\hat{\mu}^{\sigma}_{t}|_{t=0}=\mu,$
(7.46)

which by (7.12) yields that $t\mapsto\hat{\mu}_{t}^{\sigma}$ solves (6.9).
Then the proof follows by Proposition 6.1. ∎

It is a standard fact that the transition function $p^{\sigma}_{t}$ determines
the finite dimensional distributions of a Markov process,
$\mathcal{X}^{\sigma}$, with values in $\Gamma^{2}_{*}$, see [11, Theorem 1.1,
page 157]. Our aim now is to prove that it has càdlàg paths. To this end, we
employ Chentsov-like arguments, cf. [7] and [15, Proposition 7.8], and thus
the metric $\upsilon^{*}$, see (2.45), (2.46). By Lemma 2.4 it is complete.
Set

$\displaystyle
w^{\sigma}_{u}(\gamma)=\int_{\Gamma^{2}_{*}}\upsilon^{*}(\gamma,\gamma^{\prime})p^{\sigma}_{u}(\gamma,d\gamma^{\prime})$
(7.47) $\displaystyle
W^{\sigma}_{u,v}(\gamma)=\int_{\Gamma^{2}_{*}}\upsilon^{*}(\gamma,\gamma^{\prime})w^{\sigma}_{u}(\gamma^{\prime})p^{\sigma}_{v}(\gamma,d\gamma^{\prime}),\quad
u,v\geq 0.$

Thereafter, for a triple $t_{3}>t_{2}>t_{1}\geq 0$, consider

$\mathcal{W}^{\sigma}(t_{1},t_{2},t_{3})=\int_{\Gamma^{2}_{*}}W^{\sigma}_{t_{3}-t_{2},t_{2}-t_{1}}(\gamma)\hat{\mu}^{\sigma}_{t_{1}}(d\gamma)=\int_{\Gamma^{2}_{*}}W^{\sigma}_{t_{3}-t_{2},t_{2}-t_{1}}(\gamma){\mu}^{\sigma}_{t_{1}}(d\gamma).$
(7.48)

Note that this $\mathcal{W}^{\sigma}(t_{1},t_{2},t_{3})$ depends also on
$\mu=\mu^{\sigma}_{t}|_{t=0}$, see Lemma 7.4. By combining [7, Theorem 1] and
[11, Theorems 8.6 - 8.8, pages 137-139] we obtain the following statement.

######  Proposition 7.5.

Given $T>0$, $\sigma\in(0,1]$, $s\geq 0$ and $\mu\in\mathcal{P}_{\rm exp}$,
assume that there exist $C_{\sigma}>0$ and $\delta>0$ such that, for each
triple that satisfies $t_{1}\geq s$, $t_{3}\leq T$ and $t_{3}-t_{1}<\delta$,
the following holds

$\mathcal{W}^{\sigma}(t_{1},t_{2},t_{3})\leq C_{\sigma}|t_{3}-t_{1}|^{2}.$
(7.49)

Then

  * (i)

There exists a probability measure $P^{\sigma}_{s,\mu}$ on
$\mathfrak{D}_{[0,+\infty)}(\Gamma^{2}_{*})$ uniquely determined by its finite
dimensional marginals, cf. [11, eq. (1.10), page 157], expressed by the
formula

$\displaystyle
P^{\sigma}_{s,\mu}(\\{\gamma:\varpi_{t_{n}}(\gamma)\in\mathbb{A}_{n},\varpi_{t_{n-1}}(\gamma)\in\mathbb{A}_{n-1},\dots,\varpi_{t_{1}}(\gamma)\in\mathbb{A}_{1},\varpi_{0}(\gamma)\in\mathbb{A}_{0}\\})\qquad$
(7.50) $\displaystyle\
=\int_{\mathbb{A}_{n-1}}\cdots\int_{\mathbb{A}_{0}}p^{\sigma}_{t_{n}-t_{n-1}}(\gamma_{n-1},\mathbb{A}_{n})p^{\sigma}_{t_{n-1}-t_{n-2}}(\gamma_{n-2},d\gamma_{n-1})\cdots
p^{\sigma}_{t_{2}-t_{1}}(\gamma_{1},d\gamma_{2})$ $\displaystyle\ \times
p^{\sigma}_{t_{1}}(\gamma_{0},d\gamma_{1})\mu(d\gamma_{0}),$

holding for all $n\in\mathds{N}$, $t_{n}>t_{n-1}\cdots t_{1}$ and
$\mathbb{A}_{j}\in\mathcal{B}(\Gamma^{2}_{*})$, $j=0,\dots,n$.

  * (ii)

If the estimate in (7.49) holds for all $\sigma\in(0,1]$ with one and the same
$C>0$,and the family $\\{\hat{\mu}_{t}^{\sigma}\\}_{\sigma\in(0,1]}$ is tight
for all $t>0$, then the family $\\{P^{\sigma}_{s,\mu}\\}_{\sigma\in(0,1]}$ of
measures mentioned in (i) is also tight, and hence has accumulation points in
the weak topology.

Note that the tightness mentioned in item (ii) of the latter statement follows
by Prohorov’s theorem and Lemmas 6.3 and 7.4.

######  Lemma 7.6.

For every $s\geq 0$ and $\mu\in\mathcal{P}_{\rm exp}$, the estimate as in
(7.49) holds for all $\sigma\in(0,1]$ with one and the same $C>0$, dependent
on $T$ only.

###### Proof.

For convenience, we take here $s=0$ – the proof for $s>0$ is completely
analogous. Then we begin by recalling that $\delta_{\gamma}$ is in
$\mathcal{D}(L^{\dagger,\sigma})$, see Remark 7.2. Thus, by (7.11) and the
corresponding formulas, see e.g., [11, eq. (1.16), page 9], we have

$p^{\sigma}_{t}(\gamma,\cdot)=\delta_{\gamma}(\cdot)+\int_{0}^{t}L^{\dagger,\sigma}p^{\sigma}_{s}(\gamma,\cdot)ds.$
(7.51)

We use this in (7.47), which yields

$\displaystyle w^{\sigma}_{u}(\gamma)$ $\displaystyle=$ $\displaystyle
w^{\sigma}_{0}(\gamma)+\int_{0}^{u}\left(\int_{\Gamma^{2}_{*}}\upsilon^{*}(\gamma,\gamma^{\prime})L^{\dagger,\sigma}p^{\sigma}_{s}(\gamma,d\gamma^{\prime})\right)ds$
$\displaystyle=$
$\displaystyle\int_{0}^{u}\left(\int_{\Gamma^{2}_{*}}\upsilon^{*}(\gamma,\gamma^{\prime})L^{\dagger,\sigma}p^{\sigma}_{s}(\gamma,d\gamma^{\prime})\right)ds$
$\displaystyle=$
$\displaystyle\int_{0}^{u}\left(\int_{\Gamma^{2}_{*}}L^{\sigma}\upsilon^{*}(\gamma,\gamma^{\prime})p^{\sigma}_{s}(\gamma,d\gamma^{\prime})\right)ds,$

see (7.12). The second equality in (7.2) follows by the fact that
$w^{\sigma}_{0}(\gamma)=\upsilon^{*}(\gamma,\gamma)=0$ as $\upsilon^{*}$ is a
metric. The function $\gamma^{\prime}\mapsto
L^{\sigma}\upsilon^{*}(\gamma,\gamma^{\prime})=:J^{\sigma}_{\gamma}(\gamma^{\prime})$
has the following form, see (2.45), (2.46),

$\displaystyle J^{\sigma}_{\gamma}(\gamma^{\prime})$ $\displaystyle=$
$\displaystyle\sum_{x\in\gamma^{\prime}_{0}}\int_{X}a^{\sigma}_{0}(x,y)\exp\left(-\sum_{z\in\gamma_{1}}\phi_{0}(z-y)\right)\left[\upsilon^{*}(\gamma,\gamma^{\prime}\setminus
x\cup_{0}y)-\upsilon^{*}(\gamma,\gamma^{\prime})\right]dy\qquad$
$\displaystyle+$
$\displaystyle\sum_{x\in\gamma^{\prime}_{1}}\int_{X}a^{\sigma}_{1}(x,y)\exp\left(-\sum_{z\in\gamma_{0}}\phi_{1}(z-y)\right)\left[\upsilon^{*}(\gamma,\gamma^{\prime}\setminus
x\cup_{1}y)-\upsilon^{*}(\gamma,\gamma^{\prime})\right]dy.$

By the triangle inequality for the metric $\upsilon_{*}$ we then get

$\displaystyle\left|J^{\sigma}_{\gamma}(\gamma^{\prime})\right|$
$\displaystyle\leq$
$\displaystyle\sum_{x\in\gamma^{\prime}_{0}}\int_{X}a_{0}(x-y)\upsilon(\psi(\gamma^{\prime}_{0}\setminus
x\cup y),\psi\gamma_{0}^{\prime})dy$ $\displaystyle+$
$\displaystyle\sum_{x\in\gamma^{\prime}_{1}}\int_{X}a_{1}(x-y)\upsilon(\psi(\gamma^{\prime}_{1}\setminus
x\cup y),\psi\gamma_{1}^{\prime})dy.$

By (2.44) and (2.46) it follows that

$\upsilon(\psi(\gamma^{\prime}_{i}\setminus x\cup
y),\psi\gamma_{i}^{\prime})\leq\sup_{g:\|g\|_{BL}\leq
1}\left|g(x)\psi(x)-g(y)\psi(y)\right|.$ (7.55)

Proceeding similarly as in (7.39) we obtain

$\displaystyle\left|g(x)\psi(x)-g(y)\psi(y)\right|=\psi(x)\psi(y)\left|\frac{g(x)}{\psi(y)}-\frac{g(y)}{\psi(x)}\right|$
$\displaystyle\leq\psi(x)\psi(y)g(y)\left|\frac{1}{\psi(y)}-\frac{1}{\psi(x)}\right|+\psi(x)\left|g(x)-g(y)\right|$
$\displaystyle\leq\psi(x)\left[|x-y|+\sum_{l=1}^{d+1}{d+1\choose
l}|x-y|^{l}\right].$

We apply the latter in (7.55), (7.2) and then arrive at the following, see
(7.40) and (4.15),

$\left|J^{\sigma}_{\gamma}(\gamma^{\prime})\right|\leq(\alpha+\bar{\alpha})\varPsi(\gamma^{\prime}),$
(7.56)

which is uniform in $\sigma$. By (7.56) we then get from (7.2) the following
estimate

$w^{\sigma}_{u}(\gamma)\leq(\alpha+\bar{\alpha})\int_{0}^{u}\kappa^{\sigma}_{s}(\gamma)ds,\qquad\kappa^{\sigma}_{s}(\gamma):=\int_{\Gamma^{2}_{*}}\varPsi(\gamma^{\prime})p^{\sigma}_{s}(\gamma,d\gamma^{\prime}).$
(7.57)

By (7.51), similarly as in (7.2) we have

$\kappa^{\sigma}_{s}(\gamma)=\varPsi(\gamma)+\int_{0}^{s}\left(\int_{\Gamma^{2}_{*}}L^{\sigma}\varPsi(\gamma^{\prime})p^{\sigma}_{v}(\gamma,d\gamma^{\prime})\right)dv$
(7.58)

Proceeding as in (7.41) we obtain

$\left|L^{\sigma}\varPsi(\gamma^{\prime})\right|\leq\bar{\alpha}\varPsi(\gamma^{\prime}),$

by which we obtain from (7.57), (7.58) the following

$\kappa^{\sigma}_{s}(\gamma)\leq\varPsi(\gamma)+\bar{\alpha}\int_{0}^{s}\kappa^{\sigma}_{v}(\gamma)dv,$
(7.59)

which by the Grönwall inequality and (7.57) leads to

$w^{\sigma}_{u}(\gamma)\leq(\alpha+\bar{\alpha})ue^{\bar{\alpha}u}\varPsi(\gamma).$
(7.60)

Now we may pass to estimating $W^{\sigma}_{u,v}(\gamma)$. By the second line
in (7.47) and (7.60) we have

$W^{\sigma}_{u,v}(\gamma)\leq(\alpha+\bar{\alpha})ue^{\bar{\alpha}u}V^{\sigma}_{v}(\gamma),\qquad
V^{\sigma}_{v}(\gamma):=\int_{\Gamma_{*}^{2}}\upsilon^{*}(\gamma,\gamma^{\prime})\varPsi(\gamma^{\prime})p^{\sigma}_{v}(\gamma,d\gamma^{\prime}).$
(7.61)

Here we again apply (7.51) and then get, cf. (7.2),

$V^{\sigma}_{v}(\gamma)=\int_{0}^{v}\left(\int_{\Gamma_{*}^{2}}\left[L^{\sigma}\varPsi(\gamma^{\prime})\upsilon^{*}(\gamma,\gamma^{\prime})\right]p^{\sigma}_{s}(\gamma,d\gamma^{\prime})\right)ds$
(7.62)

Proceeding as in (7.2) we get, see also (7.38),

$\displaystyle\left|L^{\sigma}\varPsi(\gamma^{\prime})\upsilon^{*}(\gamma,\gamma^{\prime})\right|$
$\displaystyle\leq$
$\displaystyle\sum_{x\in\gamma^{\prime}_{0}}\int_{X}a_{0}(x-y)\bigg{|}\varPsi(\gamma^{\prime}\setminus
x\cup_{0}y)\upsilon^{*}(\gamma,\gamma^{\prime}\setminus x\cup_{0}y)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad-\varPsi(\gamma^{\prime})\upsilon^{*}(\gamma,\gamma^{\prime})\bigg{|}dy$
$\displaystyle+$
$\displaystyle\sum_{x\in\gamma^{\prime}_{1}}\int_{X}a_{1}(x-y)\bigg{|}\varPsi(\gamma^{\prime}\setminus
x\cup_{1}y)\upsilon^{*}(\gamma,\gamma^{\prime}\setminus x\cup_{1}y)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad-\varPsi(\gamma^{\prime})\upsilon^{*}(\gamma,\gamma^{\prime})\bigg{|}dy$
$\displaystyle\leq$ $\displaystyle
2\bigg{(}\sum_{x\in\gamma^{\prime}_{0}}b_{0}(x)+\sum_{x\in\gamma^{\prime}_{1}}b_{1}(x)\bigg{)}$
$\displaystyle+$
$\displaystyle\varPsi(\gamma^{\prime})\bigg{(}\sum_{x\in\gamma_{0}}\int_{X}a_{0}(x-y)\upsilon^{*}(\gamma^{\prime}\setminus
x\cup_{0}y,\gamma^{\prime})dy$
$\displaystyle\qquad+\sum_{x\in\gamma_{0}}\int_{X}a_{1}(x-y)\upsilon^{*}(\gamma^{\prime}\setminus
x\cup_{1}y,\gamma^{\prime})dy\bigg{)}$ $\displaystyle\leq$ $\displaystyle
2\bar{\alpha}\varPsi(\gamma^{\prime})+(\alpha+\bar{\alpha})\varPsi^{2}(\gamma^{\prime}),$

where we used also (7.40) and (7.2). Now we apply the latter estimate in
(7.62) and obtain

$V^{\sigma}_{v}(\gamma)\leq
2\alpha\int_{0}^{v}\kappa^{\sigma}_{s}(\gamma)ds+(\alpha+\bar{\alpha})\int_{0}^{v}K^{\sigma}_{s}(\gamma)ds,$
(7.63)

where $\kappa^{\sigma}_{s}(\gamma)$ is the same as in (7.57) and

$K^{\sigma}_{s}(\gamma)=\int_{\Gamma^{2}_{*}}\varPsi^{2}(\gamma^{\prime})p_{s}^{\sigma}(\gamma,d\gamma^{\prime}).$

By (7.51) we have, cf. (7.58), (7.59),

$K^{\sigma}_{s}(\gamma)=\varPsi^{2}(\gamma)+\int_{0}^{s}\left(\int_{\Gamma^{2}_{*}}(L^{\sigma}\varPsi^{2}(\gamma^{\prime}))p_{u}^{\sigma}(\gamma,d\gamma^{\prime})\right)du.$
(7.64)

Similarly as in (7.41) it follows that

$\left|(L^{\sigma}\varPsi^{2})(\gamma^{\prime})\right|\leq
3\bar{\alpha}\varPsi^{2}(\gamma^{\prime}),$

by which and the Grönwall inequality we get from (7.64) the following estimate

$K^{\sigma}_{s}(\gamma)\leq\varPsi^{2}(\gamma)e^{3\bar{\alpha}s}.$ (7.65)

Now we use (7.59) and (7.65) in (7.63) and arrive at

$V^{\sigma}_{v}(\gamma)\leq v\left(2\alpha
e^{\bar{\alpha}v}\varPsi(\gamma)+(\alpha+\bar{\alpha})e^{3\bar{\alpha}v}\varPsi^{2}(\gamma)\right)$
(7.66)

Now we may turn to (7.48) where we use the estimate, see (2.36),

$\int_{\Gamma^{2}_{*}}\varPsi^{n}(\gamma)\mu_{t_{1}}(d\gamma)\leq
T_{n}(\varkappa\langle\psi\rangle),\qquad n=1,2,$

and the fact that $\mu_{t_{1}}\in\mathcal{P}_{\rm exp}$ of type not exceeding
$\exp(\vartheta_{0}+\alpha t_{1})\leq\exp(\vartheta_{0}+\alpha T)=:\varkappa$,
see Proposition 4.3 and Remark 4.4. Here $e^{\vartheta_{0}}$ is the type of
$\mu$. By (7.61) and (7.66) we then conclude that
$\mathcal{W}^{\sigma}(t_{1},t_{2},t_{3})$ satisfies (7.49) with

$C=2\alpha(\alpha+\bar{\alpha})e^{2\bar{\alpha}T}T_{1}(\varkappa\langle\psi\rangle)+(\alpha+\bar{\alpha})^{2}e^{4\bar{\alpha}T}T_{2}(\varkappa\langle\psi\rangle),$

which ends the proof. ∎

##  8\. Completing the Proof

Here the hardest part is the proof of item (i), whereas the validity of (iii)
is rather standard, see cf. [9, Theorem 5.1.2, claim (iv), page 80].

###  8.1. Proving item (i)

First we prove existence by employing the fact that, for a given
$\mu\in\mathcal{P}_{\rm exp}$ and $s\geq 0$, the measure in question,
$P_{s,\mu}$, is obtained as an accumulation point of the family
$\\{P^{\sigma}_{s,\mu}\\}_{\sigma\in(0,1]}$. Our aim now is to prove that such
accumulation points have properties (a), (b), (c) mentioned in Definition 3.3.

To check (a), we note that $P_{s,\mu}\circ\varpi_{s}^{-1}=\mu$ since
$P^{\sigma}_{s,\mu}\circ\varpi_{s}^{-1}=\mu$ for all $\sigma\in(0,1]$.
Furthermore, by Lemmas 7.4 and 6.3 it follows that
$P^{\sigma}_{s,\mu}\circ\varpi^{-1}_{t}\Rightarrow\mu_{t}$ as $\sigma\to 0$,
which yields $P_{s,\mu}\circ\varpi^{-1}_{t}=\mu_{t}$, holding for all
accumulation points in view of Lemma 5.3. These facts yield the validity of
(b) of Definition 3.3.

To check (c), we take $\sf G$ as in (3.18) with fixed $t_{2}>t_{1}>s$,
$m\in\mathds{N}$ and $s_{1}<s_{2}<\cdots<s_{m}$, $s_{1}\geq s$, $s_{m}\leq
t_{1}$. Then we recall that
$\mu_{s_{1}}^{\sigma}=\hat{\mu}^{\sigma}_{s_{1}}=S^{\sigma}(s_{1}-s)\mu$, the
type of which does not exceed $e^{\vartheta(s_{1}-s)}$,
$\vartheta(t)=\vartheta_{0}+\alpha t$, see Lemma 7.4, and set
$\chi^{\sigma}_{s_{1}}=C^{-1}_{1,\sigma}F_{1}\mu^{\sigma}_{s_{1}}$, that is,

$\displaystyle\chi^{\sigma}_{s_{1}}(d\gamma)=C^{-1}_{1,\sigma}F_{1}(\gamma)\mu^{\sigma}_{s_{1}}(d\gamma),\qquad
C_{1,\sigma}:=\mu_{s_{1}}^{\sigma}(F_{1}).$ (8.1)

Note that $C_{1,\sigma}>0$ since each $F\in\widetilde{\mathcal{F}}$ is
strictly positive, see (2.49) and (6.33). By claim (d) of Proposition 3.2 it
follows that $\chi^{\sigma}_{s_{1}}\in\mathcal{P}_{\rm exp}$, and its type
does not exceed that of $\mu_{s_{1}}$, and hence $\exp(\vartheta(s_{1}-s))$.
Then we define recursively

$\displaystyle\tilde{\chi}^{\sigma}_{s_{l}}(d\gamma)=(S^{\sigma}(s_{l}-s_{l-1})\chi^{\sigma}_{s_{l-1}})(d\gamma)=\int_{\Gamma^{2}_{*}}p^{\sigma}_{s_{l}-s_{l-1}}(\gamma^{\prime},d\gamma)\chi^{\sigma}_{s_{l-1}}(d\gamma^{\prime}),$
(8.2)
$\displaystyle\chi^{\sigma}_{s_{l}}(d\gamma)=C_{l,\sigma}^{-1}F_{l}(\gamma)\tilde{\chi}^{\sigma}_{s_{l}}(d\gamma),\qquad
C_{l,\sigma}:=\tilde{\chi}^{\sigma}_{s_{l}}(F_{l}),\qquad l\leq m.$

As above, for all $l\leq m$, $\chi^{\sigma}_{s_{l}}$ is sub-Poissonian of type
$\leq\exp(\vartheta(s_{l}-s))$. Now we take $F\in\mathcal{D}(L)$, see
Definition 3.1, $t\in[s_{m},t_{2}]$, set

${\sf F}_{t}=F\circ\varpi_{t},\qquad{\sf K}_{t}=(LF)\circ\varpi_{t},\qquad{\sf
K}^{\sigma}_{t}=(L^{\sigma}F)\circ\varpi_{t},\quad\ \sigma\in(0.1],$ (8.3)

and then consider $P^{\sigma}_{s,\mu}({\sf F}_{t}{\sf G})$ with $\sf G$ as
just discussed. By (7.50) it follows that

$P^{\sigma}_{s,\mu}({\sf F}_{t}{\sf
G})=C_{\sigma}P^{\sigma}_{s,\chi^{\sigma}_{s_{m}}}({\sf
F}_{t})=C_{\sigma}P^{\sigma}_{s,\chi^{\sigma}_{s_{m}}}(F\circ\varpi_{t})=:C_{\sigma}\mu^{\sigma,s_{m}}_{t}(F),$
(8.4)

with $C_{\sigma}=P^{\sigma}_{s,\mu}({\sf G})>0$. By (7.50)

$\mu^{\sigma,s_{m}}_{t}=S^{\sigma}(t-s_{m})\chi^{\sigma}_{s_{m}},$ (8.5)

and the type of $\mu^{\sigma,s_{m}}_{t}$ is $\leq e^{\vartheta(t-s)}$. By
(8.5) it follows that

$\mu^{\sigma,s_{m}}_{t_{2}}(F)-\mu^{\sigma,s_{m}}_{t_{1}}(F)=\int_{t_{1}}^{t_{2}}\mu^{\sigma,s_{m}}_{t}(L^{\sigma}F)dt=\int_{t_{1}}^{t_{2}}P^{\sigma}_{s,\mu}({\sf
F}_{t}{\sf G})dt,$

see (8.4), which yields $P^{\sigma}_{s,\mu}({\sf H})=0$, holding for all
$\sigma\in(0,1]$.

Now let $P_{s,\mu}$ be an accumulation point of the family
$\\{P^{\sigma}_{s,\mu}\\}_{\sigma\in(0,1]}$. By Lemmas 6.3 and 7.4, all such
accumulation points have the same one dimensional marginals coinciding with
$\mu_{t}$. For this $P_{s,\mu}$, let
$\\{\sigma_{n}\\}_{n\in\mathds{N}}\subset(0,1]$, $\sigma_{n}\to 0$, be such
that $P^{\sigma_{n}}_{s,\mu}\Rightarrow P_{s,\mu}$ as $n\to+\infty$. Then set,
cf. (8.4),

$\displaystyle C_{n}=C_{\sigma_{n}}=P^{\sigma_{n}}_{s,\mu}({\sf G}),\qquad
C_{\infty}=P_{s,\mu}({\sf G}),$ (8.6)
$\displaystyle\mu^{s_{m}}_{t}(\mathbb{A})=C^{-1}_{\infty}P_{s,\mu}({\sf
G}\mathds{1}_{\mathbb{A}}\circ\varpi_{t}),\qquad\mathbb{A}\in\mathcal{B}(\Gamma^{2}_{*}),\qquad
t\in[s_{m},t_{2}].$

Let us show that the assumed convergence $P^{\sigma_{n}}_{s,\mu}\Rightarrow
P_{s,\mu}$ implies $\mu_{t}^{\sigma_{n},s_{m}}\Rightarrow\mu^{s_{m}}_{t}$, as
$n\to+\infty$. To this end, by $\tilde{\chi}_{s_{l}}$ we denote
$\mu_{s_{l}}^{s_{l-1}}$, where $\mu_{t}^{s_{l-1}}$, $t\geq s_{l-1}$ is the
solution of (1.2) with the initial condition
$\chi_{s_{l}}:=C^{-1}_{l-1,\infty}F_{l-1}\tilde{\chi}_{s_{l-1}}$, $l=2,\dots
m$, where $C_{l,\infty}=\tilde{\chi}_{s_{l}}(F_{l})$, cf. (8.2), and
$\tilde{\chi}_{s_{1}}=\mu_{s_{1}}=P_{s,\mu}\circ\varpi^{-1}_{s_{1}}$, which
solves (1.2) on $[s,s_{1}]$ with the initial condition $\mu$. The assumed
convergence of the path measures implies
$\tilde{\chi}_{s_{1}}^{\sigma_{n}}\Rightarrow\tilde{\chi}_{s_{1}}$, see (8.1).
By Lemma 6.3 this yields
$\tilde{\chi}_{s_{2}}^{\sigma_{n}}\Rightarrow\tilde{\chi}_{s_{2}}$, and thus
$\tilde{\chi}_{s_{l}}^{\sigma_{n}}\Rightarrow\tilde{\chi}_{s_{l}}$ for all
$l\leq m$. Since $\mu_{t}^{s_{m}}$ defined in (8.6) is the solution of (1.2)
on $[s_{m},t]$ with the initial condition
$\chi_{s_{m}}:=C^{-1}_{m,\infty}F_{m}\tilde{\chi}_{s_{m}}$, this yields the
convergence in question. By Proposition 6.2 this yields in turn that
$\mu^{s_{m}}_{t}\in\mathcal{P}_{\rm exp}$ and the type of $\mu^{s_{m}}_{t}$ is
$\leq\exp(\vartheta(t-s))$. Note that $C_{\infty}$ defined in (8.6) is
$C_{m,\infty}$.

Keeping the aforementioned facts in mind we write, see (3.17),

$P_{s,\mu}({\sf H})=P_{s,\mu}({\sf F}_{t_{2}}{\sf G})-P_{s,\mu}({\sf
F}_{t_{1}}{\sf G})-\int_{t_{1}}^{t_{2}}P_{s,\mu}({\sf K}_{t}{\sf G})dt,$ (8.7)

and also set

$\displaystyle a_{n}(t)=P_{s,\mu}({\sf F}_{t}{\sf
G})-P^{\sigma_{n}}_{s,\mu}({\sf F}_{t}{\sf G}),$ (8.8) $\displaystyle
b_{n}(t)=P_{s,\mu}({\sf K}_{t}{\sf G})-P^{\sigma_{n}}_{s,\mu}({\sf K}_{t}{\sf
G}),$ $\displaystyle c_{n}(t)=P^{\sigma_{n}}_{s,\mu}\left(({\sf K}_{t}-{\sf
K}^{\sigma_{n}}_{t}){\sf G}\right).$

Since $P^{\sigma}_{s,\mu}({\sf H})=0$, by (8.7) and (8.8) it follows that

$\displaystyle P_{s,\mu}({\sf H})=P_{s,\mu}({\sf
H})-P^{\sigma_{n}}_{s,\mu}({\sf H})=\left[a_{n}(t_{2})-a_{n}(t_{1})\right]$
$\displaystyle-\int_{t_{1}}^{t_{2}}b_{n}(t)dt-\int_{t_{1}}^{t_{2}}c_{n}(t)dt=:a_{n}-b_{n}-c_{n}.$

By $P_{s,\mu}^{\sigma_{n}}\Rightarrow P_{s,\mu}$ we have $a_{n}\to 0$ as
$n\to+\infty$. However, the same conclusion for $b_{n}$ and $c_{n}$ does not
follow in so simple way as $LF$ and $L^{\sigma}F$ need not be continuous. To
settle this case, by means of (8.6) we write

$\displaystyle
b_{n}=C_{n}\int_{t_{1}}^{t_{2}}\left(\mu^{s_{m}}_{t}(LF)-\mu^{\sigma_{n},s_{m}}_{t}(LF)\right)dt+(C_{\infty}-C_{n})\int_{t_{1}}^{t_{2}}\mu_{t}^{s_{m}}(LF)dt.$
(8.9)

By item (a) of Proposition 3.2, $LF$ is a bounded function; hence, the second
summand in (8.9) vanishes as $n\to+\infty$ since $C_{n}\to C_{\infty}$ by the
assumed weak convergence, see (8.6). To prove the same for the first summand –
denote it $b_{n}^{(1)}$ – we employ the fact that $\mu^{\sigma_{n},s_{m}}_{t}$
and $\mu^{s_{m}}_{t}$ are sub-Poissonian and each $F\in\mathcal{D}(L)$ can be
written as $KG$ with $G\in\widetilde{\mathcal{G}}_{\infty}$, see Proposition
5.2. Then

$\displaystyle\mu^{s_{m}}_{t}(LF)-\mu^{\sigma_{n},s_{m}}_{t}(LF)=\langle\\!\langle
k_{\mu^{s_{m}}_{t}}-k_{\mu^{\sigma_{n},s_{m}}_{t}},\widehat{L}G\rangle\\!\rangle\to
0,\qquad n\to+\infty,$ (8.10)

where we have taken into account that
$\widehat{L}G\in\widetilde{\mathcal{G}}_{\infty}$ whenever
$G\in\widetilde{\mathcal{G}}_{\infty}$, see (4.26), and also the fact that
$\mu^{\sigma_{n},s_{m}}_{t}\Rightarrow\mu^{s_{m}}_{t}$ implies the convergence
of the integrals in (8.10), see Proposition 6.2. As mentioned above, $LF$ is a
bounded function (by claim (i) of Proposition 3.2), which means that both
terms of the left-hand side of (8.10) are bounded by
$\sup_{\gamma\in\Gamma^{2}_{*}}|(LF)(\gamma)|$. Together with the convergence
$C_{n}\to C_{\infty}$ this yields $b_{n}^{(1)}\to 0$ as $n\to+\infty$.

Now we turn to $c_{n}$. By (8.3) and (8.4), and then by (6.5), we have

$\displaystyle
c_{n}(t)=C_{n}\left[\mu_{t}^{\sigma_{n},s_{m}}(LKG)-\mu_{t}^{\sigma_{n},s_{m}}(L^{\sigma_{n}}KG)\right]$
$\displaystyle=C_{n}\langle\\!\langle(L^{\Delta}-L^{\Delta,\sigma_{n}})k_{\mu_{t}^{\sigma_{n},s_{m}}},G\rangle\\!\rangle=C_{n}\langle\\!\langle\widetilde{L}^{\Delta,\sigma_{n}}k_{\mu_{t}^{\sigma_{n},s_{m}}},G\rangle\\!\rangle,$

cf. (6.21). Here we have also taken into account that
$\mu_{t}^{\sigma_{n},s_{m}}\in\mathcal{P}_{\rm exp}$, that was established
above, and the operator $\widetilde{L}^{\Delta,\sigma_{n}}$ is the same as in
(6.21). To make precise in which spaces $\mathcal{K}_{\vartheta}$ it acts, we
will take into account that
$G\in\mathcal{G}_{\infty}=\cap_{\vartheta\in\mathds{R}}\mathcal{G}_{\vartheta}$,
see Proposition 5.2, and that the type of each $\mu^{\sigma_{n},s_{m}}_{t}$
does not exceed $\exp(\vartheta(t_{2}-s))=:e^{\vartheta}$. Then we write, cf.
(6.21) and (6.22),

$\langle\\!\langle\widetilde{L}^{\Delta,\sigma_{n}}k_{\mu_{t}^{\sigma_{n},s_{m}}},G\rangle\\!\rangle=:R_{n}(t)=\sum_{j=1}^{4}R_{n,j}(t),$

where

$\displaystyle R_{n,1}(t)$ $\displaystyle=$
$\displaystyle\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\sum_{y\in\eta_{0}}\int_{X}a^{\sigma_{n}}_{0}(x,y)e(\tau^{0}_{y};\eta_{1})(\Upsilon^{0}_{y}k_{\mu_{t}^{\sigma_{n},s_{m}}})(\eta_{0}\setminus
y\cup x,\eta_{1})dx\bigg{)}$ $\displaystyle\times$ $\displaystyle
G(\eta_{0},\eta_{1})\lambda(d\eta_{0})\lambda(d\eta_{1})$ $\displaystyle=$
$\displaystyle\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}a^{\sigma_{n}}_{0}(x,y)e(\tau^{0}_{y};\eta_{1})(\Upsilon^{0}_{y}k_{\mu_{t}^{\sigma_{n},s_{m}}})(\eta_{0}\cup
x,\eta_{1})G(\eta_{0}\cup y,\eta_{1})dxdy\bigg{)}$ $\displaystyle\times$
$\displaystyle\lambda(d\eta_{0})\lambda(d\eta_{1}),$

and likewise

$\displaystyle R_{n,2}(t)$ $\displaystyle=$
$\displaystyle-\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}a^{\sigma_{n}}_{0}(x,y)e(\tau^{0}_{y};\eta_{1})(\Upsilon^{0}_{y}k_{\mu_{t}^{\sigma_{n},s_{m}}})(\eta_{0}\cup
x,\eta_{1})G(\eta_{0}\cup x,\eta_{1})dxdy\bigg{)}$ $\displaystyle\times$
$\displaystyle\lambda(d\eta_{0})\lambda(d\eta_{1}),\qquad$ $\displaystyle
R_{n,3}(t)$ $\displaystyle=$
$\displaystyle\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}a^{\sigma_{n}}_{1}(x,y)e(\tau^{1}_{y};\eta_{0})(\Upsilon^{1}_{y}k_{\mu_{t}^{\sigma_{n},s_{m}}})(\eta_{0},\eta_{1}\cup
x)G(\eta_{0},\eta_{1}\cup y)dxdy\bigg{)}$ $\displaystyle\times$
$\displaystyle\lambda(d\eta_{0})\lambda(d\eta_{1}),$ $\displaystyle
R_{n,4}(t)$ $\displaystyle=$
$\displaystyle-\int_{\Gamma_{0}}\int_{\Gamma_{0}}\bigg{(}\int_{X}\int_{X}a^{\sigma_{n}}_{1}(x,y)e(\tau^{1}_{y};\eta_{0})(\Upsilon^{1}_{y}k_{\mu_{t}^{\sigma_{n},s_{m}}})(\eta_{0},\eta_{1}\cup
x)G(\eta_{0},\eta_{1}\cup x)dxdy\bigg{)}$ $\displaystyle\times$
$\displaystyle\lambda(d\eta_{0})\lambda(d\eta_{1}).$

Now we take into account that
$k_{\mu_{t}^{\sigma_{n},s_{m}}}\in\mathcal{K}_{\vartheta}$ and
$G\in\mathcal{G}_{\infty}$, see above, employ (6.24), and then get, cf.
(6.25),

$\left|R_{n,j}(t)\right|\leq\int_{X}r^{\sigma_{n}}_{j}(y)g_{i}(y)dy,\qquad
j=1,\dots,4.$ (8.11)

with $r^{\sigma_{n}}_{j}(y)$ given in (6.26) and

$\displaystyle
g_{1}(y)=g_{2}(y)=c(\vartheta)\int_{\Gamma_{0}}\int_{\Gamma_{0}}\left|G(\eta_{0}\cup
y,\eta_{1})\right|\exp\left(\vartheta|\eta_{0}|+\vartheta|\eta_{1}|\right)\lambda(d\eta_{0})\lambda(d\eta_{1}),$
(8.12) $\displaystyle
g_{3}(y)=g_{4}(y)=c(\vartheta)\int_{\Gamma_{0}}\int_{\Gamma_{0}}\left|G(\eta_{0},\eta_{1}\cup
y)\right|\exp\left(\vartheta|\eta_{0}|+\vartheta|\eta_{1}|\right)\lambda(d\eta_{0})\lambda(d\eta_{1}),$

where $c(\vartheta)$ is the same as in (6.28). Note that the bound in (8.11)
is uniform in $t\in[s_{m},t_{2}]$, for which
$k_{\mu_{t}^{\sigma_{n},s_{m}}}\in\mathcal{K}_{\vartheta}$. Now, similarly as
in (6.2), we get

$\displaystyle\int_{X}g_{1}(y)dy\leq
c(\vartheta)e^{-\vartheta}\int_{\Gamma_{0}}\int_{\Gamma_{0}}|\eta_{0}|\left|G(\eta_{0},\eta_{1})\right|\exp\left(\vartheta|\eta_{0}|+\vartheta|\eta_{1}|\right)\lambda(d\eta_{0})\lambda(d\eta_{1})$
$\displaystyle\quad\leq
c(\vartheta)e^{-\vartheta-1-\log\varepsilon}\int_{\Gamma_{0}}\int_{\Gamma_{0}}\left|G(\eta_{0},\eta_{1})\right|\exp\bigg{(}(\vartheta+\varepsilon)\left(|\eta_{0}|+|\eta_{1}|\right)\bigg{)}\lambda(d\eta_{0})\lambda(d\eta_{1})$
$\displaystyle\quad=c(\vartheta)e^{-\vartheta-1-\log\varepsilon}|G|_{\vartheta+\varepsilon}<\infty$

that holds for all $\varepsilon>0$ as $G\in\mathcal{G}_{\infty}$. Similar
estimates can be obtained for the remaining $|R_{n,j}(t)|$. By the dominated
convergence theorem we then get that $R_{n}(t)\to 0$ as $n\to+\infty$,
uniformly in $t\in[s_{m},t_{2}]$, which together with the aforementioned
convergence $C_{n}\to C_{\infty}$, yields $c_{n}\to 0$ as $n\to+\infty$.
Therefore, for each limiting point $P_{s,\mu}$, it follows that
$P_{s,\mu}({\sf H})=0$, that yields the proof of item (c), and thus the
existence in question.

Now we turn to uniqueness. To this end we employ the following fact.

######  Proposition 8.1.

Assume that two solutions $\\{P^{(j)}_{s,\mu}:s\geq 0,\ \mu\in\mathcal{P}_{\rm
exp},\ j=1,2\\}$, see Definition 3.3, satisfy
$P^{(1)}_{s,\mu}\circ\varpi^{-1}_{t}=P^{(2)}_{s,\mu}\circ\varpi^{-1}_{t}$,
holding for all $t\geq s$, $s\geq 0$, and $\mu\in\mathcal{P}_{\rm exp}$. Then
they coincide, i.e., $P^{(1)}_{s,\mu}=P^{(2)}_{s,\mu}$ for all $s\geq 0$ and
$\mu\in\mathcal{P}_{\rm exp}$.

The proof of this statement – based on Lemma 5.3 – is completely analogous to
that of [15, Lemma 5.4], and this can be omitted here. Then the uniqueness in
question is straightforward. This completes the proof of item (i) of the
theorem.

###  8.2. Proving item (ii)

We begin by recalling that $X=\mathds{R}^{d}$. Let
$\\{r_{j}\\}_{j\in\mathds{N}}\subset\mathds{R}_{+}$ be a strictly increasing
sequence such that $\lim_{j\to+\infty}r_{j}=+\infty$. Set
$\varDelta_{j}=\\{x\in X:|x|<r_{j}\\}$ and
$\gamma_{i,j}=\gamma_{i}\cap\varDelta_{j}$, $i=0,1$, $j\in\mathds{N}$,
$\gamma=(\gamma_{0},\gamma_{1})\in\Gamma^{2}_{*}$. We also will use the
notation $\gamma_{j}$ for $(\gamma_{0,j},\gamma_{1,j})$. Then we define

$\Gamma^{2}_{*,j}=\\{\gamma\in\Gamma^{2}_{*}:\gamma_{j}\in\breve{\Gamma}^{2}\\}.$

That is, $\gamma\in\Gamma^{2}_{*}$ belongs to $\Gamma^{2}_{*,j}$ if and only
if $n_{0}(x)+n_{1}(x)=1$, holding for all $x\in p(\gamma)\cap\varDelta_{j}$,
see (1.1). By (2.31) we then have

$\breve{\Gamma}^{2}_{*}=\bigcap_{j\in\mathds{N}}\Gamma^{2}_{*,j}.$ (8.13)

Similarly as in [15, Lemma 2.7], one proves that each $\Gamma^{2}_{*,j}$ is an
open subset of $\Gamma^{2}_{*}$, see also Lemma 2.4. Define, cf. (2.24),

$\displaystyle h_{N}(x,y)$ $\displaystyle=$
$\displaystyle\psi(x)\psi(y)\min\\{N;|x-y|^{-d\epsilon}\\},\qquad
N\in\mathds{N},$ $\displaystyle H_{N}(\gamma)$ $\displaystyle=$
$\displaystyle\sum_{x\in\gamma_{0}}\sum_{y\in\gamma_{0}\setminus
x}h_{N}(x,y)+\sum_{x\in\gamma_{1}}\sum_{y\in\gamma_{1}\setminus
x}h_{N}(x,y+\sum_{x\in\gamma_{0}}\sum_{y\in\gamma_{1}}h_{N}(x,y),$

where $\epsilon\in(0,1)$ and $\psi(x)$ is as in (2.29). Now for
$\mu\in\mathcal{P}_{\rm exp}$ of type $\varkappa$, similarly as in (2.25) we
get

$\displaystyle\mu(H_{N})\leq 3\varkappa^{2}\mathcal{I}_{N},$ (8.14)

where, cf. (2.26),

$\displaystyle\mathcal{I}_{N}=\int_{X^{2}}h_{N}(x,y)dxdy\leq\int_{X}\psi(x)\left(\int_{X}\psi(y)|x-y|^{-d\epsilon}dy\right)dx$
$\displaystyle\leq\int_{X}\psi(x)\left(\int_{B_{r}}\frac{dz}{|z|^{d\epsilon}}+\frac{\langle\psi\rangle}{r^{d\epsilon}}\right)dx=\frac{c_{d}r^{d(1-\epsilon)}}{d(1-\epsilon)}\langle\psi\rangle+\frac{\langle\psi\rangle^{2}}{r^{d\epsilon}}.$

Then, similarly as in (2.27), we have that

$H(\gamma):=\lim_{N\to+\infty}H_{N}(\gamma)<\infty$

for $\mu$-almost all $\gamma\in\breve{\Gamma}^{2}_{*}$. At the same time,

$H_{j}(\gamma):=H(\gamma_{j})\leq
r^{-d\epsilon}_{\gamma_{j}}\left(|\gamma_{0}\cap\varDelta_{j}|+|\gamma_{1}\cap\varDelta_{j}|\right)^{2},$

holding for all $\gamma\in{\Gamma}^{2}_{*,j}$. Here $r_{\gamma_{j}}$ is the
minimal distance between two distinct
$x,y\in(\gamma_{0}\cap\varDelta_{j})\cup(\gamma_{0}\cap\varDelta_{j})$, which
is positive since the number of pairs of such points is finite and
$\gamma_{j}$ is simple.

Let $\\{P_{s,\mu}:s\geq 0,\ \mu\in\mathcal{P}_{\rm exp}\\}$ be the solution
which exists and is unique according to item (i). Fix some $s\geq 0$ and
$\mu\in\mathcal{P}_{\rm exp}$ of type $\varkappa$, and let $\mathcal{X}$ be
the process corresponding to $P_{s,\mu}$. For $N\in\mathds{N}$, we define the
stopping time, cf. [11, page 180],

$T_{N,j}=\inf\\{t\geq s:H_{j}(\mathcal{X}(t))>N\\}.$

Then for a fixed $j\in\mathds{N}$ and $T_{N,j}\wedge t:=\min\\{t;T_{N,j}\\}$,
we set $\mathcal{Z}(t)=\lim_{N\to+\infty}\mathcal{X}(T_{N,j}\wedge t)$ and
$T_{j}=\lim_{N\to+\infty}T_{N,j}$. Both limits exist as $T_{N,j}\leq
T_{N+1,j}$. Let $\tilde{\mu}_{t}$ be the law of $\mathcal{Z}(t)$. For
$\varPhi^{m}_{\tau}(\theta|\cdot)\in\mathcal{D}(L)$ defined in (5.24),

$\varPhi^{m}_{\tau}(\theta|\mathcal{X}(t))-\int_{s}^{t}L\varPhi^{m}_{\tau}(\theta|\mathcal{X}(u))du$
(8.15)

is a right-continuous martingale. Then, similarly as in [11, page 180], by the
optional sampling theorem, we can write

$\mathbb{E}\left[\varPhi^{m}_{\tau}(\theta|\mathcal{X}(T_{N,j}\wedge
t))\right]=\mathbb{E}\left[\varPhi^{m}_{\tau}(\theta|\mathcal{X}(s))\right]+\mathbb{E}\left[\int_{0}^{T_{N,j}\wedge
t}L\varPhi^{m}_{\tau}(\theta|\mathcal{X}(u))du\right],$

where we pass to the limit $N\to+\infty$ and get, see also (5.25),

$\displaystyle\tilde{\mu}_{t}(\varPhi^{m}_{\tau}(\theta|\cdot))$
$\displaystyle=$
$\displaystyle\mu(\varPhi^{m}_{\tau}(\theta|\cdot))+\lim_{N\to+\infty}\mathbb{E}\left[\int_{0}^{T_{N,j}\wedge
t}L\varPhi^{m}_{\tau}(\theta|\mathcal{X}(u))du\right]$ $\displaystyle\leq$
$\displaystyle\mu(\varPhi^{m}_{\tau}(\theta|\cdot))+\lim_{N\to+\infty}\mathbb{E}\left[\int_{0}^{T_{N,j}\wedge
t}\left|L\varPhi^{m}_{\tau}(\theta|\mathcal{X}(u))\right|du\right]$
$\displaystyle\leq$
$\displaystyle\mu(\varPhi^{m}_{\tau}(\theta|\cdot))+\mathbb{E}\left[\int_{0}^{t}\left|L\varPhi^{m}_{\tau}(\theta|\mathcal{X}(u))\right|du\right]$
$\displaystyle\leq$
$\displaystyle\mu(\varPhi^{m}_{\tau}(\theta|\cdot))+\mathbb{E}\left[\int_{0}^{t}\varPhi^{m}_{\tau,1}(\theta|\mathcal{X}(u))du\right]$
$\displaystyle=$
$\displaystyle\mu(\varPhi^{m}_{\tau}(\theta|\cdot))+\int_{s}^{t}\mu_{u}(\varPhi^{m}_{\tau,1}(\theta|\cdot)du.$

Here $\varPhi^{m}_{\tau,1}(\theta|\cdot)\in\mathcal{D}(L)$ is as in (5.25) and
(5.34), and $\mu_{u}=P_{s,\mu}\circ\varpi^{-1}_{u}$ is the law of
$\mathcal{X}(u)$. Similarly as in (5.42), by (8.2) we then obtain

$\tilde{\mu}_{t}(\varPhi^{m}_{\tau}(\theta|\cdot))\leq\sum_{q=0}^{\infty}\frac{(t-s)^{q}}{q!}\mu\left(\varPhi^{m}_{\tau,q}(\theta|\cdot)\right),\qquad
t-s<\log(1+\varepsilon)/c_{a}.$

Now we proceed here as in obtaining (5.3), which finally yields, see (5.50),
(5.51,

$\tilde{\mu}_{t}(\varPhi^{m}_{(}\theta|\cdot)):=\lim_{\max\\{\tau_{0},\tau_{1}\\}\to
0}\tilde{\mu}_{t}(\varPhi^{m}_{\tau}(\theta|\cdot))\leq\varkappa_{t}^{|m|}\|\theta_{0}\|^{m_{0}}_{L^{1}(X)}\|\theta_{1}\|^{m_{1}}_{L^{1}(X)},$
(8.17)

where $t-s<\log(1+\varepsilon)/c_{a}$ and $\varkappa_{t}=\varkappa
e^{(\alpha+1)(t-s)}$. By Definition 2.1 (8.17) yields
$\tilde{\mu}_{t}\in\mathcal{P}_{\rm exp}$ and hence
$\mathcal{Z}(t)\in\breve{\Gamma}^{2}_{*}$ (almost surely) for this $t$, see
(2.34). Thus, $T_{j}>t$. Now we take $\delta\in(0,1)$ and then
$s_{1}=s+\delta\log(1+\varepsilon)/c_{a}$, take into account that
$\mu_{s_{1}}=P_{s,\mu}\circ\varpi_{s_{1}}^{-1}\in\mathcal{P}_{\rm exp}$, and
repeat the above procedure with $s$ replaced by $s_{1}$. Since the type of
$\mu_{t}$ is $\varkappa e^{\alpha(t-s)}$ – and hence is finite for all $t$ –
the construction can be repeated ad infinitum to cover the whole
$[s,+\infty)$. This implies that the paths of $\mathcal{X}$ remain in
$\mathfrak{D}_{[0,+\infty)}(\Gamma^{2}_{*,j})$ for all $j\in\mathds{N}$, which
by (8.13) yields the proof of item (ii).

## Acknowledgments

This work was supported by the Deutsche Forschungsgemeinschaft through SFB
1238 “Taming uncertainty and profiting from randomness and low regularity in
analysis, statistics and their applications”, that is cordially acknowledged
by the authors.

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