$\\{\mathfrak{n}_{n}\\}_{n\in\mathbb{N}},\\{\mu_{n}\\}_{n\in\mathbb{N}}\subset\mathscr{P}(X)$
be two sequences of measures on a complete and separable metric space $(X,d)$,
such that $\\{\mathfrak{n}_{n}\\}_{n\in\mathbb{N}}$ is tight and
$\sup_{n\in\mathbb{N}}S_{N,\mathfrak{n}_{n}}(\mu_{n})<\infty$. Then
$\\{\mu_{n}\\}_{n\in\mathbb{N}}$ is tight.

Proof. First of all we observe that, being the entropy bounded, we can write
$\mu_{n}=\rho_{n}\mathfrak{n}_{n}$. Thus, a direct application of Jensen’s
inequality gives that, for every $n\in\mathbb{N}$ and for every measurable set
$E\subset X$,

$\displaystyle\frac{\mu_{n}(E)^{1-\frac{1}{N}}}{\mathfrak{n}_{n}(E)^{1-\frac{1}{N}}}\leq\frac{1}{\mathfrak{n}_{n}(E)}\int_{E}\rho_{n}^{1-\frac{1}{N}}d\mathfrak{n}_{n}\leq\frac{S_{N,\mathfrak{n}_{n}}(\mu_{n})}{\mathfrak{n}_{n}(E)}.$

The tightness of $\\{\mathfrak{n}_{n}\\}_{n\in\mathbb{N}}$ assures the
existence of a sequence of compact sets $\\{D_{l}\\}_{l\in\mathbb{N}}$ such
that $\sup_{n\in\mathbb{N}}\mathfrak{n}_{n}(X\setminus D_{l})\to 0$ as
$l\to\infty$. We write $E_{l}=X\setminus D_{l}$ and we conclude from the above
inequality that $\\{\mu_{n}\\}_{n\in\mathbb{N}}$ is tight, since

$\sup_{n\in\mathbb{N}}\mu_{n}(E_{l})^{1-\frac{1}{N}}\leq\sup_{n\in\mathbb{N}}\mathfrak{n}_{n}(E_{l})^{-\frac{1}{N}}\sup_{n\in\mathbb{N}}S_{N,\mathfrak{n}_{n}}(\mu_{n})\overset{l\to\infty}{\to}0.$

$\square$

This result can be applied to the extrinsic converging sequence
$(\mathsf{X}_{n},\mathsf{d}_{n},\mathfrak{m}_{n}^{k})\xrightarrow{n\to\infty}(\mathsf{X}_{\infty},\mathsf{d}_{\infty},\mathfrak{m}_{\infty}^{k})$,
with a straightforward normalization argument by recalling that
$\mathfrak{m}_{n}^{k}(\mathsf{X}_{n})$ approaches
$\mathfrak{m}_{\infty}^{k}(\mathsf{X}_{\infty})$, as $n\to\infty$.

######  Corollary 4.7.

Given a fixed $k\in\mathbb{N}$, and
$\\{\mu_{n}\\}_{n\in\mathbb{N}}\subset\mathscr{P}(Z)$ a sequence of
probability measures, such that
$\sup_{n\in\mathbb{N}}S_{N,\mathfrak{m}_{n}^{k}}(\mu_{n})<\infty$, then
$\\{\mu_{n}\\}_{n\in\mathbb{N}}$ is tight.

####  4.1.2 Semicontinuity properties

We present semicontinuity properties of $S_{N,\cdot}(\cdot)$ and
$T^{(t)}_{K,N}(\cdot|\cdot)$ conditioned to their domain of definition. We
start by proving the lower semicontinuity of $S_{N,\cdot}(\cdot)$ that we
anticipated in Section 3.1. This property is well known in classical
frameworks, that is, for positive values of $N$ and well-behaved reference
measures. For example, lower semicontinuity for a big class of functionals, in
which the Rényi entropy is included, was proved in [18] for locally compact
spaces endowed with reference measures having uniformly bounded volume growth.
Inspired by the techniques used in [18], we provide below a Legendre-type
representation formula for the entropy to attain our result. With this aim,
let us write

$\mathscr{P}^{\mathcal{S}}(\mathsf{X}):=\\{\mu\in\mathscr{P}_{2}(\mathsf{X})\,:\,\mu(\mathcal{S})=0\\}.$

######  Proposition 4.8.

Let $(\mathsf{X},\mathsf{d},p)$ be a pointed Polish space and
$\mathcal{S}\subset\mathsf{X}$ a closed subset with empty interior. Then the
Rényi entropy functional $S_{N,\mathfrak{n}}(\nu)$ is a lower semicontinuous
function of
$(\mathfrak{n},\nu)\in\mathscr{M}_{\mathcal{S}}(\mathsf{X})\times\mathscr{P}^{\mathcal{S}}(\mathsf{X})$.
Specifically, for sequences

$(\mathfrak{n}_{n})_{n\in\mathbb{N}\cup\\{\infty\\}}\subset\mathscr{M}_{\mathcal{S}}(\mathsf{X})\text{
and
}(\nu_{n})_{n\in\mathbb{N}\cup\\{\infty\\}}\subset\mathscr{P}^{\mathcal{S}}(\mathsf{X}),$

such that $\nu_{n}\overset{*}{\rightharpoonup}\nu$ as quasi-Radon measures and
$\nu_{n}\rightharpoonup\nu$, we have that,

$\displaystyle
S_{N,\mathfrak{m}}(\mu)\leq\liminf_{n\to\infty}S_{N,\mathfrak{m}_{n}}(\mu_{n}).$

In particular, the conclusion remains valid under $W_{2}$-convergence in the
second argument.

Proof. The semicontinuity of the Rényi entropy functional is verified by
exhibiting $S_{N,\mathfrak{m}}$ as the supremum of a set of continuous
functions on
$\mathscr{M}_{\mathcal{S}}(\mathsf{X})\times\mathscr{P}^{\mathcal{S}}(\mathsf{X})$
endowed with the corresponding product convergence. In particular we define
$\mathcal{R}^{k}:=B_{2^{k+1}}(p)\setminus\mathcal{N}_{2^{-(k+1)}}(\mathcal{S})$
and we show that, for every pair
$(\mathfrak{m},\mu)\in\mathscr{M}_{\mathcal{S}}(\mathsf{X})\times\mathscr{P}^{\mathcal{S}}(\mathsf{X})$,

$\displaystyle S_{N,\mathfrak{m}}(\mu)$ $\displaystyle=$ (4.3)
$\displaystyle\sup\bigg{\\{}\int F\mathrm{d}\mu$ $\displaystyle-\int
f^{*}(F)\mathrm{d}\mathfrak{m}\,:\,F\in C_{b}(\mathsf{X})\text{ supported on
}\mathcal{R}^{k}\text{, for some }k\in\mathbb{N}\bigg{\\}},$

where $f^{*}$ is the convex conjugate function of $f(x)=|x|^{1-\frac{1}{N}}$,
for $x\in\mathbb{R}$. That is,

$\displaystyle f^{*}:\mathbb{R}$ $\displaystyle\to$ $\displaystyle[0,\infty)$
$\displaystyle y$ $\displaystyle\mapsto$ $\displaystyle
f^{*}(y):=\sup_{x\in\mathbb{R}}(y\,x-f(x))=\left(\frac{N}{N-1}\right)^{1-N}|x|^{1-N}.$

For simplicity, let us denote the expression on the right-hand side of (4.3)
by $\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)$. And do note that,
$G(\mathfrak{m},\mu)=\int F\mathrm{d}\mu-\int f^{*}(F)\mathrm{d}\mathfrak{m}$
is indeed continuous in
$\mathscr{M}_{\mathcal{S}}(\mathsf{X})\times\mathscr{P}^{\mathcal{S}}(\mathsf{X})$
since, $f^{*}(F)\in C_{bs}(\mathsf{X})\cap C_{\mathcal{S}}(\mathsf{X})$, for
all $F\in C_{b}(\mathsf{X})$ with support in $\mathcal{R}^{k}\text{, for some
}k\in\mathbb{N}$.

We first verify that
$S_{N,\mathfrak{m}}(\mu)\geq\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)$, for
every
$(\mathfrak{m},\mu)\in\mathscr{M}_{S}(\mathsf{X})\times\mathscr{P}^{\mathcal{S}}(\mathsf{X})$.
We assume that $\mu=\rho\mathfrak{m}$ since the aforementioned inequality is
trivially satisfied when $\mu\not\ll\mathfrak{m}$. We get the desired result,
integrating the expression $f(z)\geq z\,y^{*}-f^{*}(y^{*})$ with respect to
$\mathfrak{m}$ which, by definition of $f^{*}$, holds for any
$z,y^{*}\in\mathbb{R}$, after replacing $z=\rho(x)$ and $y^{*}=F(x)$, for
$F\in C_{b}(\mathsf{X})$ with support inside some $\mathcal{R}^{k}$.

Before proceeding with the converse inequality let us point out that
$\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)=\infty$ granted
$\mu\not\ll\mathfrak{m}$. Indeed, in this case, there exists a Borel set
$A\subset\mathsf{X}$ with $\mathfrak{m}(A)=0$ and $\mu(A)>0$. At this point,
recall that every Borel finite measure in a Polish spaces is inner regular
with respect to compact sets and outer regular with respect to open sets (see
for instance [7, Theorem 7.1.7]). For this reason, since
$\mu\in\mathscr{P}^{\mathcal{S}}(\mathsf{X})$, it is possible to assume that
$A\cap S=\emptyset$, and being $A$ a compact set. Observe also that
compactness grants the existence of $k\in\mathbb{N}$ for which
$A\subset\mathcal{R}^{k}$. Since $\mathfrak{m}$ and $\mu$ restricted to
$\mathcal{R}^{k+1}$ are finite measures, there exists a sequence of compact
sets $(K_{n})_{n\in\mathbb{N}}$ and one of open sets
$(A_{n})_{n\in\mathbb{N}}$, such that $K_{n}\subset A\subset
A_{n}\subset\mathcal{R}^{k+1}$ and $(\mu+\mathfrak{m})(A_{n}\setminus
K_{n})<1/n$, for any $n\in\mathbb{N}$. Then, for $M>0$, Tietze’s Theorem
ensures the existence of a sequence of approximating functions
$\left(F^{M}_{n}\right)_{n\in\mathbb{N}}\subset C_{b}(X)$ satisfying: $0\leq
F^{M}_{n}\leq M$, $F^{M}_{n}=M$ on $K_{n}$, and $F^{M}_{n}=0$ on $X\setminus
A_{n}$. Therefore, as $n\to\infty$, the functions $F^{M}_{n}$ converge, in
$L^{1}(\mu+\mathfrak{m})$, to the scaled characteristic function $M\chi_{A}$.
On the other hand, since for every $n$, $F^{M}_{n}$ is an admissible function
for the supremum in $\tilde{\mathtt{S}}_{N,\mathfrak{m}}$, we have that

$\int F^{M}_{n}\mathrm{d}\mu-\int
f^{*}(F^{M}_{n})\mathrm{d}\mathfrak{m}\leq\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)<+\infty.$
(4.4)

for any $n\in\mathbb{N}$. Passing now to the limit as $n$ goes to infinity in
(4.4), we obtain that
$M\cdot\mu(A)\leq\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)$. The arbitrariness
of $M$ implies then that $\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)=\infty$.

We proceed now to prove that
$S_{N,\mathfrak{m}}(\mu)\leq\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)$ and we
will assume that $\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)<+\infty$, since
otherwise there is nothing to prove. The paragraph above enables us to write
$\mu=\rho\mathfrak{m}$. We then have the following expression for
$\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)$:

$\tilde{\mathtt{S}}_{N,\mathfrak{m}}(\mu)=\sup\left\\{\int
F\rho-f^{*}(F)\,\mathrm{d}\mathfrak{m}\,:F\in C_{b}(\mathsf{X})\text{
supported on }\mathcal{R}^{k}\text{, for some }k\in\mathbb{N}\right\\}.$

And, recalling that
$f(x)=(f^{*})^{*}(x)=\sup_{y\in\mathbb{R}}\\{x\,y-f^{*}(y)\\}$ because $f$ is
finite, convex and continuous, it follows that

$S_{N,\mathfrak{m}}(\mu)=\int\sup_{s^{*}\in\mathbb{Q}}\\{\rho(x)s^{*}-f^{*}(s^{*})\\}\mathrm{d}\mathfrak{m}(x).$

By fixing $\mathbb{Q}=\\{q_{n}\\}_{n\in\mathbb{N}}$, an enumeration of
rational numbers with $q_{0}=0$, we introduce the family of approximating
functionals

$\left\\{S_{N,\mathfrak{m}}^{h}(\mu):=\int\sup_{s^{*}\in\\{q_{0},\dots,q_{h}\\}}\\{\rho(x)s^{*}-f^{*}(s^{*})\\}\mathrm{d}\mathfrak{m}(x)\right\\}_{h\in\mathbb{N}}.$

Observe that the integrands are monotone increasing in $h$ and that
$0=\rho(x)q_{0}-f^{*}(q_{0})$. In particular, Beppo-Levi’s Theorem ensures
that $S_{N,\mathfrak{m}}^{h}(\mu)\to S_{N,\mathfrak{m}}(\mu)$, as
$h\to\infty$. Therefore, it suffices to show that
$S_{N,\mathfrak{m}}^{h}\leq\tilde{\mathtt{S}}_{N,\mathfrak{m}}$, for any fixed
$h\in\mathbb{N}$. To this aim, one confirms directly that

$S_{N,\mathfrak{m}}^{h}=\sup\left\\{\int(\rho
F-f^{*}(F))\mathrm{d}\mathfrak{m}\,:\,F\text{ is a step function with values
in $\\{q_{0},\dots,q_{h}\\}$}\right\\}.$ (4.5)

Note that the fact that $S$ is an $\mathfrak{m}$-null set guarantees that we
can further require that the aforementioned functions are supported in
$\mathcal{R}^{k}$ for some $k\in\mathbb{N}$ without modifying the supremum, as
an approximation argument using the Monotone Convergence Theorem shows.
Finally, since $\mathfrak{m}$ and $\mu$ are finite measures when restricted to
$\mathcal{R}^{k+1}$, every step function with support in $\mathcal{R}^{k}$ can
be obtained as the $L^{1}(\mu+\mathfrak{m})$-limit of continuous and uniformly
bounded functions implying that
$S_{N,\mathfrak{m}}^{h}\leq\tilde{\mathtt{S}}_{N,\mathfrak{m}}$, which
concludes the proof.  $\square$

A couple of observations suffice to extract a useful corollary for extrinsic
converging sequences of metric measure spaces
$\\{(\mathsf{X},\mathsf{d}_{n},\mathfrak{m}_{n})\\}_{n\in\mathbb{N}\cup\\{\infty\\}}$.
From Lemma 4.5 we deduce that, for any $k\in\mathbb{N}$, the sequence of
vertical projections $(\mathfrak{m}_{n}^{k})_{n\in\mathbb{N}}$ approximates
$\mathfrak{m}_{\infty}^{k}\in\mathscr{M}(\mathsf{X})\subset\mathscr{M}_{\emptyset}(\mathsf{X})$
in narrow convergence. Moreover, Remark 2.14 states that, granted we restrict
ourselves to the set $\mathscr{M}(Z)$ on a Polish space $(Z,\mathsf{d}_{Z})$
and set $S=\emptyset$, then weak convergence in the sense of quasi-Radon
measures coincides with the usual weak one.

######  Corollary 4.9.

Given a fixed $k\in\mathbb{N}$ and
$\\{\mu_{n}\\}_{n\in\mathbb{N}}\subset\mathscr{P}_{2}(Z)$ a sequence
converging weakly to $\mu\in\mathscr{P}_{2}(Z)$. Then it holds that

$S_{N,\mathfrak{m}^{k}_{\infty}}(\mu)\leq\liminf_{n\to\infty}S_{N,\mathfrak{m}^{k}_{n}}(\mu_{n}).$
(4.6)

We conclude the section with a corresponding continuity result for the
functional $T^{(t)}_{K,N}$. We stress that although, it would be sufficient
for the proof of Theorem 4.1 to have the upper semicontinuity of
$T^{(t)}_{K,N}$, we prefer to present a more general statement.

######  Proposition 4.10.

Let $K\geq 0$ and $N<0$ and $(\mathsf{X},\mathsf{d},\mathfrak{m})$ be a metric
measure space. Furthermore, set
$\mu_{0}=\rho_{0}\mathfrak{m},\mu_{1}=\rho_{1}\mathfrak{m}\in\mathscr{P}(\mathsf{X})$
to be absolutely continuous with respect to the quasi-Radon reference measure
$\mathfrak{m}$, with
$S_{N,\mathfrak{m}}(\mu_{0}),S_{N,\mathfrak{m}}(\mu_{1})<\infty$. Consider a
sequence
$(\pi_{n})_{n\in\mathbb{N}}\subset\mathscr{P}(\mathsf{X}\times\mathsf{X})$,
weakly converging to $\pi\in\mathscr{P}(\mathsf{X}\times\mathsf{X})$ and such
that

$(\mathsf{p}_{1})_{\\#}\pi_{n}=\mu_{0}\quad\text{and}\quad(\mathsf{p}_{1})_{\\#}\pi_{n}=\mu_{1}\qquad\text{for
every }n\in\mathbb{N}.$

Then, for $t\in[0,1]$, it holds that

$\lim_{n\to\infty}T^{(t)}_{K,N}(\pi_{n}|\mathfrak{m})=T^{(t)}_{K,N}(\pi|\mathfrak{m}).$

Additionally, the conclusion remains valid for $K<0$, granted $\rm
diam(\mathsf{X})<\pi\sqrt{\frac{N-1}{K}}$.

Proof. Set $t\in(0,1)$, since the statement is clearly ture for the remaining
values. We only proceed to prove that,

$\lim_{n\to\infty}\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N}}\mathrm{d}\pi_{n}=\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N}}\mathrm{d}\pi,$
(4.7)

since the other term of $T^{(t)}_{K,N}(\cdot|\mathfrak{m})$ can be treated
analogously. Notice that, since $S_{N,\mathfrak{m}}(\mu_{0})<\infty$, then
$\rho_{0}(x)^{-1/N}\in L^{1}(\mu_{0})$ thus, according to Lemma 2.12, for
every fixed $\varepsilon>0$ there exists $f^{\varepsilon}\in
C_{b}(\mathsf{X})$ such that
$\lVert\rho_{0}^{-1/N}-f^{\varepsilon}\rVert_{L^{1}(\mu_{0})}<\varepsilon$.
Moreover, notice that the coefficients
$\tau_{K,N}^{(1-t)}(\cdot|\mathfrak{m})$ are bounded and continuous. Indeed,
this is always the case for $K\geq 0$, and since $\rm
diam(X)<\pi\sqrt{\frac{N-1}{K}}$ is bounded by our assumptions, this holds as
well for $K<0$. Therefore,

$\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)$

is itself bounded and continuous. Consequently, the weak convergence
$(\pi_{n})_{n}\rightharpoonup\pi$ shows that,

$\lim_{n\to\infty}\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}\pi_{n}=\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}\pi.$

Furthermore, the boundedness of $\tau_{K,N}^{(1-t)}$ allows to deduce the
following estimate

$\begin{split}\limsup_{n\to\infty}\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N}}\mathrm{d}\pi_{n}&\leq\lim_{n\to\infty}\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}\pi_{n}+\varepsilon\lVert\tau_{K,N}^{(1-t)}\rVert_{L^{\infty}}\\\
&=\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}\pi+\varepsilon\lVert\tau_{K,N}^{(1-t)}\rVert_{L^{\infty}}\\\
&\leq\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N}}\mathrm{d}\pi+2\varepsilon\lVert\tau_{K,N}^{(1-t)}\rVert_{L^{\infty}}.\end{split}$

Analogously, it follows that

$\liminf_{n\to\infty}\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N}}\mathrm{d}\pi_{n}\geq\int_{X\times
X}\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N}}\mathrm{d}\pi-2\varepsilon\lVert\tau_{K,N}^{(1-t)}\rVert_{L^{\infty}},$

and since $\varepsilon>0$ and can be chosen arbitrarily, equation (4.7) holds
true. We conclude by recalling the arbitrariness of $t$.
$\square$

###  4.2 Proof of the Approximate CD Condition

The objective of this section is to prove the next partial result

######  Theorem 4.11.

Let $K\in\mathbb{R}$, $N\in(\infty,0)$, and
$\big{\\{}(\mathsf{X}_{n},\mathsf{d}_{n},\mathfrak{m}_{n},\mathcal{S}_{\mathfrak{m}_{n}},p_{n})\big{\\}}_{n\in\mathbb{N}\cup\\{\infty\\}}\subset\mathcal{M}^{qR}_{\bar{k}}$
be a sequence of metric measure spaces satisfying the assumptions of the
Stability Theorem 4.1, for some $\bar{k}\in\mathbb{N}$. Then
$(\mathsf{X}_{\infty},\mathsf{d}_{\infty},\mathfrak{m}_{\infty})$ is a
$\mathsf{CD}^{a}(K,N)$ space.

We follow the plan discussed at the beginning of the section and argue, in
steps, using vertical approximations. Specifically, Steps 1 and 2 serve the
purpose of constructing useful approximations of the marginals $\mu_{0}$ and
$\mu_{1}$. Next, we follow the argumentation of Sturm in [34] in Steps 3 to 6,
to exhibit the upper semicontinuity of $T^{(t)}_{K,N}$ along a sequence of
optimal couplings, provided by the curvature-dimension assumption.
Additionally, we demonstrate the existence of a favorable limiting optimal
coupling. Step 7 focuses in proving the convergence of inner points of a
vertical sequence of Wasserstein geodesics, as well as, the lower
semicontinuity of the Renyi entropy along this sequence.

Let us fix first the notation. Set $k\geq\bar{k}$ and assume that
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}^{ac}(\mathsf{X}_{\infty})$ have supports
satisfying
$\text{supp}(\mu_{0}),\,\text{supp}(\mu_{1})\subset\mathcal{R}_{\infty}^{k-1}$.
We denote by $\rho_{i}$ the density of $\mu_{i}$ with respect to
$\mathfrak{m}$, for $i=\\{0,1\\}$. Define the set,

$I:=\\{N^{\prime}\in[N,0)\,:\,S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{0}),S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{1})<\infty\\},$

and observe that $I$ is an interval, as a consequence of Jensen’s inequality.
Then, surely, we are able to assume that (and set)

$(\,M:=\,)\;\max\left\\{S_{N,\mathfrak{m}_{\infty}}(\mu_{0}),\,S_{N,\mathfrak{m}_{\infty}}(\mu_{1})\right\\}<\infty,$

and that, for every $q\in\mathsf{Opt}(\mu_{0},\mu_{1})$,

$T^{(t)}_{K,N^{\prime}}(q|\mathfrak{m}_{\infty})<\infty,$ (4.8)

since the $\mathsf{CD}$-condition is trivial in failure of any of these
inequalities. Therefore, the arbitrariness of $k$ and initial measures shows
that, in order to demonstrate Theorem 4.11, we are required to validate the
$\mathsf{CD}$-inequality (3.4), for every $N^{\prime}\in I$, and every
$t\in[0,1]$.

Additionally, we fix as before an optimal coupling
$p_{n}\in\mathsf{Opt}\left(\bar{\mathfrak{m}}_{\infty}^{k},\bar{\mathfrak{m}}_{n}^{k}\right)$
between the normalized $k$-cuts of the reference measures, for every
$n\in\mathbb{N}$. And, as defined in Section 4.1, we consider
$\\{P_{n}(x)\\}_{x\in X^{\infty}}\in\mathscr{P}(\mathsf{X}_{n})$ and
$\\{P^{\prime}_{n}(y)\\}_{y\in X_{n}}\in\mathscr{P}(\mathsf{X}_{\infty})$ the
disintegration kernels of $p_{n}$ with respect to the projections
$\mathsf{p}_{1}$ and $\mathsf{p}_{2}$ respectively, and consider the map
$P^{\prime}_{n}\colon\mathscr{P}_{2}(\mathsf{X}_{\infty},\mathsf{d}_{\infty},\mathfrak{m}_{\infty}^{k})\to\mathscr{P}_{2}(\mathsf{X}_{n},\mathsf{d}_{n},\mathfrak{m}_{n}^{k})$.
Note that, in contrast to Section 4.1, here we have omitted the dependence on
the number $k$, since it is fixed for now.

STEP 1: Horizontal approximation with bounded densities
During the argument it proves useful to work with bounded-density measures.
Therefore we construct here a horizontal approximation of $\mu_{0}$ and
$\mu_{1}$, for which its elements enjoy this property.
For the construction, we fix an arbitrary optimal coupling
$\tilde{q}\in\mathsf{Opt}(\mu_{0},\mu_{1})$ and define, for every $r>0$,

$E_{r}:=\\{(x_{0},x_{1})\in X_{\infty}\times
X_{\infty}\,:\,\rho_{0}(x_{0})<r,\rho_{1}(x_{1})<r\\}$ (4.9)

and consequently, for sufficiently large $r$,

$\tilde{q}^{(r)}:=\alpha_{r}^{-1}\tilde{q}(\,.\cap E_{r}),$

where $\alpha_{r}:=\tilde{q}(E_{r})$. The measure
$\tilde{q}^{(r)}\in\mathscr{P}(\mathsf{X}_{\infty}\times\mathsf{X}_{\infty})$
has marginals given by

$\mu_{0}^{(r)}:=(\mathsf{p}_{1})_{\sharp}\,\tilde{q}^{(r)}\quad\text{ and
}\quad\mu_{1}^{(r)}:=(\mathsf{p}_{2})_{\sharp}\,\tilde{q}^{(r)}.$ (4.10)

Notice that both $\mu_{0}^{(r)}$ and $\mu_{1}^{(r)}$ have bounded densities
and that $\mu_{i}^{(r)}$ converges to $\mu_{i}$ in
$(\mathscr{P}^{2}(\mathsf{X}_{\infty}),W_{2})$, for $i=0,1$. Moreover, notice
that $S_{N,\mathfrak{m}_{\infty}}(\mu_{i}^{(r)})\to
S_{N,\mathfrak{m}_{\infty}}(\mu_{i})$ as $r\to\infty$, for $i=0,1$. Then we
fix $\varepsilon>0$ and find $r=r(\varepsilon)$ such that $\alpha_{r}\geq
1-\varepsilon$ and that the following estimates holds:

$\max_{i\in\\{1,2\\}}W_{2}(\mu_{i},\mu_{i}^{(r)})\leq\varepsilon\quad\text{
and
}\quad\max_{i\in\\{1,2\\}}S_{N,\mathfrak{m}_{\infty}^{k}}(\mu_{i}^{(r)})=\max_{i\in\\{1,2\\}}S_{N,\mathfrak{m}_{\infty}}(\mu_{i}^{(r)})\leq
M+\frac{1}{2}.$ (4.11)

We point out that the parameter $r$ depends on $\varepsilon$, but we won’t be
explicit on this dependence for the sake of the presentation.

STEP 2: Vertical approximation
Once we have identified the horizontal approximations $\mu_{0}^{(r)}$ and
$\mu_{1}^{(r)}$, we may proceed to their vertical approximation. First of all,
observe that $\mu_{0}^{(r)}$ and $\mu_{1}^{(r)}$ are absolutely continuous
with respect to the normalized reference measure
$\bar{\mathfrak{m}}_{\infty}^{k}$, so we denote by $\tilde{\rho}_{0}^{(r)}$
and $\tilde{\rho}_{1}^{(r)}$ their bounded densities. Then, for every
$n\in\mathbb{N}$, we define
$\mu_{0,n},\,\mu_{1,n}\in\mathscr{P}_{2}(\mathsf{X}_{n},\mathsf{d}_{n},\mathfrak{m}_{n}^{k})$
as

$\mu_{i,n}:=P_{n}^{\prime}(\mu_{i}^{(r)})=\rho_{i,n}\bar{\mathfrak{m}}_{n}^{k},$
(4.12)

where
$\rho_{i,n}(y)=\int\tilde{\rho}_{i}^{(r)}(x)P_{n}^{\prime}(y)({\mathrm{d}}x)$.
Notice that $\mu_{0,n}$ and $\mu_{1,n}$ depend on $r$ (and ultimately on
$\varepsilon$), but, once again, we prefer not to make this dependence
explicit, in order to maintain an easy notation in the following. Anyway, we
invite the reader to keep in mind that every object we are going to define
depends only on $\varepsilon$. Now, since
$\mathfrak{m}^{k}_{n}(\mathsf{X}_{n})\to\mathfrak{m}^{k}_{\infty}(\mathsf{X}_{\infty})$,
observe that Lemma 4.5 guarantees the existence of an $\bar{n}\in\mathbb{N}$,
such that if $n\geq\bar{n}$ it holds that

$\max_{i\in\\{1,2\\}}W_{2}(\mu_{i}^{(r)},\mu_{i,n})\leq\varepsilon.$ (4.13)

and that

$\max_{i\in\\{1,2\\}}S_{N,\mathfrak{m}_{n}}(\mu_{i,n})\leq\max_{i\in\\{1,2\\}}S_{N,\mathfrak{m}_{n}^{k}}(\mu_{i,n})\leq
M+1.$ (4.14)

Moreover, according to Lemma 4.5, for every $N^{\prime}\in I$, it holds that

$S_{N^{\prime},\mathfrak{m}_{n}}(\mu_{i,n})\leq
S_{N^{\prime},\mathfrak{m}^{k}_{n}}(\mu_{i,n})\leq\bigg{[}\frac{\mathfrak{m}^{k}_{n}(\mathsf{X}_{n})}{\mathfrak{m}^{k}_{\infty}(\mathsf{X}_{\infty})}\bigg{]}^{-\frac{1}{N^{\prime}}}S_{N^{\prime},\mathfrak{m}^{k}_{\infty}}(\tilde{\mu}^{(r)}_{i})<\infty.$

Therefore, for every $n\in\mathbb{N}$ large enough, since
$(\mathsf{X}_{n},\mathsf{d}_{n},\mathfrak{m}_{n})$ is a $\mathsf{CD}(K,N)$
space, there exist an optimal plan
$\pi_{n}\in\mathsf{Opt}(\mu_{0,n},\mu_{1,n})$ and a 2-Wasserstein geodesic
$(\mu_{t,n})_{t\in[0,1]}\subset\mathscr{P}_{2}^{ac}(\mathsf{X}_{n})$
connecting $\mu_{0,n}$ and $\mu_{1,n}$, for which,

$S_{N^{\prime},\mathfrak{m}_{n}}(\mu_{t,n})\leq
T_{K,N^{\prime}}^{(t)}(\pi_{n}|\mathfrak{m}_{n})$ (4.15)

holds, for every $t\in[0,1]$ and every $N^{\prime}\in I$. Note that Remark 3.4
together with the assumption that $\sup_{i\in\mathbb{N}\cup\\{\infty\\}}\rm
diam(\mathsf{X}_{i},\mathsf{d}_{i})<\pi|K|^{-1/2}$, if $K<0$, assures that the
geodesic $\mu_{t,n}$ is absolutely continuous with respect to
$\mathfrak{m}_{n}$.

STEP 3: Estimate for $T_{K,N^{\prime}}^{(t)}$
In this step we start the proof of the upper semicontinuity of the functional
$T_{K,N^{\prime}}^{(t)}$. In particular, we fix $N^{\prime}\in[N,0)$ and a
time $t\in[0,1]$ and we call $Q_{n}$ and $Q^{\prime}_{n}$ be the
disintegrations of $\pi_{n}$ with respect to $\mu_{0,n}$ and $\mu_{1,n}$
respectively. Then we define the following two functions

$v_{0}(y_{0})=\int_{\mathsf{X}_{n}}\tau^{(1-t)}_{K,N^{\prime}}(\mathsf{d}(y_{0},y_{1}))Q_{n}(y_{0},dy_{1})$

and

$v_{1}(y_{1})=\int_{\mathsf{X}_{n}}\tau^{(t)}_{K,N^{\prime}}(\mathsf{d}(y_{0},y_{1}))Q^{\prime}_{n}(y_{1},dy_{0}).$

A direct application of Jensen’s theorem leads to

$\begin{split}T_{K,N^{\prime}}^{(t)}(\pi_{n}|\bar{\mathfrak{m}}_{n}^{k})&=\displaystyle\sum_{i=0}^{1}\,\int_{\mathsf{X}_{n}}\rho_{i,n}(y_{i})^{1-1/N^{\prime}}\cdot
v_{i}(y_{i})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{n}(y_{i})\\\
&=\sum_{i=0}^{1}\,\int_{\mathsf{X}_{n}}\bigg{[}\int_{\mathsf{X}_{\infty}}\tilde{\rho}^{(r)}_{i}(x_{i})P^{\prime}_{n}(y_{i},{\mathrm{d}}x_{i})\bigg{]}^{1-1/N^{\prime}}\cdot
v_{i}(y_{i})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{n}(y_{i})\\\
&\leq\sum_{i=0}^{1}\,\int_{\mathsf{X}_{n}}\int_{\mathsf{X}_{\infty}}\tilde{\rho}^{(r)}_{i}(x_{i})^{1-1/N^{\prime}}P^{\prime}_{n}(y_{i},{\mathrm{d}}x_{i})\cdot
v_{i}(y_{i})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{n}(y_{i})\\\
&=\sum_{i=0}^{1}\,\int_{\mathsf{X}_{\infty}}\tilde{\rho}^{(r)}_{i}(x_{i})^{1-1/N^{\prime}}\bigg{[}\int_{\mathsf{X}_{n}}v_{i}(y_{i})P_{n}(x_{i},{\mathrm{d}}y_{i})\bigg{]}\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{\infty}(x_{i}).\end{split}$

At this point we see that

$\displaystyle\displaystyle\int_{\mathsf{X}_{n}}v_{0}(y_{0})$ $\displaystyle
P_{n}(x_{0},{\mathrm{d}}y_{0})=\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(y_{0},y_{1}))Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})$
$\displaystyle=\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(y_{0},y_{1}))\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})$
$\displaystyle\leq\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\big{[}\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x_{0},x_{1}))+C\cdot|\mathsf{d}(y_{0},y_{1})-\mathsf{d}(x_{0},x_{1})|\big{]}$
$\displaystyle\hskip
128.0374pt\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})$
$\displaystyle\leq\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\big{[}\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x_{0},x_{1}))+C\cdot(\mathsf{d}(x_{0},y_{0})+\mathsf{d}(x_{1},y_{1}))\big{]}$
$\displaystyle\hskip
128.0374pt\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0}),$

and analogously that

$\displaystyle\displaystyle\int_{\mathsf{X}_{n}}v_{1}(y_{1})P_{n}(x_{1},{\mathrm{d}}y_{1})$
$\displaystyle\leq\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\big{[}\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x_{0},x_{1}))+C\cdot(\mathsf{d}(x_{0},y_{0})+\mathsf{d}(x_{1},y_{1}))\big{]}$
$\displaystyle\hskip
128.0374pt\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})}{\rho_{0,n}(y_{0})}P_{n}^{\prime}(y_{0},{\mathrm{d}}x_{0})Q_{n}(y_{1},{\mathrm{d}}y_{0})P_{n}(x_{1},{\mathrm{d}}y_{1}),\vspace{-0.3cm}$

where
$C:=\max_{\theta\in[0,M],s\in[0,1]}\frac{\partial}{\partial\theta}\tau_{K,N^{\prime}}^{(s)}(\theta)$
and $M$ is the maximum between $\mathsf{d}(x_{0},x_{1})$ and
$\mathsf{d}(x_{1},y_{1})$. Observe that the constant $C$ is indeed finite
because we know that
$\sup_{i\in\mathbb{N}\cup\\{\infty\\}}\text{diam}(\mathsf{X}_{i},\mathsf{d}_{i})<\pi\sqrt{\frac{1}{|K|}}$,
if $K<0$, from assumption (iii) in Theorem 4.11.
Moreover we notice that

$\begin{split}\int_{\mathsf{X}_{\infty}}\tilde{\rho}^{(r)}_{i}(x_{0})^{1-1/N^{\prime}}&\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{0},y_{0})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{\infty}(x_{0})\\\
&=\int_{\mathsf{X}_{\infty}}\tilde{\rho}^{(r)}_{i}(x_{0})^{1-1/N^{\prime}}\int_{\mathsf{X}_{n}}\mathsf{d}(x_{0},y_{0})P_{n}(x_{0},{\mathrm{d}}y_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{\infty}(x_{0})\\\
&\leq
r^{1-1/N^{\prime}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{0},y_{0})\mathrm{d}p_{n}^{k}(x_{0},y_{0})\leq
r^{1-1/N^{\prime}}W_{2}(\bar{\mathfrak{m}}_{n}^{k},\bar{\mathfrak{m}}_{\infty}^{k})\end{split}$
(4.16)

and

$\begin{split}\int_{\mathsf{X}_{\infty}}\tilde{\rho}^{(r)}_{0}(x_{0})^{1-1/N^{\prime}}&\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{\infty}(x_{0})\\\
&\leq
r^{-1/N^{\prime}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad
Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})\tilde{\rho}^{(r)}_{0}(x_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}^{k}_{\infty}(x_{0})\\\
&=r^{-1/N^{\prime}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})\mu_{0,n}(\mathrm{d}y_{0})\\\
&=r^{-1/N^{\prime}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{1},y_{1})\tilde{\rho}_{1}^{(r)}(x_{1})P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})\mathrm{d}\bar{\mathfrak{m}}_{n}^{k}(y_{1})\\\
&\leq
r^{1-1/N^{\prime}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{\infty}}\mathsf{d}(x_{1},y_{1})\mathrm{d}p_{n}^{k}(x_{1},y_{1})\leq
r^{1-1/N^{\prime}}W_{2}(\bar{\mathfrak{m}}_{n}^{k},\bar{\mathfrak{m}}_{\infty}^{k}),\end{split}$
(4.17)

where the last inequality in both chains follows by the Jensen’s inequality.
Consequently, for every $n\in\mathbb{N}$, we define a – not necessarily
optimal – coupling
$\bar{q}_{n}^{(r)}\in\mathsf{Adm}(\mu_{0}^{(r)},\mu_{1}^{(r)})$ by imposing
that

$\begin{split}{\mathrm{d}}\bar{q}_{n}^{(r)}(x_{0},x_{1})&=\displaystyle\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{0,n}(y_{0})\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})P_{n}^{\prime}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})\\\
&=\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q_{n}(y_{0},{\mathrm{d}}y_{1})P_{n}(x_{0},{\mathrm{d}}y_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}_{\infty}^{k}(x_{0})\\\
&=\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{0,n}(y_{0})}P_{n}^{\prime}(y_{1},{\mathrm{d}}x_{1})Q^{\prime}_{n}(y_{1},{\mathrm{d}}y_{0})P_{n}(x_{0},{\mathrm{d}}y_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}_{\infty}^{k}(x_{1}).\end{split}$

With this definition of $\bar{q}^{(r)}$ and keeping in mind (4.16) and (4.17),
we end up with

$T_{K,N^{\prime}}^{(t)}(\pi_{n}|\bar{\mathfrak{m}}_{n}^{k})\leq
T^{(t)}_{K,N^{\prime}}(\bar{q}^{(r)}|\bar{\mathfrak{m}}_{\infty}^{k})+4Cr^{1-1/N^{\prime}}W_{2}(\bar{\mathfrak{m}}_{\infty}^{k},\bar{\mathfrak{m}}_{n}^{k}).$

Now, up to taking a greater $\bar{n}$, we can require that for every
$n\geq\bar{n}$ it holds that

$W_{2}(\bar{\mathfrak{m}}_{\infty}^{k},\bar{\mathfrak{m}}_{n}^{k})\leq\dfrac{\varepsilon}{4Cr^{\frac{N^{\prime}-1}{N^{\prime}}}},$

for every $N^{\prime}\in[N,\varepsilon)$. As a consequence we obtain that

$T_{K,N^{\prime}}^{(t)}(\pi_{n}|\bar{\mathfrak{m}}^{k}_{n})\leq
T_{K,N}^{(t)}(\bar{q}_{n}^{(r)}|\bar{\mathfrak{m}}^{k}_{\infty})+\varepsilon,$
(4.18)

for every $n\geq\bar{n}$ and every $N^{\prime}\in[N,\varepsilon)$.

STEP 4: $q^{(r)}_{n}$ converges to an optimal plan
The objective now is to prove that

$\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}\bar{q}^{(r)}_{n}(x_{0},x_{1})\to
W^{2}_{2}(\mu_{0}^{(r)},\mu_{1}^{(r)})\qquad\text{as }n\to\infty.$ (4.19)

First of all notice that, since every $\bar{q}^{(r)}_{n}$ is an admissible
plan between $\mu_{0}^{(r)}$ and $\mu_{1}^{(r)}$, then for every
$n\in\mathbb{N}$ it holds that

$\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}\bar{q}^{(r)}_{n}(x_{0},x_{1})\geq
W^{2}_{2}(\mu_{0}^{(r)},\mu_{1}^{(r)}).$ (4.20)

On the other hand the triangular inequality ensures that

$\mathsf{d}(x_{0},x_{1})\leq\mathsf{d}(x_{0},y_{0})+\mathsf{d}(y_{0},y_{1})+\mathsf{d}(x_{1},y_{1})$

and consequently, since
$\mathsf{d}(y_{0},y_{1})<\text{diam}(\mathcal{R}^{k}_{n})\leq 2^{k+2}$ for
$\pi_{n}$-almost every pair $(y_{0},y_{1})$, we have that

$\mathsf{d}^{2}(x_{0},x_{1})-\mathsf{d}^{2}(y_{0},y_{1})\leq
2\mathsf{d}^{2}(x_{0},y_{0})+2\mathsf{d}^{2}(x_{1},y_{1})+2^{k+3}\mathsf{d}(x_{0},y_{0})+2^{k+3}\mathsf{d}(x_{1},y_{1})$

for $\pi_{n}$-almost every pair $(y_{0},y_{1})$. It is then possible to
perform the following estimate

$\displaystyle\int_{\mathsf{X}_{\infty}\times\mathsf{X}_{\infty}}\mathsf{d}^{2}(x_{0},x_{1})$
$\displaystyle\mathrm{d}\bar{q}^{(r)}_{n}(x_{0},x_{1})$
$\displaystyle=\int_{\mathsf{X}_{\infty}\times\mathsf{X}_{\infty}}\mathsf{d}^{2}(x_{0},x_{1})\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{0,n}(y_{0})\rho_{1,n}(y_{1})}P^{\prime}_{n}(y_{1},{\mathrm{d}}x_{1})P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$
$\displaystyle\leq\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\mathsf{d}^{2}(y_{0},y_{1})\mathrm{d}\pi_{n}(y_{0},y_{1})$
$\displaystyle\quad+\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2\mathsf{d}^{2}(x_{0},y_{0})\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})}{\rho_{0,n}(y_{0})}P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$
$\displaystyle\quad+\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2^{k+3}\mathsf{d}(x_{0},y_{0})\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})}{\rho_{0,n}(y_{0})}P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$
$\displaystyle\quad+\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2\mathsf{d}^{2}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P^{\prime}_{n}(y_{1},{\mathrm{d}}x_{1})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$
$\displaystyle\quad+\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2^{k+3}\mathsf{d}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P^{\prime}_{n}(y_{1},{\mathrm{d}}x_{1})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$

We can now consider one term at the time and start by noticing that, according
to Lemma 4.5,

$\displaystyle\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2\mathsf{d}^{2}(x_{0},y_{0})\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})}{\rho_{0,n}(y_{0})}$
$\displaystyle
P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$
$\displaystyle=\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}}2\mathsf{d}^{2}(x_{0},y_{0})\tilde{\rho}_{0}^{(r)}(x_{0})P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\bar{\mathfrak{m}}_{n}^{k}(y_{0})$
$\displaystyle=\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}}2\mathsf{d}^{2}(x_{0},y_{0})\tilde{\rho}_{0}^{(r)}(x_{0})\,\mathrm{d}p^{k}_{n}(x_{0},y_{0})\to
0,$

and similarly

$\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2\mathsf{d}^{2}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P^{\prime}_{n}(y_{1},{\mathrm{d}}x_{1})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})\to
0.$

Moreover Hölder’s inequality ensures that

$\displaystyle\int_{\mathsf{X}_{\infty}}$
$\displaystyle\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2^{k+3}\mathsf{d}(x_{0},y_{0})\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})}{\rho_{0,n}(y_{0})}P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})$
$\displaystyle\leq
2^{k+3}\left[\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\mathsf{d}^{2}(x_{0},y_{0})\dfrac{\tilde{\rho}_{0}^{(r)}(x_{0})}{\rho_{0,n}(y_{0})}P^{\prime}_{n}(y_{0},{\mathrm{d}}x_{0})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})\right]^{\frac{1}{2}}\to
0,$

and analogously that

$\int_{\mathsf{X}_{\infty}}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}2^{k+3}\mathsf{d}(x_{1},y_{1})\dfrac{\tilde{\rho}_{1}^{(r)}(x_{1})}{\rho_{1,n}(y_{1})}P^{\prime}_{n}(y_{1},{\mathrm{d}}x_{1})\,{\mathrm{d}}\pi_{n}(y_{0},y_{1})\to
0.$

Therefore, putting together the estimates on every term, we can conclude that

$\displaystyle\limsup_{n\to\infty}\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}\bar{q}^{(r)}_{n}(x_{0},x_{1})$
$\displaystyle\leq\limsup_{n\to\infty}\int_{\mathsf{X}_{n}\times\mathsf{X}_{n}}\mathsf{d}^{2}(y_{0},y_{1})\mathrm{d}\pi_{n}(y_{0},y_{1})$
$\displaystyle=\limsup_{n\to\infty}W^{2}_{2}(\mu_{0,n},\mu_{1,n})=W^{2}_{2}(\mu_{0}^{(r)},\mu_{1}^{(r)}),$

where we used that $\pi_{n}$ is an optimal plan and that
$\mu_{0,n}\xrightarrow{W_{2}}\mu_{0}^{(r)}$,
$\mu_{1,n}\xrightarrow{W_{2}}\mu_{1}^{(r)}$. This last inequality, combined
with (4.20), allows us to conclude (4.19).

STEP 5: Definition of approximating plan with fixes marginals
We have shown the existence of $n(\varepsilon)\geq\bar{n}$ such that

$\bigg{|}\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}\bar{q}^{(r)}_{n(\varepsilon)}(x_{0},x_{1})-W^{2}_{2}(\mu_{0}^{(r)},\mu_{1}^{(r)})\bigg{|}<\varepsilon.$
(4.21)

Recalling the properties of $\bar{n}$ proven in the previous steps, we also
know that

$T_{K,N^{\prime}}^{(t)}(\pi_{n(\varepsilon)}|\bar{\mathfrak{m}}^{k}_{n(\varepsilon)})\leq
T_{K,N^{\prime}}^{(t)}(\bar{q}_{n(\varepsilon)}^{(r)}|\bar{\mathfrak{m}}^{k}_{\infty})+\varepsilon,$
(4.22)

for every $n\geq\bar{n}$ and every $N^{\prime}\in[N,\varepsilon)$. At this
point, using $\bar{q}_{n(\varepsilon)}^{r}$, we define a coupling
$q^{\varepsilon}$ between $\mu_{0}$ and $\mu_{1}$ by

$q^{\varepsilon}(.):=\alpha_{r}\bar{q}_{n(\varepsilon)}^{(r)}+\tilde{q}(.\cap(\mathsf{X}_{\infty}^{2}\setminus
E_{r})).$ (4.23)

First of all, notice that

$\bigg{|}\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}\bar{q}^{(r)}_{n(\varepsilon)}(x_{0},x_{1})-\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}q^{\varepsilon}(x_{0},x_{1})\bigg{|}\leq(1-\alpha_{r})\text{diam}(\mathcal{R}^{k}_{\infty})^{2}\leq\varepsilon
2^{2k+4}.$

Consequently, putting together this last estimate with (4.11) and (4.21), we
can conclude that

$\int\mathsf{d}^{2}(x_{0},x_{1})\mathrm{d}q^{\varepsilon}(x_{0},x_{1})=W^{2}_{2}(\mu_{0},\mu_{1})+O(\varepsilon).$
(4.24)

On the other hand, it is immediate from the definition of $q^{\varepsilon}$
that

$(1-\varepsilon)^{1-1/N^{\prime}}T^{(t)}_{K,N^{\prime}}(\bar{q}^{r}_{n(\varepsilon)}|\bar{\mathfrak{m}}^{k}_{\infty})\leq\alpha_{r}^{1-1/N^{\prime}}T^{(t)}_{K,N^{\prime}}(\bar{q}^{r}_{n(\varepsilon)}|\bar{\mathfrak{m}}^{k}_{\infty})\leq
T^{(t)}_{K,N^{\prime}}(q^{\varepsilon}|\bar{\mathfrak{m}}^{k}_{\infty}).$
(4.25)

STEP 6: Convergence of plans
We turn now to prove the weak convergence of the plans $q^{\varepsilon}$
introduced in the previous step, as $\varepsilon\to 0$, and consequently the
upper semicontinuity of $T^{(t)}_{K,N^{\prime}}$.
We first note that, since for every $\varepsilon>0$, it holds that
$q^{\varepsilon}\in\mathsf{Adm}(\mu_{0},\mu_{1})$, then the family
$(q^{\varepsilon})_{\varepsilon>0}$ is tight and Prokhorov Theorem ensures the
existence of a sequence $(\varepsilon_{m})_{m\in\mathbb{N}}$ converging to $0$
such that $q^{\varepsilon_{m}}\rightharpoonup
q\in\mathsf{Adm}(\mu_{0},\mu_{1})$. Equation (4.24) ensures the optimality of
$q\in\mathsf{Opt}(\mu_{0},\mu_{1})$. Furthermore, putting together the
estimates (4.22) (that holds definitely for every $N^{\prime}\in[N,0)$) and
(4.25), we conclude that, for every $N^{\prime}\in[N,0)$ and $t\in[0,1]$,

$\begin{split}\limsup_{m\to\infty}T_{K,N^{\prime}}^{(t)}(\pi_{n(\varepsilon_{m})}|\mathfrak{m}_{n(\varepsilon_{m})})&\leq\limsup_{m\to\infty}T_{K,N^{\prime}}^{(t)}(\pi_{n(\varepsilon_{m})}|\mathfrak{m}_{n(\varepsilon_{m})}^{k})\\\
&=\limsup_{m\to\infty}\frac{1}{\mathfrak{m}_{n(\varepsilon_{m})}^{k}(\mathsf{X}_{n})^{-1/N^{\prime}}}\cdot
T_{K,N^{\prime}}^{(t)}(\pi_{n(\varepsilon_{m})}|\bar{\mathfrak{m}}_{n(\varepsilon_{m})}^{k})\\\
&=\frac{1}{\mathfrak{m}_{\infty}^{k}(\mathsf{X}_{\infty})^{-1/N^{\prime}}}\limsup_{m\to\infty}T_{K,N^{\prime}}^{(t)}(\pi_{n(\varepsilon_{m})}|\bar{\mathfrak{m}}_{n(\varepsilon_{m})}^{k})\\\
&\leq\frac{1}{\mathfrak{m}_{\infty}^{k}(\mathsf{X}_{\infty})^{-1/N^{\prime}}}\limsup_{m\to\infty}T_{K,N^{\prime}}^{(t)}(\bar{q}_{n(\varepsilon_{m})}^{(r)}|\bar{\mathfrak{m}}^{k}_{\infty})+\varepsilon_{m}\\\
&=\frac{1}{\mathfrak{m}_{\infty}^{k}(\mathsf{X}_{\infty})^{-1/N^{\prime}}}\limsup_{m\to\infty}T_{K,N^{\prime}}^{(t)}(\bar{q}_{n(\varepsilon_{m})}^{(r)}|\bar{\mathfrak{m}}^{k}_{\infty})\\\
&\leq\frac{1}{\mathfrak{m}_{\infty}^{k}(\mathsf{X}_{\infty})^{-1/N^{\prime}}}\limsup_{m\to\infty}\frac{1}{(1-\varepsilon_{m})^{1-1/N^{\prime}}}T^{(t)}_{K,N^{\prime}}(q^{\varepsilon_{m}}|\bar{\mathfrak{m}}^{k}_{\infty})\\\
&=\limsup_{m\to\infty}T^{(t)}_{K,N^{\prime}}(q^{\varepsilon_{m}}|\mathfrak{m}^{k}_{\infty}).\end{split}$
(4.26)

Now, notice that every $q^{\varepsilon_{m}}$ has as marginals $\mu_{0}$ and
$\mu_{1}$, which are supported on $\mathcal{R}^{k-1}_{\infty}$ and therefore

$T^{(t)}_{K,N^{\prime}}(q^{\varepsilon_{m}}|\mathfrak{m}^{k}_{\infty})=T^{(t)}_{K,N^{\prime}}(q^{\varepsilon_{m}}|\mathfrak{m}_{\infty}).$

Thus we can apply Proposition 4.10 to
$T^{(t)}_{K,N^{\prime}}(q^{\varepsilon_{m}}|\mathfrak{m}_{\infty})$, for
$N^{\prime}\in I$, which together with the above estimate guarantees that

$\limsup_{m\to\infty}T_{K,N^{\prime}}^{(t)}(\pi_{n(\varepsilon_{m})}|\mathfrak{m}_{n(\varepsilon_{m})})\leq\limsup_{m\to\infty}T^{(t)}_{K,N^{\prime}}(q^{\varepsilon_{m}}|\mathfrak{m}_{\infty})=T^{(t)}_{K,N^{\prime}}(q|\mathfrak{m}_{\infty})$
(4.27)

holds for every $N^{\prime}\in I$ and $t\in[0,1]$.

STEP 7: Convergence of midpoints
The goal of this step is to show the existence of a limit geodesic
$\\{\mu_{t}\\}_{t\in[0,1]}$, such that for any $t\in[0,1]$,
$\mu_{t,n(\varepsilon_{m})}$ $W_{2}$-converges (up to subsequences) to
$\mu_{t}$, as $m\to\infty$. Furthermore, we are going to prove a suitable
lower semicontinuity of the Renyi entropies that will allow us to pass to the
limit the $\mathsf{CD}$ inequality. In order to ease the notation we will
denote the Renyi entropy $S_{N,\mathfrak{m}_{n(\varepsilon_{m})}}$ by
$S_{N,n(\varepsilon_{m})}$.

######  Claim 1.

For every fixed $t\in[0,1]$, the sequence
$(\mu_{t,n(\varepsilon_{m})})_{m\in\mathbb{N}}$ converges (up to subsequences)
to a measure $\mu_{t}\in\mathscr{P}(\mathsf{X}_{\infty})$.

First of all, notice that estimate (4.27), the $\mathsf{CD}$-condition (4.15)
and assumption (4.8) together ensure that the entropies
$S_{N,n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})})$ are uniformly bounded
above by a constant $M^{\prime}$, for $m\in\mathbb{N}$. Moreover, for every
$n(\varepsilon_{m})$, the approximation Lemma 4.4 provides the existence of a
sequence $(\mu_{t,n(\varepsilon_{m})}^{l})_{l\in\mathbb{N}}$, that
$W_{2}$-converges to $\mu_{t,n(\varepsilon_{m})}$, as $l\to\infty$, and such
that
$\text{supp}(\mu_{t,n(\varepsilon)}^{l})\subseteq\mathcal{R}_{n(\varepsilon)}^{l}$.
From the proof of Lemma 4.4 we recall that
$\mu_{t,n(\varepsilon_{m})}^{l}=c^{l}\,f^{l}\,\mu_{t,n(\varepsilon_{m})}$ so,
we can easily notice that,

$(c^{l})^{-1}=\int
f^{l}\,\mathrm{d}\mu_{t,n(\varepsilon_{m})}\geq\mu_{t,n(\varepsilon_{m})}(\mathcal{R}^{l-1}_{n(\varepsilon_{m})})\geq
1-\omega(k,l-1,M+1).$

Consequently, for sufficiently large $l$ it holds that, as measures,

$\mu_{t,n(\varepsilon_{m})}^{l}\leq\frac{1}{1-\omega(k,l-1,M+1)}\cdot\mu_{t,n(\varepsilon_{m})}.$
(4.28)

Notice that we took into account that
$\text{supp}(\mu_{0,n(\varepsilon_{m})}),\text{supp}(\mu_{1,n(\varepsilon_{m})})\subseteq\mathcal{R}_{n(\varepsilon_{m})}^{k}$
and we have used the $\omega$-uniform convexity assumption, keeping in mind
that $M+1$ bounds from above the terminal entropies (4.14). In turn,
inequality (4.28) implies that,

$S_{N,n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}^{l})\leq\frac{1}{(1-\omega(k,l-1,M+1))^{1-1/N}}S_{N,n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}).$
(4.29)

Moreover, the fact that
$\text{supp}(\mu_{0,n(\varepsilon_{m})}),\text{supp}(\mu_{1,n(\varepsilon_{m})})\subseteq\mathcal{R}_{n(\varepsilon_{m})}^{k}$
shows that the measure $\mu_{t,n(\varepsilon_{m})}$ has bounded support. In
particular, for every $l\in\mathbb{N}$ (and every $m\in\mathbb{N}$)

$\text{supp}(\mu_{t,n(\varepsilon_{m})}^{l})\subseteq\text{supp}(\mu_{t,n(\varepsilon_{m})})\subseteq
B(p_{n(\varepsilon_{m})},2^{k+2}).$

As a consequence of this bound, it’s easy to deduce that,

$W_{2}^{2}(\mu_{t,n(\varepsilon_{m})}^{l},\mu_{t,n(\varepsilon_{m})})\leq(2\cdot
2^{2k+2})^{2}\,\omega(k,l-1,M+1),$ (4.30)

because $\mu_{t,n(\varepsilon_{m})}\leq\mu_{t,n(\varepsilon_{m})}^{l}$ when
restricted to $\mathcal{R}_{n(\varepsilon_{m})}^{l-1}$, and
$\mu_{t,n(\varepsilon_{m})}(X\setminus\mathcal{R}_{n(\varepsilon_{m})}^{l-1})\geq\omega(k,l-1,M+1)$,
by $\omega$-uniform convexity. Now, for every fixed $l$ sufficiently large,
such that $\omega(k,l-1,M+1)<1$, observe that, according to (4.29), the
entropies $S_{N,n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}^{l})$ are
uniformly bounded above by the constant

$\frac{1}{(1-\omega(k,l-1,M+1))^{1-1/N}}M^{\prime},$

for all $m\in\mathbb{N}$. Notice also that, since
$\mu_{t,n(\varepsilon_{m})}^{l}$ is supported on
$\mathcal{R}_{n(\varepsilon_{m})}^{l}$, it holds that

$S_{N,\mathfrak{m}^{l+1}_{n(\varepsilon_{m})}}(\mu_{t,n(\varepsilon_{m})}^{l})=S_{N,n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}^{l}).$

Therefore, Corollary 4.7 shows that $\mu_{t,n(\varepsilon_{m})}^{l}$ weakly
converges to some $\mu_{t}^{l}\in\mathscr{P}(X_{\infty})$ as $m\to\infty$, for
some choice of a subsequence. Moreover, we extract the bound
$S_{N,\mathfrak{m}_{\infty}^{l+1}}(\mu_{t}^{l})<\infty$ from Corollary 4.9,
which guarantees the lower semicontinuity of $S_{N,\cdot}(\cdot)$ along our
sequence. Consequently, this implies that the support of $\mu_{t}^{l}$ is
contained in $\mathcal{R}_{\infty}^{l+1}$. Finally, note that then the
sequence of measures $(\mu_{t,n(\varepsilon_{m})}^{l})_{m\in\mathbb{N}}$ is
supported in a uniformly bounded set, since
$\text{supp}(\mu_{t,n(\varepsilon_{m})}^{l})\subseteq
B\left(p_{n(\varepsilon_{m})},2^{k+2}\right)$ and $p_{n(\varepsilon_{m})}\to
p_{\infty}$, as $m\to\infty$. Thus, we are able to conclude, up to picking
again subsequence, that for every sufficiently large $l\in\mathbb{N}$

$\mu_{t,n(\varepsilon_{m})}^{l}\xrightarrow{W_{2}}\mu_{t}^{l}\quad\text{ as
}\quad m\to\infty.$

As a matter of fact, we can show using inequality (4.30) that,

$W_{2}^{2}(\mu_{t}^{i},\mu_{t}^{j})\leq
2^{2k+7}\big{[}\omega(k,i-1,M+1))+\omega(k,j-1,M+1))\big{]},$

for every (large enough) $i,j\in\mathbb{N}$. Then, our assumption on $\omega$
ensures that $(\mu_{t}^{l})_{l\in\mathbb{N}}$ is a Cauchy sequence, which
therefore $W_{2}$-converges to $\mu_{t}\in\mathscr{P}(\mathsf{X}_{\infty})$.
We conclude by noting that the uniform estimate (4.30) guarantees that
$\mu_{t,n(\varepsilon_{m})}\to\mu_{t}$.

######  Claim 2.

For every $t\in[0,1]$ the measure $\mu_{t}$ does not give mass to the set $S$
of singular points.

For every $m\in\mathbb{N}$ and every $l\in\mathbb{N}$ sufficiently large, let
us introduce the measures

$\tilde{\mu}_{t,n(\varepsilon_{m})}^{l}=[1-\omega(k,l-1,M+1)]\mu_{t,n(\varepsilon_{m})}^{l},$

and notice that, for every $l\in\mathbb{N}$ sufficiently large,

$\tilde{\mu}_{t,n(\varepsilon_{m})}^{l}\rightharpoonup\tilde{\mu}_{t}^{l}:=[1-\omega(k,l-1,M+1)]\mu_{t}^{l}.$

Observe also that all measures $\tilde{\mu}_{t,n(\varepsilon_{m})}^{l}$ have
total mass equal to $[1-\omega(k,l-1,M+1)]$, as $m$ varies. Thus,
$\tilde{\mu}_{t}^{l}$ also has total mass equal to $[1-\omega(k,l-1,M+1)]$. On
the other hand it follows from the uniform convexity properties (and in
particular from (4.28)) that for every $m\in\mathbb{N}$ and every
$l\in\mathbb{N}$ sufficiently large, there exists a positive measure
$\bar{\mu}_{t,n(\varepsilon_{m})}^{l}$ such that

$\mu_{t,n(\varepsilon_{m})}=\tilde{\mu}_{t,n(\varepsilon_{m})}^{l}+\bar{\mu}_{t,n(\varepsilon_{m})}^{l}.$

Notice that, since the sequences $\mu_{t,n(\varepsilon_{m})}$ and
$(\tilde{\mu}_{t,n(\varepsilon_{m})}^{l})_{m\in\mathbb{N}}$ are weakly
converging, the sequence $\bar{\mu}_{t,n(\varepsilon_{m})}^{l}$ is also weakly
converging to a (positive) measure $\bar{\mu}_{t}^{l}$, such that

$\mu_{t}=\tilde{\mu}_{t}^{l}+\bar{\mu}_{t}^{l}.$

As pointed out before $\mu_{t}^{l}$ is supported on
$\mathcal{R}_{\infty}^{l+1}$, thus the same holds for $\tilde{\mu}_{t}^{l}$,
and therefore

$\mu_{t}(\mathcal{R}_{\infty}^{l+1})\geq 1-\omega(k,l-1,M+1).$

Finally observe that this is sufficient to prove the claim, because of the
arbitrariness of $l$.

######  Claim 3.

The lower semicontinuity of the Renyi entropies holds, that is for every
$N^{\prime}\in[N,0)$

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t})\leq\liminf_{m\to\infty}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}).$
(4.31)

First of all notice that, the result of Claim 2 combined with Proposition 4.8
yields that

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t})\leq\liminf_{l\to\infty}S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}^{l}).$
(4.32)

On the other hand Lemma 4.9 ensures that for every $l\in\mathbb{N}$ large
enough

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}^{l})=S_{N^{\prime},\mathfrak{m}_{\infty}^{l+2}}(\mu_{t}^{l})\leq\liminf_{m\to\infty}S_{N^{\prime},\mathfrak{m}^{l+2}_{n(\varepsilon_{m})}}(\mu_{t,n(\varepsilon_{m})}^{l})=\liminf_{m\to\infty}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}^{l}).$
(4.33)

Moreover, we deduce as in Claim 1 the following estimate for every
$N^{\prime}\in[N,0)$

$S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}^{l})\leq\frac{1}{(1-\omega(k,l-1,M+1))^{1-1/N^{\prime}}}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})})$

and consequently for every $l\in\mathbb{N}$

$\begin{split}\liminf_{m\to\infty}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})})&\geq\liminf_{m\to\infty}(1-\omega(k,l-1,M+1))^{1-1/N^{\prime}}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}^{l})\\\
&\geq(1-\omega(k,l-1,M+1))^{1-1/N^{\prime}}S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}^{l}),\end{split}$

where the last passage follows from (4.33). Then, since this last inequality
holds for every $l\in\mathbb{N}$, we can conclude that

$\begin{split}\liminf_{m\to\infty}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})})&\geq\liminf_{l\to\infty}(1-\omega(k,l-1,M+1))^{1-1/N^{\prime}}S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}^{l})\\\
&\geq S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}),\end{split}$

where we used (4.32). This is exactly what we wanted to prove.

CONCLUSION
So far we were able to prove that for every fixed $t\in[0,1]$, the sequence
$(\mu_{t,n(\varepsilon_{m})})_{m\in\mathbb{N}}$ converges (up to subsequences)
to a measure $\mu_{t}\in\mathscr{P}_{2}(\mathsf{X}_{\infty})$ and

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t})\leq\liminf_{m\to\infty}S_{N^{\prime},n(\varepsilon_{m})}(\mu_{t,n(\varepsilon_{m})}),$
(4.34)

for every $N^{\prime}\in[N,0)$. Now, a diagonal argument ensures that, by
selecting a suitable subsequence (that we do not rename for sake of
simplicity),

$(\mu_{t,n(\varepsilon_{m})})\overset{W_{2}}{\longrightarrow}\mu_{t},$

and that estimate (4.34) holds for every $t\in[0,1]\cap\mathbb{Q}$. Our
approximation ensures also that
$\mu_{0,n(\varepsilon_{m})}\overset{W_{2}}{\longrightarrow}\mu_{0}$ and
$\mu_{1,n(\varepsilon_{m})}\overset{W_{2}}{\longrightarrow}\mu_{1}$ therefore,
since $\mu_{t,n(\varepsilon_{m})}$ is a $t$-midpoint of
$\mu_{0,n(\varepsilon_{m})}$ and $\mu_{1,n(\varepsilon_{m})}$, for every
$t\in[0,1]\cap\mathbb{Q}$ the limit point $\mu_{t}$ is a $t$-midpoint of
$\mu_{0}$ and $\mu_{1}$. Now it is easy to realize that we can extend by
continuity $\mu_{t}$ to a Wasserstein geodesic (connecting $\mu_{0}$ and
$\mu_{1}$) on the whole interval $[0,1]$, obtaining also that

$(\mu_{t,n(\varepsilon_{m})})\overset{W_{2}}{\longrightarrow}\mu_{t},\qquad\text{for
every }t\in[0,1].$

Moreover, we know from the proof of Claim 2, that for every $l\in\mathbb{N}$

$\mu_{t}(\mathcal{R}_{\infty}^{l+1})\geq 1-\omega(k,l-1,M+1),$

for every $t\in[0,1]\cap\mathbb{Q}$. Then by continuity we can conclude the
same inequality for every $t\in[0,1]$, and consequently we know that $\mu_{t}$
gives no mass to the set of singular points.
Finally, inequality (4.34), combined with (4.27), allows to pass to the limit
as $m\to\infty$ inequality (4.15) at every rational time and obtaining that

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t})\leq
T^{(t)}_{K,N^{\prime}}(q|\mathfrak{m}_{\infty})$

holds for every $t\in[0,1]\cap\mathbb{Q}$ and every $N^{\prime}\in I$.
Finally, the lower semicontinuity of the entropy (ensured by the fact that
$\mu_{t}$ gives no mass to the set of singular points) and the continuity of
$T_{K,N^{\prime}}^{(t)}(q|\mathfrak{m}_{\infty})$ in $t$ (which is a
straightforward consequence of the dominated convergence theorem), allow to
extend this last inequality to every $t\in[0,1]$, concluding the proof of the
approximate $\mathsf{CD}$-condition.

###  4.3 Proof of the CD Condition

This final section is dedicated to the proof of our main result, that is
Theorem 4.2.. As already mentioned, the proof of the approximate
$\mathsf{CD}$-condition and the approximation argument that made it possible
are the foundation to prove the $\mathsf{CD}$-condition. As the reader will
notice, we are going to use basically the same techniques, but refining them a
little bit to achieve the more general result. We specify that we could prove
the $\mathsf{CD}$-condition directly, but we preferred to divide the proof in
order to be clearer.

Before going on, we prove a preliminary lemma, that will help us in the
following. Notice that a result of this type is now needed because the
marginals may not have bounded support.

######  Lemma 4.12.

Given a metric space $(X,d)$, for every $n\in\mathbb{N}$ let
$(\nu_{t}^{n})_{t\in[0,1]}\subset\mathscr{P}_{2}(X)$ be a Wasserstein
geodesic. Assume that for every $t\in[0,1]$ the family
$(\nu_{t}^{n})_{n\in\mathbb{N}}$ is tight and that there exist
$\nu_{0},\nu_{1}\in\mathscr{P}_{2}(X)$ such that

$\nu_{0}^{n}\overset{W_{2}}{\longrightarrow}\nu_{0}\quad\text{ and
}\quad\nu_{1}^{n}\overset{W_{2}}{\longrightarrow}\nu_{1}\qquad\text{ as
}n\to\infty.$

Then there exists a Wasserstein geodesic
$(\nu_{t})_{t\in[0,1]}\subset\mathscr{P}_{2}(X)$ connecting $\nu_{0}$ and
$\nu_{1}$ such that, up to subsequences,

$\nu_{t}^{n}\rightharpoonup\nu_{t}\qquad\text{as }n\to\infty\text{, for every
$t\in[0,1]\cap\mathbb{Q}$}.$

Proof. First of all, notice that applying Prokhorov theorem we deduce that,
for every fixed $t$, the sequence $(\nu_{t}^{n})_{n\in\mathbb{N}}$ is weakly
convergent, up to subsequences. Thus the diagonal argument ensures that, up to
take a suitable subsequence which we do not recall for simplicity, for every
$t\in[0,1]\cap\mathbb{Q}$ there exists $\nu_{t}\in\mathscr{P}_{2}(X)$ such
that

$\nu_{t}^{n}\rightharpoonup\nu_{t}\qquad\text{as }n\to\infty\text{, for every
$t\in[0,1]\cap\mathbb{Q}$}.$

It is well-known that the Wasserstein distance is lower semicontinuous with
respect to the weak convergence (see for example Proposition 7.1.3 in [1],
then

$W_{2}(\nu_{0},\nu_{t})\leq\liminf_{n\to\infty}W_{2}(\nu_{0}^{n},\nu_{t}^{n})=\liminf_{n\to\infty}t\cdot
W_{2}(\nu_{0}^{n},\nu_{1}^{n})=t\cdot W_{2}(\nu_{0},\nu_{1})$

and analogously

$W_{2}(\nu_{t},\nu_{1})\leq(1-t)\cdot W_{2}(\nu_{t},\nu_{1}).$

Combining this two inequalities with the triangular inequality we deduce that

$W_{2}(\nu_{0},\nu_{t})=t\cdot W_{2}(\nu_{0},\nu_{1})\quad\text{and}\quad
W_{2}(\nu_{t},\nu_{1})=(1-t)\cdot W_{2}(\nu_{t},\nu_{1}),$

which means that $\nu_{t}$ is a $t$-midpoint of $\nu_{0}$ and $\nu_{1}$. The
lower semicontinuity of the Wasserstein distance also ensures that for every
$s,t\in[0,1]\cap\mathbb{Q}$ it holds that

$W_{2}(\nu_{t},\nu_{s})\leq\liminf_{n\to\infty}|t-s|\cdot
W_{2}(\nu_{t}^{n},\nu_{s}^{n})=|t-s|\cdot\liminf_{n\to\infty}W_{2}(\nu_{0}^{n},\nu_{1}^{n})=|t-s|\cdot
W_{2}(\nu_{0},\nu_{1}).$

Finally, since for every $r\in[0,1]\cap\mathbb{Q}$ $\nu_{r}$ is an
$r$-midpoint of $\nu_{0}$ and $\nu_{1}$, the triangular inequality allow us
conclude that

$W_{2}(\nu_{t},\nu_{s})=|t-s|\cdot W_{2}(\nu_{0},\nu_{1}),\qquad\text{for
every $s,t\in[0,1]\cap\mathbb{Q}$},$

then we can extend $\nu_{t}$ to the whole interval $[0,1]$, finding a
Wasserstein geodesic $(\nu_{t})_{t\in[0,1]}$ connecting $\nu_{0}$ and
$\nu_{1}$.  $\square$

Now that we have this last result at our disposal we can proceed to the proof
of Theorem 4.1. To this aim, we fix
$\mu_{0},\mu_{1}\in\mathscr{P}_{2}^{ac}(\mathsf{X}_{\infty})$. In analogy with
the previous section, we can assume that
$S_{N,\mathfrak{m}_{\infty}}(\mu_{0}),S_{N,\mathfrak{m}_{\infty}}(\mu_{1})<\infty$
and introduce the constant

$M:=\max\\{S_{N,\mathfrak{m}_{\infty}}(\mu_{0}),S_{N,\mathfrak{m}_{\infty}}(\mu_{1})\\}.$

We can also define the interval

$I:=\\{N^{\prime}\in[N,0)\,:\,S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{0}),S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{1})<\infty\\},$

in particular we will need to prove (3.4) for every $N^{\prime}\in I$ and
every $t\in[0,1]$. Now, according to Lemma 4.4 there exist two sequences
$(\mu_{0}^{l})_{l\in\mathbb{N}}$ and $(\mu_{1}^{l})_{l\in\mathbb{N}}$,
$W_{2}$-converging to $\mu_{0}$ and $\mu_{1}$ respectively and such that

$\text{supp}(\mu_{0}^{l}),\text{supp}(\mu_{1}^{l})\subseteq\mathcal{R}_{\infty}^{l-1}\qquad\text{for
every }l\in\mathbb{N}.$

Moreover, keeping in mind the definition of $\mu_{0}^{l}$ and $\mu_{1}^{l}$,
it is easy to realize that for $l$ sufficiently large

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{0}^{l}),S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{1}^{l})<\infty\quad\text{
for every $N^{\prime}\in I$}$

and that the dominated convergence theorem ensures that

$\lim_{l\to\infty}S_{N,\mathfrak{m}_{\infty}}(\mu_{0}^{l})=S_{N,\mathfrak{m}_{\infty}}(\mu_{0})\qquad\text{
and
}\qquad\lim_{l\to\infty}S_{N,\mathfrak{m}_{\infty}}(\mu_{1}^{l})=S_{N,\mathfrak{m}_{\infty}}(\mu_{1}).$

Thus, for every $l$ large enough

$S_{N,\mathfrak{m}_{\infty}}(\mu_{0}^{l}),S_{N,\mathfrak{m}_{\infty}}(\mu_{1}^{l})\leq\max\big{\\{}S_{N,\mathfrak{m}_{\infty}}(\mu_{0}),S_{N,\mathfrak{m}_{\infty}}(\mu_{1})\big{\\}}+1:=M+1$

and then we can apply the argument presented in the last section and deduce
the existence of an optimal plan
$q^{l}\in\mathsf{Opt}(\mu^{l}_{0},\mu^{l}_{1})$ and of a Wasserstein geodesic
$(\mu^{l}_{t})_{t\in[0,1]}$ connecting $\mu_{0}^{l}$ and $\mu_{1}^{l}$, such
that

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}^{l})\leq
T_{K,N^{\prime}}^{(t)}(q^{l}|\mathfrak{m}_{\infty})$ (4.35)

holds for every $t\in[0,1]$ and every $N^{\prime}\in I$. Now, we divide the
proof in two steps, the first dedicated to the convergence of the plans
$(q^{l})_{l\in\mathbb{N}}$ and to the upper semicontinuity of
$T_{K,N^{\prime}}^{(t)}$, the second dedicated to the convergence of the
measures $(\mu_{t}^{l})_{l\in\mathbb{N}}$ and the lower semicontinuity of
$S_{N^{\prime},\mathfrak{m}_{\infty}}$.

Step 1: Upper semicontinuity for $T_{K,N^{\prime}}^{(t)}$

Notice that $(q^{l})_{l\in\mathbb{N}}$ is a sequence of probability measures
having as marginals two sequences of converging, and thus tight, probability
measures. As a consequence the sequence $(q^{l})_{l\in\mathbb{N}}$ is itself
tight, then up to subsequences it weakly converges to a plan
$q\in\mathscr{P}(\mathsf{X}_{\infty}\times\mathsf{X}_{\infty})$. It is well
know and easy to prove that $q\in\mathsf{Opt}(\mu_{0},\mu_{1})$. We are now
going to prove that

$\limsup_{l\to\infty}T_{K,N^{\prime}}^{(t)}(q^{l}|\mathfrak{m}_{\infty})\leq
T_{K,N^{\prime}}^{(t)}(q|\mathfrak{m}_{\infty})$ (4.36)

for every $t\in[0,1]$ and every $N^{\prime}\in I$. The argument we are going
to use is essentially the same as the one explained in the proof of
Proposition 4.10, nevertheless we briefly recall it for the sake of
completeness, avoiding to repeat all the details.
In particular, for every $l\in\mathbb{N}$ let us call $\rho_{0}^{l}$ and
$\rho_{1}^{l}$ the densities of $\mu_{0}^{l}$ and $\mu_{1}^{l}$ with respect
to the reference measure $\mathfrak{m}_{\infty}$, we just need to prove that

$\lim_{l\to\infty}\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}^{l}(x)^{-\frac{1}{N^{\prime}}}\mathrm{d}q^{l}=\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N^{\prime}}}\mathrm{d}q.$

Notice that, the particular definition of $\mu_{0}^{l}$ (check Lemma 4.4),
ensures that the density $\rho_{0}^{l}$ is a suitable renormalization of
$f^{l}\rho_{0}$, then for a fixed $\varepsilon>0$ we can find
$\bar{l}\in\mathbb{N}$ such that

$\lVert\big{(}\rho_{0}^{l}\big{)}^{-1/N^{\prime}}-\rho_{0}^{-1/N^{\prime}}\rVert_{L^{1}(\mu_{0}^{l})}<\varepsilon\,\,\text{
for every }l\geq\bar{l}.$

Furthermore, for the same reason combined with Lemma 2.12 (up to possibly
change $\bar{l}$) we can find $f^{\varepsilon}\in C_{b}(X)$ such that

$\lVert\rho_{0}^{-1/N^{\prime}}-f^{\varepsilon}\rVert_{L^{1}(\mu_{0})}<\varepsilon\qquad\text{and}\qquad\lVert\rho_{0}^{-1/N^{\prime}}-f^{\varepsilon}\rVert_{L^{1}(\mu_{0}^{l})}<\varepsilon\,\,\,\text{
for every }l\geq\bar{l}.$

Putting together this last two estimates we end up proving that

$\lVert\big{(}\rho_{0}^{l}\big{)}^{-1/N^{\prime}}-f^{\varepsilon}\rVert_{L^{1}(\mu_{0}^{l})}<2\varepsilon\,\,\text{
for every }l\geq\bar{l}.$

On the other hand, since the function

$\tau_{K,N}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)$

is bounded and continuous, the weak convergence of $(q^{l})_{l}$ to $q$ yields
that

$\lim_{l\to\infty}\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}q^{l}=\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}q.$

Then, since definitely $l\geq\bar{l}$, we can deduce the following estimate

$\begin{split}\limsup_{l\to\infty}\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}^{l}(x)^{-\frac{1}{N^{\prime}}}\mathrm{d}q^{l}&\leq\lim_{l\to\infty}\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}q^{l}+2\varepsilon\lVert\tau_{K,N^{\prime}}^{(1-t)}\rVert_{L^{\infty}}\\\
&=\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))f^{\varepsilon}(x)\mathrm{d}q+2\varepsilon\lVert\tau_{K,N^{\prime}}^{(1-t)}\rVert_{L^{\infty}}\\\
&\leq\int\tau_{K,N^{\prime}}^{(1-t)}(\mathsf{d}(x,y))\rho_{0}(x)^{-\frac{1}{N^{\prime}}}\mathrm{d}\pi+3\varepsilon\lVert\tau_{K,N^{\prime}}^{(1-t)}\rVert_{L^{\infty}}.\end{split}$

and since $\varepsilon>0$ can be chosen arbitrarily, (4.36) holds true.

Step 2: Lower semicontinuity for $S_{N^{\prime},\mathfrak{m}_{\infty}}$
In this second step we prove an additional property on $\mu^{l}_{t}$, which is
fundamental to prove the $\mathsf{CD}$-condition. Let us start with a
preliminary lemma.

######  Lemma 4.13.

Fix $k\leq h\in\mathbb{N}$ and let
$\nu\in\mathscr{P}_{ac}(\mathsf{X}_{\infty},\mathfrak{m}_{\infty})$ with
bounded density be such that
$\text{supp}(\nu)\subseteq\mathcal{R}_{\infty}^{k-1}$. Then, for every
$\epsilon>0$, there exists $\tilde{n}\in\mathbb{N}$ large enough such that,
$P^{\prime}_{n,h}(\nu)(\mathcal{R}_{n}^{k+1})\geq 1-\epsilon$ for every
$n\geq\tilde{n}$.

Proof. Notice that, accordingly to Lemma 4.5, both the sequences
$(P^{\prime}_{n,k}(\nu))_{n\in\mathbb{N}}$ and
$(P^{\prime}_{n,h}(\nu))_{n\in\mathbb{N}}$ $W_{2}$-converge to $\nu^{k}$.
Assume that $P^{\prime}_{m,h}(\nu)(\mathcal{R}_{m}^{k+1})<1-\epsilon$ for some
$m\in\mathbb{N}$. Observe that

$\inf\big{\\{}\mathsf{d}(x,y)\,:\,x\in\mathcal{R}_{m}^{k},\,y\in\mathcal{(}R_{m}^{k+1})^{c}\big{\\}}=2^{-(k+2)},$

as a consequence, since $P^{\prime}_{m,k}(\nu^{k})$ is supported on
$\mathcal{R}^{k}_{m}$, we obtain that

$W_{2}^{2}(P^{\prime}_{m,k}(\nu),P^{\prime}_{m,h}(\nu))\geq\epsilon\cdot
2^{-(2k+4)}.$ (4.37)

On the other, since the sequences
$(P^{\prime}_{n,k}(\nu^{k}))_{n\in\mathbb{N}}$ and
$(P^{\prime}_{n,h}(\nu^{k}))_{n\in\mathbb{N}}$ have the same limit, it holds
that

$W_{2}^{2}(P^{\prime}_{n,k}(\nu^{k}),P^{\prime}_{n,h}(\nu^{k}))\to
0\qquad\text{as }n\to\infty.$

Then definitely (4.37) cannot hold, proving the desired result.  $\square$

Fix $\epsilon>0$ and take $k(\epsilon)\in\mathbb{N}$ such that
$\mu_{0}(\mathcal{R}_{\infty}^{k(\epsilon)-1}),\mu_{1}(\mathcal{R}_{\infty}^{k(\epsilon)-1})>1-\frac{\epsilon}{2}$.
Then we take $l>k(\epsilon)$ and repeat the argument of the previous section
on $\mu_{0}^{l}$ and $\mu_{1}^{l}$. We are also going to use the same
notation, forgetting for the moment the dependence on $l$. It is easy to
realize that there exist two measures $\nu^{k(\epsilon)}_{0}$ and
$\nu_{1}^{k(\epsilon)}$ with
$\text{supp}(\nu^{k(\epsilon)}_{0}),\text{supp}(\nu^{k(\epsilon)}_{1})\subseteq\mathcal{R}_{\infty}^{k(\epsilon)-1}$
and
$\nu^{k(\epsilon)}_{0}(\mathsf{X}_{\infty}),\nu^{k(\epsilon)}_{1}(\mathsf{X}_{\infty})>1-\epsilon$,
such that for $r$ sufficiently large (and thus for $\varepsilon$ sufficiently
small) $\tilde{\mu}_{0}^{(r)}\geq\nu^{k(\epsilon)}_{0}$ and
$\tilde{\mu}_{1}^{(r)}\geq\nu^{k(\epsilon)}_{1}$ (in particular this tells us
that $\nu^{k(\epsilon)}_{0}$ and $\nu_{1}^{k(\epsilon)}$ have bounded
density). Then we can apply Lemma 4.13 to the probability measures

$\frac{1}{\nu^{k(\epsilon)}_{0}(\mathsf{X}_{\infty})}\nu^{k(\epsilon)}_{0}\quad\text{and}\quad\frac{1}{\nu^{k(\epsilon)}_{1}(\mathsf{X}_{\infty})}\nu^{k(\epsilon)}_{1},$

obtaining that for $m$ sufficiently large (in particular such that
$n(\varepsilon_{m})\geq\tilde{n}$) it holds that

$\mu_{0,n(\varepsilon_{m})}(\mathcal{R}_{n(\varepsilon_{m})}^{k(\epsilon)+1}),\mu_{1,n(\varepsilon_{m})}(\mathcal{R}_{n(\varepsilon_{m})}^{k(\epsilon)+1})\geq(1-\epsilon)^{2}\geq
1-2\epsilon.$

Consequently our uniform convexity assumption ensures that

$\mu_{t,n(\varepsilon_{m})}(\mathcal{R}_{n(\varepsilon_{m})}^{h})\geq
1-\Omega(k(\epsilon)+1,h,M+2,2\epsilon),$

for every $t\in[0,1]$ and every $h\in\mathbb{N}$. Proceeding as in Step 7 and
Conclusion of the previous section we can actually conclude that

$\mu_{t}^{l}(\mathcal{R}_{\infty}^{h+1})\geq
1-\Omega(k(\epsilon)+1,h-1,M+2,2\epsilon),$ (4.38)

for every $t\in[0,1]$ and every $h\in\mathbb{N}$ sufficiently large.

######  Claim 4.

For a fixed $t>0$, the family $(\mu_{t}^{l})_{l\in\mathbb{N}}$ is tight.

Given a fixed $\delta>0$, we have to find a compact set $K_{\delta}$, such
that $\mu_{t}^{l}(K_{\delta})\geq 1-\delta$ for every $l\in\mathbb{N}$. To
this aim we take suitable $\epsilon$ and $h$ such that (4.38) ensures that

$\mu_{t}^{l}(\mathcal{R}_{\infty}^{h+1})\geq 1-\frac{\delta}{2}.$ (4.39)

Moreover, combining the result of Step 1 (that is (4.36)) with (4.35), we
conclude that $S_{N,\mathfrak{m}_{\infty}}(\mu_{t}^{l})$ is definitely
bounded. Then, since $\mathfrak{m}_{\infty}|_{\mathcal{R}_{\infty}^{h+1}}$ is
a finite Radon measure, we can argue as in the proof of Lemma 4.6 and prove
the tightness of the family of measure
$(\mu_{t}^{l}|_{\mathcal{R}_{\infty}^{h+1}})_{l\in\mathbb{N}}$. As a
consequence, keeping in mind (4.39), there exists a compact set $K_{\delta}$
such that

$\mu_{t}^{l}(K_{\delta})\geq\mu_{t}^{l}|_{\mathcal{R}_{\infty}^{h+1}}(K_{\delta})\geq
1-\delta,$

proving the claim.

Now, we can apply Lemma 4.12 and find a Wasserstein geodesic
$(\mu_{t})_{t\in[0,1]}\subset\mathscr{P}(\mathsf{X}_{\infty})$ connecting
$\mu_{0}$ and $\mu_{1}$ such that (up to subsequences)

$\mu_{t}^{l}\rightharpoonup\mu_{t}\in\mathscr{P}(\mathsf{X}_{\infty})\qquad\text{as
$l\to\infty$ for every $t\in[0,1]\cap\mathbb{Q}$}$

Then, since the bound (4.38) is uniform in $l$, we can conclude that

$\mu_{t}(\mathcal{R}_{\infty}^{h+1})\geq
1-\Omega(k(\epsilon)+1,h-1,M+2,2\epsilon),$ (4.40)

for every $t\in[0,1]\cap\mathbb{Q}$ and every $h\in\mathbb{N}$ sufficiently
large. Moreover, by continuity we can deduce (4.40) for every time $t\in[0,1]$
(and every $h\in\mathbb{N}$ sufficiently large). This is sufficient to
conclude that $\mu_{t}$ gives no mass to the set $\mathcal{S}$ of singular
points, for every $t\in[0,1]$. In fact, assume by contradiction that
$\mu_{t}(\mathcal{S})=\delta>0$, then condition (3.11) on $\Omega$ ensures
that there exist $\epsilon$ and $h\in\mathbb{N}$ such that

$\Omega(k(\epsilon)+1,h-1,M+2,2\epsilon))<\delta,$

and consequently

$\mu_{t}(\mathcal{R}_{\infty}^{h+1})\geq 1-\delta,$

which contradicts $\mu_{t}(\mathcal{S})=\delta$. At his point we know from
Proposition 4.8 that

$\liminf_{l\to\infty}S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}^{l})\leq
S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t}),$ (4.41)

for every $t\in[0,1]\cap\mathbb{Q}$ and $N^{\prime}\in[N,0)$.

Finally, we can use (4.41) and (4.36) to pass at the limit as $l\to\infty$ the
inequality (4.35) and deduce that

$S_{N^{\prime},\mathfrak{m}_{\infty}}(\mu_{t})\leq
T_{K,N^{\prime}}^{(t)}(q|\mathfrak{m}_{\infty})$ (4.42)

holds for every $t\in[0,1]\cap\mathbb{Q}$ and every $N^{\prime}\in I$. Then
the lower semicontinuity of $S_{N^{\prime},\mathfrak{m}_{\infty}}$ (granted by
(4.40)) and the continuity of
$T_{K,N^{\prime}}^{(t)}(q|\mathfrak{m}_{\infty})$ in $t$ (which is a
straightforward consequence of the dominated convergence theorem), allow to
conclude (4.42) for every $t\in[0,1]$, finishing the proof.

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