Particular neighborhoods of analytical sets

Question

Let $X$ be a standard Borel space: a topological space isomorphic to a Borel subset of a complete separable metric space. Denote by $\mathcal P(X)$ the set of all Borel probability measures over $X$ endowed with the topology of weak convergence. Let $\rho(p,q)$ denote the total variation distance between probability measures $p,q\in \mathcal P(X)$.

Let $\Gamma\subseteq X\times \mathcal P(X)$ be an analytical set, and let us denote its $x$-section at $x\in X$ by $$ \Gamma_x := \{p\in \mathcal P(X):(x,p)\in \Gamma\}. $$ For some $\varepsilon > 0$ consider another set $\Gamma^\varepsilon\subseteq X\times \mathcal P(X)$ defined by its sections as follows: $$ \Gamma^\varepsilon_x:=\{q\in \mathcal P(X):\exists p\in \Gamma_x \text{ such that }\rho(p,q)\leq \varepsilon\} \qquad \forall x\in X. $$ What can be said about the measurability of $\Gamma^\varepsilon$? Perhaps, for any $\delta>0$ there exists an analytical set $\Gamma'$ such that $\Gamma^\varepsilon\subseteq \Gamma'\subseteq \Gamma^{\varepsilon+\delta}$?

If needed, we can use here that $\Gamma_x\neq\emptyset$ for any $x\in X$ and that $\Gamma$ contains a graph of a Borel map. Perhaps, a slight variation when $\Gamma^\varepsilon_x$ is defined using the strict inequality $\rho(p,q)<\varepsilon$ is easier to deal with? Any hint is appreciated.

Answer 1

If I am correct, one can proceed as follows. Consider a set $A\subset X\times \mathcal P(X)^2$ given by $$ A = \{(x,p,q):(x,p)\in \Gamma,\rho(p,q)\leq\varepsilon\} $$ then we obtain $\Gamma^\varepsilon = \pi(A)$ where $\pi:(x,p,q)\mapsto (x,q)$. We obtain $$ A = \left(\Gamma\times \mathcal P(X)\right)\cap \left(X\times \{(p,q):\rho(p,q)\leq\varepsilon\}\right). $$ The first set is analytic, and the second is Borel, hence $A$ is an analytic set, so that its projection $\Gamma^\varepsilon$.