Weak convergence to product measure form conditional convergence of marginals

Question

$\newcommand\Ac{\mathcal A}$ $\newcommand\BL{\operatorname{BL}}$ $\newcommand\reals{\mathbb R}$ $\newcommand\eps{\varepsilon}$ $\newcommand\pr{\mathbb P}$ $\newcommand\ex{\mathbb E}$ $\newcommand\given{\,|\,}$ $\newcommand\convd{\rightsquigarrow}$

Let $\BL_1(\reals)$ the class of 1-bounded 1-Lipschitz functions on $\reals$, that is, the set of functions $h : \reals \to [-1,1]$ such that $|h(x) - h(y)| \le |x-y|$ for all $x,y \in \reals$.

Consider random variables $X_{ni}, i \in [K] := \{1,\dots,K\}$ and $Y_n$ and assume that $\{X_{ni}, i \in [K]\}$ are independent conditional on $Y_n$. In addition, we have \begin{align*} \sup_{h \,\in\, \BL_1(\reals)} | \ex [h(X_{ni}) \given Y_n] - \ex h(Z)| \cdot 1\{Y_n \in \Ac_n\} \;\le\; \eps_n ,\quad i \in [K] \end{align*} for some sequence of events $\Ac_n$ and a deterministic sequence of $\eps_n > 0$, and some random variable $Z \sim \mu$. Assume that $\eps_n \to 0$ and $\pr(Y_n \in \Ac_n^c) \to 0$ as $n\to \infty$. Then is it true that $$ (X_{n1},\dots,X_{nK}) \convd \mu^{\otimes K} $$ where $\convd$ denotes the convergence in distribution?

PS. There is really another question which perhaps I should ask separately, but would appreciate if there is a short answer to it. Would the answer change if we replace $\BL_1(\reals)$ above with the set of indicators of half intervals $\{t \mapsto 1\{t \le x\}:\; x \in \reals\}$?

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I believe this is true and have a proof. I appreciate if someone can verify that I not missing anything. Here is the proof: Let $Z_i, i \in [K]$ be i.i.d. draws from $\mu$. By Corollary 1.4.5 van der Vaart and Wellner, it is enough to show that \begin{align*} \ex \Bigl[\prod_{i=1}^K f_i(X_{ni})\Bigr] \to \ex\Bigl[\prod_{i=1}^K f_i(Z_i)\Bigr] = \prod_i \ex f_i(Z_i) \end{align*} for any collection of $f_1,\dots,f_K \in \BL_1(\reals)$.

Let us fix one such collection and, for simplicity, write $ W_n = \prod_i f_i(X_{ni})$ and $C = \prod_i \ex f_i(Z_i)$. We want to show $\ex[W_n| \to C$. From the assumption, it follows that \begin{align*} \bigl| \ex [f_i(X_{ni}) \given Y_n] - \ex f_i(Z_i) \bigr| \;\le\; \eps_n + 2 \cdot 1\{Y_n \in \Ac_n^c\}. \end{align*} By conditional independence, $\ex [W_n \given Y_n] = \prod_i \ex[f_i(X_{ni})\given Y_n]$. Then, using $|\prod_{i=1}^K a_i - \prod_{i=1}^K b_i| \le K \max_i |a_i - b_i|$, which holds if $a_i, b_i \in [-1,1]$ for all $i$, we have \begin{align*} \bigl| \ex [W_n \given Y_n ] - C\bigr| \;\le\; K \eps_n + 2 K \cdot 1\{Y_n \in \Ac_n^c\}. \end{align*} Then, we have \begin{align*} | \ex [W_n] - C| &= | \ex[\ex[W_n \given Y_n]] - C| \\ &\le \ex \bigl|\ex[W_n \given Y_n]] - C\bigr| \le K \eps_n + 2 K \pr(Y_n \in \Ac^c). \end{align*} Letting $n \to \infty$, the desired result follows from the assumptions.

Answer 1

Your proof is fine (however, asking to check proofs is discouraged on MathOverflow).

As for your other question, the answer to it is yes. Indeed, you can use almost the same proof, with the only difference that now you will have $f_i=1_{(-\infty,x_i]}$ for all real $x_i$, and the convergence in your formula $$E\prod_{i=1}^K f_i(X_{ni})\to E\prod_{i=1}^K f_i(Z_i) = \prod_i Ef_i(Z_i)$$ for all such functions $f_i$ will be the pointwise convergence of the joint cdf of $(X_{n1},\dots,X_{nK})$ to the joint cdf of $(Z_1,\dots,Z_K)$, and the latter convergence will of course imply the convergence of $(X_{n1},\dots,X_{nK})$ to $(Z_1,\dots,Z_K)$ in distribution.