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

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 $\{t \mapsto 1\{t \le x\}:\; x \in \reals\}$?

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.