Let $X_n=(X_{n,1}, \ldots, X_{n,n})$ be sequences of random vectors with increasing dimension. Precisely, $X_{n,1}, \ldots, X_{n,n}$ are independent and identically distributed with absolutely continuous distribution $\mu_n$, a probability measure on $\mathbb{R}$. Thus, $X_n$ takes values in $\mathbb{R}^n$ and has joint probability measure $\times_{i=1}^n\mu_n$, the $n$-fold product measure of $\mu_n$.
Let $Y_n$ be a sequence of random variables, i.e. with values in $\mathbb{R}$, dependent on $X_n$, which also have absolutely continuous distribution.
Assume that:
(A.1) for a given bounded set $B\subset \mathbb{R}$ and a constant $c \in B$, $P(Y_n \in B)=1$ and $Y_n \to c$ in probability;
(A.2) for $i=1, \ldots,n$, $X_{i,n}\overset{d}{\to} Z$, where $Z$ is a nondegenerate tight random variable. Equivalently, $\mu_n $ weakly converges to $\mu$, the probability measure of $Z$, as $n \to \infty$.
Finally, define $\overline{Y}_n= \sum_{j=1}^{k_n}y_{n,j}1(Y_n \in B_{n,j})$, where $(B_{n,j})_{j=1}^{k_n}$ represents nested partitions of $B$, which become finer and finer as $n \to \infty$, i.e. $k_n\to \infty$, and $y_{n,j}\in B_{n,j}$, $j=1,\ldots,k_n$.
Question: Let $E_n$ be a sequence of measurables sets in $\mathbb{R}^n$ and, for each $b\in \mathbb{R}$, let us use the convention $X_n - b=(X_{n,1}-b, \ldots, X_{n,n}-b)$. Is it true that $$ P(X_n-Y_n \in E_n)-P(X_n-\overline{Y}_n \in E_n)\to 0, \quad \textbf{(E.1)} $$ as $n \to \infty$?
Comment: In the simplest case of $X_n$ being a sequence of random variables, by Slutsky's lemma we would have $X_n-Y_n\overset{d}{\to} Z- c$ and $X_n-\overline{Y}_n\overset{d}{\to} Z- c$, thus, I would say that in such a case the result in $\textbf{(E.1)}$ holds true.
How about the increasing dimensional case described above? Would some sort of infinite dimensional version of Slutsky's lemma do the job? Would it be true that $X_n \overset{d}{\to} (Z_i)_{i=1}^\infty$, where $(Z_i)_{i=1}^\infty$ is a countable collection of independent replicates of $Z$ (i.e. $Z_i \overset{d}{=}Z$)? Does Slutsky's lemma still apply? As $(Z_i)_{i=1}^\infty$ and $X_n$ are not of the same dimension, this last question is probably not very precise, to be formal the dimension of $X_n$ should be augmented by binding $X_n$ to some suitable countable collection of random variables, possibly some copies of $Z$ indepdendent from $X_n$, obtaining $X_n'$, with values in $\mathbb{R}^\infty$ and whose first $n$ components coincide with $X_n$. A formally more correct conjecture to be verified would then be whether $X_n' \overset{d}{\to}(Z_{i})_{i=1}^\infty$.