Let $X_1,\ldots,X_n$ be a sequence of independent and identically distributed random variables, with joint probability measure $\mu^{(n)}$. Let $f_n:\mathbb{R}^n \to \mathbb{R}$ and $g_n:\mathbb{R}^n \to \mathbb{R}$ be two sequences of measruable maps, such that the random variables $$ Y_n=f_n(X_1,\ldots,X_n), \quad Z_n=g_n(X_1,\ldots,X_n). $$ have absolutely continuous distributions. Clearly, $Y_n$ and $Z_n$ may be dependent. Also, define $Z_n'=Z_n1(Z_n \in B_n)$, where $1(Z_n \in B_n)$ equals one if $Z_n$ lies in $B_n$, a measurable set, and zero otherwise.
Question: Is it true that, given a sequence of measurable sets $E_n$, $$ \lim_{n \to \infty} \sup_{s \in B_n}P(Y_n -s\in E_n)=0 \implies \lim_{n \to \infty}P(Y_n-Z_n'\in E_n) =0 $$ or in order this to be true we need further conditions?
My preliminary remarks: We have that $$ P(Y_n-Z_n'\in E_n)=\int_{B_n}P(Y_n -s \in E_n|Z_n=s)d\mu^{(n)}\circ g_n^{-1}(s)\\ \leq \sup_{s\in B_n}P(Y_n -s \in E_n|Z_n=s) $$ so I was wondering whether there are conditions under which $P(Y_n -s \in \cdot|Z_n=s)$ is absolutely continuous with respect to $P(Y_n-s \in \cdot)$ and such that this "dominance" relation is preserved at the limit.
Additional comments: I'm wondering about the need of additional conditions because, as far as I can understand, in general $P(Y_n-s\in E_n)=0$ only entails that $P(Y_n -s \in \cdot|Z_n=z)=0$ for almost every $z$ (i.e. on a set of $z$'s having probability one with respect to the law of $Z_n$), but $z=s$ needs not be in such a set. Also I'm trying to figure out an answer which could work as well in the cases where $Y_n$ and $Z_n$ are random vectors of dimensions increasing with $n$, so I'd look for arguments not relying on the dimensionality. Also, I would avoid the cases where $B_n$ is chosen such that $P(Z_n\in B_n)=0$, or the cases where $Z_n$ is a trivial transformation of $Y_n$, e.g. $Z_n=-Y_n$. Moreover, I'm wondering whether an argument through conditional probabilities is unnecessary complicated and there's a more direct way of reasoning, maybe by contradiction.