Convergence of conditional probability when sequences of random variables converge in probability Let $\{X_{n}\}_{n=1}^{\infty}$ be a sequence of non-negative random variables that converges to $X$ in probability. Also, let $\{Y_{n}\}_{n=1}^{\infty}$ be a sequence of non-negative random variables such that $Y_{n} \overset{p}{\rightarrow} c$, for some non-negative constant $c$. I would like to prove if
$$ P(X_{n}\leq Y_{n}|Y_{n}) \overset{p}{\rightarrow} P(X \leq c)$$
is true. Please note that $X_{n}$ and $Y_{n}$ are not independent. 
I spent few days but still having a trouble to get an answer. I tried to prove this using Dominated Convergence Theorem such that
$$ \lim_{n \rightarrow \infty} E\{1(X_{n} \leq Y_{n})|Y_{n}\} = 
E\{\lim_{n \rightarrow \infty} 1(X_{n} \leq Y_{n})|Y_{n}\} \\ =
E\{1(X \leq c)\},$$
but I am not sure if that is correct because of the conditioning. It would be much appreciated if I can get any help.
 A: This is not true in general. Suppose that $X_n\equiv 1$ for all $n\ge 1$ and $Y_n=1-n^{-1}$. Then
$$
\mathsf{P}(X_n\le Y_n\mid Y_n)=\mathbf{1}\{1\le 1-n^{-1}\}=0.
$$
However,
$$
\mathsf{P}(X\le 1)=\mathbf{1}\{1\le 1\}=1.
$$

Let $F_n$ be the conditional cdf of $X_n$ given $Y_n$ and let $F(x)=\mathsf{P}(X\le x)$. Then for any $\epsilon>0$,
\begin{align}
|\mathsf{P}(X_n\le Y_n\mid Y_n)-\mathsf{P}(X\le c)| &\le |\mathsf{P}(X_n\le Y_n\mid Y_n)-\mathsf{P}(X_n\le c\mid Y_n)| \\
&+|\mathsf{P}(X_n\le c\mid Y_n)-\mathsf{P}(X_n\le c)| \\
&\le |F_n(c+\epsilon)-F_n(c-\epsilon)|+\mathbf{1}\{|Y_n-c|> \epsilon\} \\
&+|F_n(c)-F(c)| \\
&\le 2\sup_{x\in [c-\epsilon,c+\epsilon]}|F_n(x)-F(x)|+\mathbf{1}\{|Y_n-c|> \epsilon\} \\
&+|F(c+\epsilon)-F(c-\epsilon)|.
\end{align}
Therefore, the RHS converges to $0$ (in prob.) if


*

*$F$ is continuous at $c$;

*$\mathsf{E}\sup_{x\in [c-\epsilon,c+\epsilon]}|F_n(x)-F(x)|\to 0$ for some $\epsilon>0$.


Condition (1) and (2) hold, for example, when $F$ is continuous on some neighbourhood $\mathcal{N}_c$ of $c$ and $F_n(x)\xrightarrow{p} F(x)$ for every $x\in \mathcal{N}_c$.
