If $|X_n - Y_n| \overset{P}{\longrightarrow} 0$, does $|F_{X_{n}}(t) - F_{Y_n}(t)| \to 0$ for all $t \in \mathbb R\,$? Suppose that $(X_n)$ and $(Y_n)$ are sequences of random variables, all defined on the probability space $(\Omega, \mathcal F, P)$.
Suppose the difference of the random variables converges in probability to $0$, i.e.
$$
|X_n - Y_n| \overset{P}{\longrightarrow} 0 \quad \text{as $n \to \infty$.}
$$
Does the difference of their c.d.f.'s then converge pointwise to zero? That is: Does the convergence
$$
\big| F_{X_n}(t) - F_{Y_n}(t) \big| = \big| P(X_n \leq t) - P(Y_n \leq t)\big| \to 0 \quad \text{as $n \to \infty$}
$$
hold for all $t \in \mathbb R\,$?
Thoughts
Intuitively, this seems reasonable, but I'm not sure how to show this.
On can reason that
$$
\big| P(X_n \leq t) - P(Y_n \leq t)\big| 
\leq P(|X_n - Y_n| > 0) = \lim_{\varepsilon \downarrow 0} P(|X_n - Y_n| > \varepsilon ),
$$
so
$$
\limsup_{n \to \infty} \big| P(X_n \leq t) - P(Y_n \leq t)\big| \leq \limsup_{n \to \infty} \lim_{\varepsilon \downarrow 0} P(|X_n - Y_n| > \varepsilon ).
$$
If we could swap the order of the limits on the right hand side of the inequality above, we would be finished. I don't see why that would be justified, though.
 A: We give the proof of the following:
$$ X_n \rightsquigarrow X \text { and } d\left(X_n, Y_n\right) \stackrel{\mathrm{P}}{\rightarrow} 0 \text {, then } Y_n \rightsquigarrow X $$
Where $\rightsquigarrow$ indicates convergence in distribution.
Proof:
We give the result for random variables only. The proof for the vector case is similar. For every $\varepsilon>0$,
$$
\begin{aligned}
\mathrm{P}\left(Y_n \leq x\right) & \leq \mathrm{P}\left(Y_n \leq x, d\left(X_n, Y_n\right) \leq \varepsilon\right)+\mathrm{P}\left(d\left(X_n, Y_n\right)>\varepsilon\right) \\
& \leq \mathrm{P}\left(X_n \leq x+\varepsilon\right)+o(1) .
\end{aligned}
$$
If $x+\varepsilon$ is a continuity point of the distribution function of $X$, then the right side converges to $\mathrm{P}(X \leq x+\varepsilon)$ and we conclude that $\lim \sup \mathrm{P}\left(Y_n \leq x\right) \leq$ $\mathrm{P}(X \leq x+\varepsilon)$. This is true for all $\varepsilon>0$ except at most the countably many values such that $x+\varepsilon$ is a jump point of $x \rightarrow \mathrm{P}(X \leq x)$. In particular, it is true for a sequence $\varepsilon_m \downarrow 0$ and we conclude that
$$
\lim \sup \mathrm{P}\left(Y_n \leq x\right) \leq \lim _{m \rightarrow \infty} \mathrm{P}\left(X \leq x+\varepsilon_m\right)=\mathrm{P}(X \leq x) .
$$
This gives one half of the proof. By arguing in an analogous manner, we can prove that limsup $\mathrm{P}\left(Y_n>x\right) \leq \mathrm{P}(X>x-\varepsilon)$ for every $x$ and $\varepsilon>0$ and hence that $\lim \sup \mathrm{P}\left(Y_n>x\right) \leq \mathrm{P}(X \geq x)$. For $x$ a continuity point of the distribution function of $X$, the right side is equal to $\mathrm{P}(X>x)$. Taking complements we obtain that $\lim \inf \mathrm{P}\left(Y_n \leq x\right) \geq \mathrm{P}(X \leq x)$.
The two inequalities combined yield that $\mathrm{P}\left(Y_n \leq x\right) \rightarrow \mathrm{P}(X \leq x)$ for every continuity point of the distribution function of $X$.
From A. Van Der Vaart - Asymptotic Statistics
A: This is not true. For a simple counterexample, take $X_n=1/n$, $Y_n=0$, and $t=0$. Something that is true, and is similar, is that if $|X_n-Y_n|\to0$ in probability and $f$ is bounded and uniformly continuous, then
$$\tag{$\ast$}\label{ast}
  |E[f(X_n)] - E[f(Y_n)]| \to 0.
$$
If $f(x)$ is a uniformly continuous approximation of $1_{(-\infty,t]}(x)$, then you can get close to what you want.
To see that \eqref{ast} is true, first note that
\begin{align}
  |E[f(X_n)] - E[f(Y_n)]| &= |E[f(X_n) - f(Y_n)]|\\
  &\le E|f(X_n) - f(Y_n)|.
\end{align}
By the hypothesis, every subsequence of $\{X_n-Y_n\}$ has a subsequence that converges to 0 a.s. So by dominated convergence, we can conclude that every subsequence of the left-hand side of \eqref{ast} has a subsequence that converges to 0. Thus, the whole sequence converges to 0.
