Problem on weak convergence If $X_n \implies X$, $Y_n \implies X$ and $P(X_n -Y_n \leq -\epsilon) \to 0$ then $X_n -Y_n \implies 0$.
My attempt :
I proved $\limsup (X_n - Y_n) > -\epsilon $ almost surely,for all $\epsilon > 0$. It's sufficient to prove $P(X_n-Y_n < \epsilon) \to 1$ for all $\epsilon > 0$, this part I am couldn't prove. Any hints are appreciated.
$\implies$ Denotes weak convergence.
 A: Let $M$ be any positive number, $\varphi(x)=(x\vee(-M))\wedge M$ , that is,
\begin{cases}
\varphi(x)=M &\text{if } x>M\\
\varphi(x)=x &\text{if } |x|\le M\\
\varphi(x)=-M &\text{if } x<-M\\
\end{cases}
One fact about $\varphi$ is that it is increasing(*), thus
$$\begin{align}
|\varphi(Y_n)-\varphi(X_n)|+\varphi(Y_n)-\varphi(X_n)&=2(\varphi(Y_n)-\varphi(X_n))_+ \\
&\stackrel{(*)}{=} 2\big( \varphi(X_n+(Y_n-X_n)_+)-\varphi(X_n) \big)
\end{align}$$
Besides, the hypothesis on $X_n-Y_n$ implies that: $(Y_n-X_n)_+ \xrightarrow[n \rightarrow \infty]{P} 0$, thus
$$\left(X_n,(Y_n-X_n)_+\right) \xrightarrow[n \rightarrow \infty]{(d)} (X,0) $$
So by definition of weak convergence, we imply that:
$$\lim \Bbb{E}(|\varphi(Y_n)-\varphi(X_n)|) =0$$
In addition, by the way we define $\varphi$, we have:
$$0 \le \min( |X_n-Y_n|,1) \Bbb{1}_{ |X_n|,|Y_n|<M} \le |\varphi(Y_n)-\varphi(X_n)| $$
Thus
$$\lim \Bbb{E}( \min( |X_n-Y_n|,1) \Bbb{1}_{ |X_n|,|Y_n|<M} ) =0$$
From which, we imply that:
$$
\begin{align}
\limsup \Bbb{E}( \min( |X_n-Y_n|,1) ) &\le \limsup \Bbb{E}( 1-\Bbb{1}_{ |X_n|,|Y_n|<M}) \\
&\le \limsup \Bbb{P}( |X_n| \ge M) + \limsup \Bbb{P}( |Y_n| \ge M)\\
&\le 2\Bbb{P}(|X| \ge M)
\end{align} $$
where the last inequality is due to Portemanteau's theorem. Now, let $M$ converge to infinity, the  previous inequality implies that:
$$\lim \Bbb{E}( \min( |X_n-Y_n|,1) )=0$$
Or $X_n-Y_n$ converges to $0$ in probability.
