# Show that $P(X_n \leq x_n) \rightarrow P(X\leq x)$

Suppose $X_n$ converges in distribution to $X$, $x_n\rightarrow x$ and the cumulative distribution function for $X$ is continuous at $x$. Show that $P(X_n \leq x_n) \rightarrow P(X\leq x)$.

I tried to use the fact that $F_n(x_n) \subset [0,1]$ and according to compactness it has at least one convergence subsequence i.e. $F_{n_k}(x_{n_k})$ converges to $y$. Is this the right path? I was stuck here.

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Let $\left(y_k\right)_{k\geqslant 1}$ be a decreasing sequence of continuity points of the cumulative distribution function of $X$, and convergent to $x$. Let $k\geqslant 1$ be fixed. Then for $n$ large enough we have $x_n\leqslant y_k$, hence $\mathbb P(X_n\leqslant x_n)\leqslant \mathbb P\left(X_n\leqslant y_k\right)$, and taking $\limsup_{n\to +\infty}$, $$\limsup_{n\to +\infty}\mathbb P(X_n\leqslant x_n)\leqslant \mathbb P\left(X\leqslant y_k\right).$$ Now, letting $k$ going to infinity, we derive that $$\limsup_{n\to +\infty}\mathbb P(X_n\leqslant x_n)\leqslant \mathbb P(X\leqslant x).$$ By a similar reasoning (using this time an increasing sequence $\left(z_k\right)_{k\geqslant 1}$ of continuity points of the cumulative distribution function of $X$, and convergent to $x$), we get $$\mathbb P(X\leqslant x)\leqslant \liminf_{n\to +\infty}\mathbb P(X_n\leqslant x_n).$$