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

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What have you tried so far? What are you struggling with? –  Jonathan Christensen Jan 23 '13 at 16:26
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 –  ie86 Jan 23 '13 at 16:36

Hint Remember the Kolmogorov's Axioms of Probability and definition convergence of sequence of sets type $[X\leq x_n]$.