# Convergence in distribution of random variables question

$X, X_1, X_2,\ldots$ are real random variables with $\mathbb{P}(X_n\leq x)\to \mathbb{P}(X\leq x)$ whenever $\mathbb{P}(X=x)=0$.

Why does $X_n\stackrel{L}{\to} X$? At the least, where would I begin?

A sequence $X_1,X_2,\ldots$ of random variables is said to converge in distribution, or converge weakly, or converge in law to a random variable $X$ if $$\lim_{n\to\infty}F_n(x)=F(x)$$ for every number $x\in\mathbb R$ at which $F$ is continuous, where $F_n(x)=\mathbb P(X_n\le x)$ and $F(x)=\mathbb P(X\le x)$.
Thus, we need to show that $F(x)$ is continuous at $x$ if and only if $\mathbb P(X=x)=0$. $F$ is continuous from the right, so we need to investigate continuity from the left. Since $$\mathbb P(X=x)=F(x)-\lim_{y\uparrow x}F(y),$$ we have that $\mathbb P(X=x)=0$ if and only if $F(x)=\lim_{y\uparrow x}F(y)$.
• Nevermind my comment, I took it for granted that the definition of convergence in distribution was ${\rm E}[f(X_n)]\to{\rm E}[f(X)]$ for continuous, bounded $f$. This doesn't seem to be the definition the OP is working with. Sorry about the fuzz. Mar 6, 2014 at 9:43
Start with the definition of what is the convergence. There is also a theorem (http://fr.wikipedia.org/wiki/Convergence_en_loi, can't find the english version) that says that you need to prove that $P(X_n \in A) -> P(X \in A)$ for any A an open