A sequence of random variables which converges in distributon converges "to" some random variable Let $(X_n)$ be a sequence of random variables on a probability space $\Omega$, with distribution functions $F_n$. Suppose $F_n \rightarrow F$ in distribution for some distribution function $F$. Must there be a random variable $X$ on $\Omega$ with $X_n \rightarrow X$ in distribution? 
This seems like the sort of result one would want to be true, but I couldn't think of a way to prove it. To do so, one would need to somehow define $X$ "backwards" given its distribution $F$. Does anyone have any ideas?
 A: For every distribution function $F$ there exists random variables $X$ such that $F=F_X$. A way to exhibit such a random variable is to set $X=G(U)$ where $U$ is uniform on $(0,1)$ and $G$ is an (extended) inverse of $F$. The function
$$
G:u\mapsto\inf\{x\mid F(x)\geqslant u\}
$$
might do the job but you should check the inequality sign.
Edit: For a counterexample, one should consider $\Omega$ such that no uniform random variable $U$ as above exists on $\Omega$. Consider $\Omega=\mathbb N$ with the sigma-algebra $2^\Omega$ and $\mathbb P(\{n\})=p_n$. Call $S\subseteq\mathbb N$ the set $S=\{s(K)\mid K\subseteq\mathbb N\}$ where $s(K)=\sum\limits_{n\in K}p_n$ for $K\subseteq\mathbb N$. Choose $(s_n)\subseteq S$ such that $s_n\to s$. Then $X_n=\mathbb 1_{K_n}$, with $s_n=s(K_n)$, converges in distribution to a Bernoulli random variable $X$ with parameter $s$, hence if $s$ is not in $S$, $X$ cannot be defined on $\Omega$. To complete the proof, one must exhibit $(p_n)$ such that the set $S$ is not closed. At the moment, I do not know if this is possible...
