TITLE: Weak convergence of random variables in $L^2$ and vague convergence
QUESTION [5 upvotes]: Dumb question: Let $X_n:\Omega \to \mathbf{R}$ be a sequence of $L^2(\Omega,\Sigma,\mathbf{P})$ random variables that has a weak limit $X$ in $L^2$.
Suppose also that $\mu_n$, the distributions of $X_n$ on $\mathbf{R}$ converge vaguely to $\mu$. 
Does the distribution of $X$ have to be $\mu$? If the $X_n$ are uniformly bounded by constants, does this have to be true?

REPLY [7 votes]: No, it is not true. Take $\Omega = [0,1]$ equipped with the Lebesgue measure and let $(X_n)_{n \in \mathbb{N}}$ be the sequence of Rademacher functions -- all of these have the same distribution (uniform on $\{-1,1\}$) but they converge weakly in $L^{2}[0,1]$ to $0$, since they form an orthonormal system. They are also uniformly bounded.