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?



Weak convergence of random variables in $L^2$ and vague convergence
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.


