Clarification about existence of limit random variable (for convergence in distribution). Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $(X_n)_n$ a sequence of random variables on this space. Let $\mu: \mathbb{B}(\mathbb{R}) \to [0,1]$ a probability distribution
Suppose that $\mathbb{P}_{X_n} \stackrel{w}\to \mu $. Does there exist a random variable $Y$ on $(\Omega, \mathcal{F}, \mathbb{P})$ with $\mu = \mathbb{P}_Y?$
I looked here:
Weak convergence of random variables
The answer in this question seems to indicate that the answer is true. However, one of the comments says that this is false. 
What is the right answer?
I can't find a mistake in either of them. 
Is it maybe that the random variables in the answer itself are allowed to take the values $\pm \infty$? Does this make the difference?
 A: The answer of Will Nelson in the thread you linked looks fine to me. The argument where „$\Omega$ is countable and $(X_n)_{n\in\mathbb N}$ is a sequence of random variables such that their pushforward measures converge weakly to a continuous measure“ is flawed. The reason is that in this case there must be an $\mathcal F$-atom $A$ of positive $\mathbb P$-measure and then when $x_n$ is the value of $X_n$ on $A$, there must be a limit point $y\in[-\infty, +\infty]$ of the sequence $(x_n)_{n\in\mathbb N}$. It is then not difficult to check that $\mu(\{y\})\geq\mathbb P(A)>0$ and thus $\mu$ cannot be continuous. 
Edit: With an $\mathcal F$-atom I meant an atom of $\mathcal F$ understood as a boolean algebra. In this case this means some $A\neq\emptyset$ in $\mathcal F$ so that if $B\subsetneq A$ and $B\in\mathcal F$ then $B=\emptyset$. Such an atom $A$ with positive $\mu$-measure most exist, which can be seen as follows: Since $\Omega$ is countable  and $\mathcal F$ is a $\sigma$-algebra, for every $\omega\in\Omega$ there is a unique $\mathcal F$-atom $A_\omega$ with $\omega\in\Omega$. This can be seen by an application of Zorn's lemma for example. Consider $Z=\{B\in\mathcal F\mid \omega\in B\}$ ordered by $\supseteq$. As $\Omega$ is countable, any chain in $(Z, \supseteq)$ is countable and as $\mathcal F$ is closed under countable intersections, every chain there has an upper bound (namely the intersection over all elements of the chain). Thus there is a maximal element in $Z$, it is now easy to see that this is what we were looking for. 
Now finally, $\Omega=\bigcup_{\omega\in\Omega} A_\omega$ and as $\Omega$ is countable, we may use countable additivity of $\mu$ to see that 
$$1=\mu(\Omega)\leq\Sigma_{\omega\in\Omega}\mu(A_\omega)$$
hence there must be some $\omega$ so that $A_\omega$ has positive measure.
