Weak convergence of random variables Suppose we have an arbitrary probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a sequence of random variables $X_n:\Omega\to\mathbb{R}$ such that the pushforward measures $(X_n)_*(\mathbb{P})$ converge weakly to a probability $\tilde{\mathbb{P}}$. 
Can we always construct a random variable $X:\Omega\to\mathbb{R}$ such that $\tilde{\mathbb{P}}=X_*(\mathbb{P})$?
 A: If I didn't make a mistake, such a random variable must always exist.
Let $[-\infty,+\infty]$ be the set of real numbers along with the two infinities $\pm\infty$ in the order topology and with the Borel $\sigma$-algebra.  Throughout this proof, random variables will be allowed to take values in $[\infty,+\infty]$.
Terminology: Suppose that for each $n\ge 1$, $\pi_n$ is a partition of some set fixed set $V$.  We say that $\pi_n$ is an "increasingly fine" sequence iff $\pi_{n+1}$ is finer than $\pi_n$ for all $n$.
Recall the proposed proposition: Suppose $(\Omega,\Sigma,\mathbb{P})$ is a probability space, $X_n$ is a sequence of random variables, the cumulative distribution function of $X_n$ is $F_n$, and $F$ is some cumulative distribution function such that $F_n \to F$. (That is, at each point of continuity $x$ of $F$, $\lim_{n\to \infty} F_n(x) = F(x)$.) 
Claim There is a random variable $X$ on the same probability space $(\Omega,\Sigma,\mathbb{P})$ with cumulative distribution function $F$.
Proof
For each integer $n\ge 1$, let $\mathcal{I}_n$ be the finite partition of $[-\infty,+\infty]$ consisting of
the set $[-\infty,-n)$, the set $[n,+\infty]$, and the set $\left[i/2^n,(i+1)/2^n\right)$ for each integer $i\in [- 2^n n , 2^n n - 1]$. Loosely speaking, this is just a sequence of increasingly fine finite partitions that becomes arbitrarily fine at any fixed real number.
For each integer $n \ge 1$, let $\mathcal{P}_n$ be the finite partition of $\Omega$ consisting of sets of the form $X_1^{-1}(I_1)\cap \ldots \cap X_n^{-1}(I_n)$ where $I_1\in\mathcal{I}_1,\ldots,I_n\in\mathcal{I}_n$. Observe that $\mathcal{P}_n$ is a sequence of increasingly fine partitions of $\Omega$, each $\mathcal{P}_n$ is finite, and any $P\in\mathcal{P}_n$ is measurable.
We will consider the "limit" $\mathcal{P}$ of the sequence $\mathcal{P}_n$, which is the coarsest partition of $\Omega$ that is finer than any $\mathcal{P}_n$. The family $\mathcal{P}$ consists of all non-empty sets of the form
$$
(1) \ \ \ \bigcap_{n\ge 1} P_n
$$
for some sequence $P_n$ where $P_n\in\mathcal{P}_n$ for all $n$. That is,
$$
\mathcal{P} = \left\{ \bigcap_{n\ge 1} P_n : P_n\in\mathcal{P}_n  \mbox{ for all } n  \right\}\setminus\{\emptyset\}.
$$
Now let $\mathcal{A}$ consist of sets in $\mathcal{P}$ with positive measure:
$$
\mathcal{A} = \{P\in\mathcal{P}: \mathbb{P}(P) > 0\}.
$$
Loosely speaking, the sets in $\mathcal{A}$ are "atoms": nowhere in this proof will we encounter a nonempty proper subset of an atom. The number of atoms must be countable (otherwise, we'd have infinite measure), so we will write
$$
\mathcal{A} = \{A_1, A_2,\ldots\}.
$$
Next, define $A$ and $S$ as
$$
A = \bigcup_{n\ge 1} A_n
$$
and
$$
S = \Omega \setminus A.
$$
$A$ and $S$ are measurable. Basically, any random variable on $(\Omega,\Sigma,\mathbb{P})$ can be continuous on $S$ but $A$ will be responsible for jumps in the RV's cdf.
For each $n\ge 1$, construct the partition $\mathcal{S}_n$ of $S$ defined by
$$
\mathcal{S}_n = \{ S \cap P : P\in \mathcal{P}_n \}.
$$
Observe $\mathcal{S}_n$ is a sequence of increasingly fine finite partitions of $S$. Also, for any sequence $S_n$ with $S_n\in\mathcal{S}_n$,
$$
(*) \ \ \ \ \mathbb{P}\left(\bigcap_{n\ge 1} S_n \right) = 0.
$$
This is a key fact used below.  To prove it, note
$$
\mathbb{P}\left(\bigcap_{n\ge 1} S_n \right) = \mathbb{P}\left(S\cap \bigcap_{n\ge 1} P_n \right)
$$
for some sequence $P_n\in\mathcal{n}$. If $\mathbb{P}\left(\bigcup_{n\ge 1} P_n\right) = 0$, then $(*)$ obviously holds. If $\bigcup_{n\ge 1} P_n$ has positive probability, then it must be $A_k$ for some $k$, by definition. But $S \cap A_k=\emptyset$, so $(*)$ again holds.
Now that some basic constructions and notation are in place, let's start proving things!
Lemma 1. For any $A_k$, there is an $\alpha_k\in [-\infty,+\infty]$ and a subsequence $X_{m_1},X_{m_2},\ldots$ of the sequence $X_1,X_2,\ldots$ such that
$$
\lim_{n\to \infty} X_{m_n}(\omega) = \alpha_k \mbox{  for all } \omega\in A_k.
$$
Proof. Fix $A_k$ and choose a single $x_n\in X_n(A_k)$ for each $n$. Since $[-\infty,+\infty]$ is compact, the sequence $x_1,x_2,\ldots$ must have a limit point $\alpha_k$ in $[-\infty,+\infty]$. Thus, there is a subsequence $x_{m_1}, x_{m_2},\ldots$ such that
$$
(2)\ \  \lim_{i\to\infty} x_{m_i} = \alpha_k.
$$
Recall from $(1)$ that there are sets $I_1\in\mathcal{I}_1,I_2\in\mathcal{I}_2,\ldots$ such that
$$
A_k = \bigcap_{n\ge 1} X_n^{-1}(I_n),
$$
so $x_n\in X_n(A_k)\subset I_n$ for all $n$. Suppose $\alpha_k$ is real and fix some $\omega\in A_k$.  By $(2)$ and the construction of sets in $\mathcal{I}_n$, there has to be an $N$ such that $I_n$  is an interval of length $2^{-n}$ for all $n>N$. Thus,
$$
|X_{m_n}(\omega) - x_{m_n}| \le 2^{-m_n}
$$
for $n$ big enough. In light of this fact and (2), we have $X_{m_n}(\omega) \to \alpha_k$.
A similar argument shows the same limit holds if $\alpha_k\in\{\pm\infty\}$.
Lemma 2. There are constants $\alpha_k\in[-\infty,+\infty]$ for $k\ge 1$ and a subsequence $X_{m_1},X_{m_2},\ldots$ of $X_1,X_2,\ldots$ such that for all $k$,
$$
\lim_{n\to\infty} X_{m_n}(\omega) = \alpha_k \mbox{  for all  } \omega\in A_k.
$$
Proof Use the previous lemma and a diagonalization argument. First choose a subsequence $X_{11}, X_{12},\ldots$ of $X_1,X_2,\ldots$ such that $X_{1n}(\omega)\to\alpha_1$ for some $\alpha_1$ and all $\omega\in A_1$. Then choose a subsequence $X_{11}, X_{22}, X_{23}, \ldots$ of $X_{11},X_{12},\ldots$ such that
$X_{2n}(\omega)\to\alpha_2$ for some $\alpha_2$ and all $\omega\in A_2$. At stage $k$, choose a subsequence $X_{11}, X_{22},\ldots, X_{kk},X_{k,k+1},\ldots$ such that $X_{kn}(\omega)\to \alpha_k$ for all $\omega\in A_k$. The subsequence $X_{11},X_{22},X_{33},\ldots$ is well-defined and has the desired property.
Important From now on, we will assume without loss of generality that for all $k$,
$$
(3) \ \ \ \ \ \lim_{n\to\infty} X_n(\omega) = \alpha_k \mbox{  for all  } \omega\in A_k.
$$
We will now begin to construct a random variable $X$ with cumulative distribution function $F$. To do so, first define the following functions:
\begin{equation}
(4) \ \ \ F_A(t) = \sum_{n\ge 1,\ \ \alpha_n \le t} \alpha_n \\
(5) \ \ \ F_S(t) = F(t) - F_A(t)
\end{equation}
for all $t\in[-\infty,+\infty]$. We will construct random variables $X_A$ and $X_S$ such that
$$
(6) \ \ \ \ \mathbb{P}(A \cap X_A \le t) = F_A(t) \\
(7) \ \ \ \ \mathbb{P}(S \cap X_S \le t) = F_S(t).
$$
At that point, the proof will be complete, since we can let $X = 1_A X_A + 1_S X_S$. The cumulative distribution
function of $X$ will then be $F_A + F_S = F$.
Let $X_A$ be a random variable with
\begin{equation}
X_A(\omega) = \alpha_k \mbox{  for all } k \mbox{ and for all } \omega\in A_k.
\end{equation}
The value of $X_A$ on $S$ is unimportant.  One can set $X_A(S) = 0$, say, to ensure $X_A$ is measurable.
By construction of $X_A$, condition $(6)$ holds. Now now consider $X_S$.
Lemma 3  $F_S$ is increasing, right continuous, $F_S(-\infty) \ge 0$, and $F_S(\infty) = \mathbb{P}(S) $.
Proof
By definition of $F_S$ and $F_A$, $F_S(-\infty) = 0$ and $F_S(+\infty) = \mathbb{P}(S)$. Consider $F(t)$. Since $X_n$ converges in distribution to the cumulative distribution function $F$, we can write
$$
F(t) = \lim_{\epsilon\downarrow 0}\liminf_n \mathbb{P}(X_n \le t+\epsilon).
$$
(Ask me why if you're not sure.) We can then write
$$
F(t) = \lim_{\epsilon\downarrow 0}\liminf_n \mathbb{P}(A \cap X_n\le t+\epsilon) +
\lim_{\epsilon\downarrow 0}\liminf_n \mathbb{P}(S \cap X_n\le t+\epsilon).
$$
By $(3)$ and $(4)$, the first term is exactly $F_A(t)$. Thus,
$$
F_S(t) = \lim_{\epsilon\downarrow 0}\liminf_n \mathbb{P}(S \cap X_n\le t+\epsilon).
$$
This function is obviously increasing and right continuous.
I will use the following Lemma, which is really pretty easy to prove by direct construction. The proof is a little too tedious for me to include here. If you need more info, let me know.
Lemma 4  Suppose $U\subset\Omega$ is measurable and $\pi_1,\pi_2,\ldots$ is a sequence of increasingly fine finite partitions of $U$.  Suppose for each $\epsilon>0$, there is an $n$ such that $\mathbb{P}(B)<\epsilon$ for all $B\in\pi_n$.  Suppose $F:[-\infty,+\infty]\to[0,1]$ is increasing, right continuous, and $F(+\infty)=\mathbb{P}(U)$. Then there is a random variable $X$ on $(\Omega,\Sigma,\mathbb{P})$ such that $\mathbb{P}(U \cap X\le t) = F(t)$ for all $t\in[-\infty,+\infty]$.
Now apply Lemma 4 to construct $X_S$:
Lemma 5  There is a random variable $X_S$ such that $(7)$ holds, i.e., $\mathbb{P}(S \cap X_S \le t) = F_S(t)$.
Proof We will apply Lemma 4 with the set $S$, the sequence of partitions $\mathcal{S}_n$, and the function $F_S$. By Lemma 3, the conditions on $F_S$ required by Lemma 4 are met. We already observed $\mathcal{S}_n$ is a sequence of increasingly fine finite partitions. It only remains to prove that for any $\epsilon>0$, there is an $n$ such that $\mathbb{P}(B)<\epsilon$ for all $B\in\mathcal{S_n}$.
Fix $\epsilon>0$. For each $B\in\mathcal{S}_n$, let $c(B)$ represent the number of elements in the set
$$
\{B'\in\mathcal{S}_{n+p} : B' \subset B,\ \ \mathbb{P}(B')>\epsilon, \ \ \ \ p\ge 0\}.
$$
Informally, $c(B)$ is the number of elements "below" $B$ in the sequence of partitions $\mathcal{S}_n$ with probability greater than $\epsilon$.  Observe $c(B)=0$ if and only if $\mathbb{P}(B)\le\epsilon$. Also,  for any set $B\in\mathcal{S}_n$, if $\mathbb{P}(B) > \epsilon$, then
$$
(8)\ \ \ c(B) = 1 + \sum_{B'\in\mathcal{S}_{n+1},\ \ B'\subset B} c(B').
$$
Fix some $B_0\in\mathcal{n}$ and suppose $c(B_0)=+\infty$.  By $(8)$, there is a $B_1\in\mathcal{S}_{n+1}$ with $B_1\subset B_0$ and $c(B_1)=+\infty$. Continuing this process, we see there is a sequence $B_0 \supset B_1 \supset B_2 \supset\cdots$ such that $B_p\in\mathcal{S}_{n+p}$ and $\mathbb{P}(B_p) > \epsilon$ for all $p\ge 0$. But then
$$
\mathbb{P}\left(\bigcap_{p\ge 0} B_{n+p}\right) \ge \epsilon > 0.
$$
This contradicts $(*)$.  Thus, $c(B)$ is finite for all $B\in \mathcal{S}_n$. By $(8)$, it is clear that there is an $n$ with $c(B)=0$ for all $B\in\mathcal{S}_n$. This proves the result.
