# Limit of pushforward measures of random variables is “represented” by a random variable

Suppose we have an arbitrary probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a sequence of real 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 $\tilde{\mathbb{P}}=X_*(\mathbb{P})$? (i.e. such that $X_n\to X$ in distribution?)

Yes I think so. We may assume that $\mathcal F$ is the $\sigma$-algebra generated by sets of the form $X_n^{-1}(a,\infty)$ as $n$ runs over the positive integers and $a$ runs over the reals as $\mathcal F$ already contains these sets. In so doing, we might be making $\mathcal F$ smaller, but that only makes the problem harder.

Let $\mathcal P_1,\mathcal P_2,\mathcal P_3,\ldots$ be a sequence of refining finite partitions of $\mathbb R$ such that the intersection of an element of $\mathcal P_n$ for each $n$ consists of at most one point. For example, $\mathcal P_n$ could consist of intervals $[n,\infty)$, $(-\infty,-n)$ and half-open dyadic intervals of length $1/2^n$ between $-n$ and $n$.

Then let $\mathcal F_n$ be the finite sub-algebra of $\mathcal F$ generated by the sets $X_i^{-1}A$ for $1\le i\le n$ and $A$ running over $\mathcal P_n$. Write $\mathcal F_n=\{B^n_1,\ldots B^n_{k_n}\}$. Now there is a measurable map $\Phi$ from $\Omega$ to $\Xi=\prod_{n=1}^\infty\{1,\ldots,k_n\}$ sending $\omega$ to sequence of partition elements that it lies in.

Equip $\Xi$ with the lexicographic ordering. We can then check that $\mathbb P$ induces a measure $\mu$ on $\Xi$. There are at most countably many atoms in $\Omega$ with respect to $\mathcal F$. Let the set of atoms be $A$. We can map each of these atoms disjointly to atoms of the same mass (at least) in the limit measure $\tilde {\mathbb P}$ (some justification needed here, but I'm pretty confident).

First we couple the atoms in both $\tilde{\mathbb P}$ and $\Omega$. That is we define $X$ restricted to the atoms. To finish, use quantile coupling to couple what's left. Given an $\omega$, set $t(\omega)=\mu(\{\omega'\in \Omega\setminus A\colon \Phi(\omega')\le \Phi(\omega)\})$ and finally, define for $\omega$ not in an atom, $X(\omega)=\inf\{s\colon \big(\tilde{\mathbb P}(-\infty,s])-\sum\text{(atoms of$\tilde{\mathbb P}\le s$)} \big)\ge t\}$.

• Basically what you said can be recast as follows: first, there are countably many atoms in $\mathcal F,P$ and we can pass to a subsequence of $X_k$ converging almost surely on each atom, which will imply that the push-forwards also converge weakly to some part of the limiting measure. After that, we are left with non-atomic measure and $\sigma$-field, which can carry a function with any distribution of equal total mass, so you choose one that fits ignoring $X_k$ completely. Neat! Note that I made the same mistake as Ofer and thought that we can change the probability but not the set. – fedja Oct 19 '13 at 12:55
• I like the idea of passing to a subsequence taking care of the atoms. I think this simplifies things a bit. – Anthony Quas Oct 19 '13 at 18:43
• Very nice! Thanks to fedja too, for his idea. When you talk about atoms, you are identifying atoms such that their symmetric difference is a null set, right? – Mizar Oct 19 '13 at 22:36
• Yeah, it's indeed always possible to construct such an RV. I posted a thorough solution over at math.stackexchange.com/questions/519999/…. Ultimately, very similar ideas to what you wrote about. And yes, I passed to a subsequence to ensure convergence on atoms.Great question. I learned a lot proving it. – Will Nelson Oct 20 '13 at 19:06