# Weak Convergence - similar to continuous mapping theorem

I recently revisited the following textbook exercise (Probability and Measures by Billingsley, Problem 25.8 in the third edition):

Let $$X_n$$ and $$X$$ be real-valued random variables and $$h_n, h: \mathbb{R} \to \mathbb{R}$$ be Borel functions. Let $$E$$ be the set of $$x \in \mathbb{R}$$ such that $$h_n(x_n) \to h(x)$$ fails for some sequence $$x_n \to x$$. Suppose that $$X_n$$ converges to $$X$$ weakly (a.k.a. in distribution), denoted by $$X_n \Rightarrow X$$, and that $$E$$ is measurable with $$P(X\in E) = 0$$. Show that $$h_n(X_n) \Rightarrow h(X)$$.

This exercise can be solved quite easily using Skorohad's theorem: on some probability space, there exist real-valued random variables $$Y_n$$ and $$Y$$ with the same distribution as $$X_n$$ and $$X$$ satisfying $$Y(\omega) \to Y(\omega)$$ pointwise.

But now I want to prove it without using Skorohad's theorem. I wonder if I can do so by using the common characterization in Portmanteau theorem: $$\limsup_n \mu(C) \to \mu(C)$$ for any closed set $$C$$ if and only if $$\mu_n \Rightarrow \mu$$.

My efforts:

Let $$\mu_n:= P \circ X_{n}^{-1}$$ and $$\mu:= P \circ X^{-1}$$ denote the probability measures induced by $$X_n$$ and $$X$$. Also, we let $$\nu_n := P \circ X_n^{-1} \circ h_n^{-1}$$ and $$\nu:= P \circ X^{-1} \circ h^{-1}$$ denote the probability measures induced by $$h_n(X_n)$$ and $$h(X)$$. The goal is to show $$\nu_n \Rightarrow \nu$$.

Let $$C$$ be an arbitrary closed subset of $$\mathbb{R}$$. Then we have $$\limsup_n \nu_n(C) = \limsup_n \mu_n (h_n^{-1}(C) ) \leq \limsup_n \mu_n( \overline{ h_n^{-1}(C)} ),$$ where $$\overline{ h_n^{-1}(C)}$$ denote the closure of $$h_n^{-1}(C)$$. I would like to use Portmanteau theorem on $$\mu_n \Rightarrow \mu$$, but the set $$\overline{ h_n^{-1}(C)}$$ also depends on $$n$$. Is there a way to bypass this issue?

Here is a proof using Portmanteau. It is adapted from Billingsley's Convergence of Probability Measures.

Claim: Let $$G$$ be an open set. Then $$h^{-1}(G) = E \cup \bigcup_k \operatorname{int}(T_k)$$ where $$\operatorname{int}(A)$$ denotes the interior of set $$A$$ and we define $$T_k:=\bigcap_{i\ge k}\left\{ x:h_i(x)\in G\right\}.\tag1$$

Proof: Suppose $$x\in h^{-1}(G)$$, i.e., $$h(x)\in G$$. Since $$G$$ is open, there exists $$\epsilon>0$$ such that $$\text{y\in G whenever |y-h(x)|<\epsilon}.\tag2$$ Suppose also that $$x$$ is not in $$E$$. By definition of $$E$$, this means $$\text{\exists k\exists\delta>0 such that |h_i(x')-h(x)|<\epsilon whenever i\ge k and |x'-x|<\delta.}\tag3$$ By (2), this implies $$\text{\exists k\exists\delta>0 such that h_i(x')\in G whenever i\ge k and |x'-x|<\delta.}$$ Equivalently, $$\text{\exists k\exists\delta>0 such that x'\in T_k whenever |x'-x|<\delta,}$$ which is to say there exists $$k$$ such that $$x$$ is an interior point of $$T_k$$.

To finish the proof, let $$G$$ be open. We show $$\liminf P(h_n(X_n)\in G)\ge P(h(X)\in G)$$. By the Claim, $$P(h(X)\in G)\le P(X\in E) + P\left(\bigcup_k \left\{X\in \operatorname{int} (T_k)\right\}\right).$$ The first term on the RHS is zero, while $$P\left(\bigcup_k \left\{X\in \operatorname{int} (T_k)\right\}\right)=\lim_k P(X\in \operatorname{int}(T_k))\tag a$$ by continuity of $$P$$ from below (note the $$\operatorname{int}(T_k)$$ are increasing). Let $$\epsilon>0$$. By (a), there exists a (large) $$k$$ for which $$P(h(X)\in G)\le P(X\in \operatorname{int}(T_k)) +\epsilon.\tag b$$ Portmanteau implies $$P(X\in \operatorname{int}(T_k))\le\liminf _nP(X_n\in\operatorname{int}(T_k)).\tag c$$ But $$\operatorname{int}(T_k)\subset h_n^{-1}(G)$$ for all large $$n$$, by definition (1) of $$T_k$$. Putting it all together we've shown $$P(h(X)\in G)\le \liminf P(h_n(X_n)\in G)+\epsilon.$$ Since $$\epsilon>0$$ is arbitrary, this completes the proof.

• I guess the key idea here is to consider the set like $\{h_i \in G \ ev.\}$. Then we modify it by taking interior inside the union. It is a clever idea.
– L.Z.
Commented Dec 10, 2020 at 18:17

First of all, recall that weak convergence is equivalent to $$\int f\;d\mu_n\to \int f\;d\mu$$ for every bounded Lipschitz function $$f.$$ With this in mind, our goal reduces to proving that for every bounded Lipschitz function $$f,$$ we have $$\int f\circ h_n d\mu_n\to f\circ h\;d\mu.$$ With some simple algebra, it can be reduced to proving that $$\int (f\circ h_n-f\circ h)d\mu_n\to 0.$$

To this end, we use Erogoff's theorem. Fix $$\epsilon>0,$$ and let $$A$$ be a closed set such that $$h_n\to h$$ uniformly on $$A$$ and $$P(A^c)<\epsilon/||f||_{\infty}.$$ Now note that \begin{align} \int |f\circ h_n-f\circ h|d\mu_n &=\int_A |f\circ h_n(X_n)-f\circ h(X_n)|d\mathbb{P}+ \int_{A^c}|f\circ h_n(X_n)-f\circ h(X_n)|d\mathbb{P}\\ &\le ||f||_{\text{Lip}}||h_n-h||_{\infty, A}\mathbb{P}(A)+2||f||_{\infty}\mathbb{P}(A^{c})\\ &\le ||f||_{\text{Lip}}||h_n-h||_{\infty, A}+2\epsilon. \end{align} Since $$h_n\to h$$ unfiromly on $$A,$$ the norm $$||h_n-h||_{\infty, A}\to 0$$ as $$n\to \infty.$$

• I agree that $h_n \to h \ \mu$-a.s. so that we can apply Egorov's theorem to obtain a set $A \subset \mathbb{R}$ with $\mu(A)$ small. But your proof seems to claim that $\sup_n \mu_n(A)$ can also be shown to be small, which is not obvious to me. This is kind of similar to my original question about how to control a set over the sequence of measures $\mu_n$.
– L.Z.
Commented Dec 5, 2020 at 4:17
• That can be done, but I am not using that (directly) in my proof. Precisely for this reason, I change only work with the reference probability measure $P.$ And I have chosen $A$ such that its complement has small $P$ measure. Commented Dec 5, 2020 at 5:06
• I see your intention, but do we know $X_n(\omega) \in A$ for all $\omega \in \{X \in A\}$? By Skorohod's theorem, we can assume this is true. Nevertheless, I don't think it is always true for general $X_n \Rightarrow X$.
– L.Z.
Commented Dec 5, 2020 at 7:18