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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?

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2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – L.Z.
    Commented Dec 10, 2020 at 18:17
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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 $n\to \infty.$

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  • $\begingroup$ 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 $\mu_n$. $\endgroup$
    – L.Z.
    Commented Dec 5, 2020 at 4:17
  • $\begingroup$ 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. $\endgroup$
    – Raghav
    Commented Dec 5, 2020 at 5:06
  • $\begingroup$ 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 $X_n \Rightarrow X$. $\endgroup$
    – L.Z.
    Commented Dec 5, 2020 at 7:18

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