If $Y_n$ are random variables and $Y_n \to Y_\infty$ a.s. and $G$ is an open set, then $\lim \inf_{n \to \infty} 1_G(Y_n) \geq 1_G(Y_\infty)$

Question

From Rick Durret's book "Probability: Theory and Examples," 5th edition, chapter 3, on Theorem 3.2.11 in the course of proving that if $X_n$ are random variables converging weakly to $X_\infty$, then for all open sets $G$, $\lim \inf \mathbb P(X_n \in G) \geq \mathbb P(X_\infty \in G)$, he mentions the following:

Let $Y_n$ have the same distribution as $X_n$ and $Y_n \to Y_\infty$ almost surely. Since $G$ is open, $\lim \inf_{n \to \infty} 1_G(Y_n) \geq 1_G(Y_\infty)$.

However, I am having trouble understanding why this has to be true, and why the fact that $G$ is open has anything to do with this.

Answer 1

Let $\Omega_{0}=\{\omega\in\Omega\mid Y_{n}(\omega)\rightarrow Y_{\infty}(\omega)\}$. By assumption, $P(\Omega_{0})=1$. Let $\omega\in\Omega_{0}$. If $Y_{\infty}(\omega)\notin G$, we have that $1_{G}(Y_{\infty}(\omega))=0$. Since $1_{G}(Y_{n}(\omega))\geq0$ for each $n$, it follows that $\liminf_{n\rightarrow\infty}1_{G}(Y_{n}(\omega))\geq1_{G}(Y_{\infty}(\infty))$. If $Y_{\infty}(\omega)\in G$, then there exists $N$ such that $Y_{n}(\omega)\in G$ whenever $n\geq N$ (because $G$ is open and $Y_{n}(\omega)\rightarrow Y_{\infty}(\omega)$). Hence, for each $n\geq N$, $\inf_{k\geq n}1_{G}(Y_{k}(\omega))=1$ because $1_{G}(Y_{k}(\omega))=1$ for each $k\geq n$. It follows that $\liminf_{n\rightarrow\infty}1_{G}(Y_{n}(\omega))=\lim_{n\rightarrow\infty}\inf_{k\geq n}1_{G}(Y_{k}(\omega))=1=1_{G}(Y_{\infty}(\omega))$.

This shows that $\liminf_{n\rightarrow\infty}1_{G}(Y_{n})\geq1_{G}(Y_{\infty})$ on $\Omega_{0}$.