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

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.

• Just consider the cases when RHS is $0$ and when RHS is $1$. Nov 20, 2022 at 4:33

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 $$\Omega_{0}$$.