I am trying to understand the proof of Thm 3.2.11 in Durrett's Probability Theory and Examples (5th edition), however, I cannot understand (i) -> (ii)
I can understand that the (baby) Skorokhod's theorem gives $Y_n$ converges almost surely to $Y_\infty$ having the same distribution as $X_n$ and $X$ respectively if $X_n$ converges in distribution to $X_\infty$.
In this situation, how the limit supremum of $1_G(Y_n)$ is larger than $1_G(Y_\infty)$? I learned that almost sure convergence does not mean point-wise convergence. Does the openness of $G$ playing an important role?
Best Answer
$\lim \inf 1_G(Y_n(\omega)) \geq 1_G(Y_{\infty}(\omega))$ for any $\omega$ for which $Y_n(\omega) \to Y_{\infty} (\omega)$: Fix any such $\omega$ and suppose $1_G(Y_{\infty}(\omega))=1$. Then $Y_{\infty}(\omega) \in G$ and $G$ is open. Hence, $Y_{n}(\omega) \in G$ for all $n$ sufficiently large which gives $\lim \inf 1_G(Y_n(\omega)) \geq 1$. In the case when $1_G(Y_{\infty}(\omega))=0$ the inequality trivially holds. Hence $\lim \inf 1_G(Y_n) \geq 1_G(Y_{\infty}))$ almost surely. Note that $E(\lim \inf 1_G(Y_n)) \geq E(1_G(Y_{\infty})))=P(Y_{\infty} \in G)$ and Fatou's Lemma now gives $\lim \inf E(1_G(Y_n) \geq E(1_G(Y_{\infty})=P(Y_{\infty} \in G)$. This gives i) implies ii).