Q:

Fatou's Lemma application

Above is a theorem from Durrett.  Could someone explain how the line "so Fatou's lemma applies"?

A:

To see this in full transparency, first we need to show $\liminf_{n\to\infty} 1_G(Y_n) = 1_{\liminf_{n\to\infty}\,[Y_n\in G]}$. Simplify this a bit by noting that $1_G(Y_n) = 1_{[Y_n\in G]}$. Make one more simplification by replacing the event $[Y_n\in G]$ with an arbitrary event $E_n$, and we are reduced to showing $$\liminf_{n\to\infty} 1_{E_n} = 1_{\liminf_{n\to\infty}E_n}.$$
The event $\liminf_{n\to\infty}E_n$ is defined as
$$
\liminf_{n\to\infty}\,E_n = \bigcup_{n=1}^\infty\bigcap_{k=n}^\infty E_k.
$$
So a point $\omega$ belongs to $\liminf_{n\to\infty}\,E_n$ if and only if $\omega$ eventually belongs to every $E_k$. Thus if $\omega$ belongs to $\liminf_{n\to\infty}E_n$, then $1_{E_n}(\omega) = 1$ for all $n$ sufficiently large, so $\liminf_{n\to\infty}1_{E_n}(\omega) = 1$. If $\omega\notin\liminf_{n\to\infty}E_n$, then $\omega$ belongs to infinitely many $E_n^c$. Hence $1_{E_n}(\omega) = 0$ for infinitely many $n$, so $\liminf_{n\to\infty}1_{E_n}(\omega) = 0$. Hence we have shown
$$
\liminf_{n\to\infty}1_{E_n}(\omega) =
\begin{cases}
1, & \text{if $\omega\in \liminf_{n\to\infty} E_n$}\\
0, & \text{if $\omega\notin\liminf_{n\to\infty} E_n$,}
\end{cases}
$$
so our claim is proved.
Since
$$
1_G(Y_\infty) \le \liminf_{n\to\infty} 1_G(Y_n)=1_{\liminf_{n\to\infty}[Y_n\in G]} \implies P(Y_\infty \in G) \le P\big(\liminf_{n\to\infty}\,[Y_n\in G]\big),
$$
Fatou's lemma implies that
$$
P\big(\liminf_{n\to\infty}\,[Y_n\in G]\big) \le \liminf_{n\to\infty}P(Y_n\in G).
$$

