Let $\{E_n\}$ be a collection of measurable sets. Define $$\lim_{n\rightarrow \infty} \inf E_n=\bigcup_{n=1}^{\infty} \left(\bigcap_{k=n}^{\infty} E_k\right).$$ How does one show that
$$ \mu\left(\lim_{n\rightarrow \infty} \inf E_n\right) \leq \lim_{n\rightarrow \infty} \inf ~\mu\left(E_n\right).$$
This is my attempt.
Since $\bigcap_{k=n}^{\infty} E_k\subseteq E_n~~n=1,2,3,\ldots~~,$ by monotonicity, we have $$ \mu\left(\bigcap_{k=n}^{\infty} E_k\right)\leq \mu\left(E_n\right).$$
This is all I have for now. I'll add more when I make progress.
Best Answer
This can be proven without reference to Fatou's Lemma or integration.
You had the right idea. Let $F_n = \bigcap_{k=n}^\infty E_k$, and note that $F_n \subseteq F_{n+1}$ for all $n \geq 1$. Because of this nesting, it follows** that $$\mu(\liminf E_n) = \mu\left(\bigcup_{n=1}^\infty F_n \right) = \lim_{n\to\infty} \mu(F_n).$$
On the other hand, as you noted, $F_n \subseteq E_k$ for every $k \geq n$, and so $$\mu(F_n) \leq \inf_{k \geq n}\ \mu(E_k).$$
Now take limits of both sides.
** Here, I am using the fact (which you are encouraged to prove yourself) that if $E_n \subset E_{n+1}$ for every $n \geq 1$, then $\mu(\bigcup_{n=1}^\infty E_n) = \lim \mu(E_n)$.