Limit inferior of the Lebesgue measure 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.
 A: 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 $\mu(\bigcup_{n=1}^\infty E_n) = \lim \mu(E_n)$.
A: Note that 
$1_{\liminf_n E_n}=\liminf_n 1_{E_n},$
where $1_{E_n}$ is the characteristic function of $E_n$. Therefore this is precisely Fatou's lemma applied to the sequence of functions $(1_{E_n})_n$.
A: Well, it is simplier than Fatou's Lemma and can be proved in a simplier way.
The only fact that we will use is The Continuity Theorem for Increasing Sets/Events. It says that when the sets are bigger and bigger (included in each other), then the measure of the limit is the limit of the measures.
Notice, that when you analyse a sequence of finite sums instead of the infinite one, you get bigger and bigger sets -  that is what we call 'increasing sets'. Such a finite sum is actually its last element (think about it for a minute if it is not clear immediately).
$S_N=\bigcup_{n=1}^{N} \left(\bigcap_{k=n}^{\infty} E_k\right)=\bigcap_{k=N}^{\infty} E_k$
$\mu(S_N) \le \mu(E_N)$ (why?)
by taking $\liminf$ on both sides, noticing that $\lim=\liminf$ on the left and using the continuity theorem you finish the proof.
EDIT: too late:)
