Measure and limsup and liminf 
Given a ring of set $R$, $\mu$ a measure on $R$ and if $\{E_n\}$ is a sequence of sets in $R$ for which
$\bigcap_{i=n}^{\infty}E_i$ for $n=1,2,...$ and $\lim \inf E_n \in R$ then $\mu(\lim \inf E_n)\leq \lim \inf \mu(E_n)$
$\bigcup_{i=n}^{\infty} E_i$ for $n=1,2,...$ and $\lim \sup E_n \in R$ then $\mu(\lim \sup E_n)\geq \lim\sup  \mu(E_n)$

How can I prove these results? I don't know how to begin. I know that it is a direct use of the definition but I only know how to work with $\lim\sup$ for set point by point, not as a whole.
When I posted this question I had no idea about how to begin the problem. Now, I have resolved my doubts (someone in this forum give me an advice), but I can not post what I have done if nothing have been done, not because I was not thinking in the problem, but because nothing come to my mind to solve it.
In this kind of cases (in which you really don't have any idea about how to attack the problem) I am not sure how to post here. Since I need a hint, but could happen someone thinks that I want that others solve my homework (which is not my case).
Thanks!
 A: Defining $F_n=\bigcap_{i=n}^\infty E_i$ we can see that $F_1\subset F_2\subset \cdots $ then $\{F_n\}$ is an incresing sequence of set. Which implies that 
$$\mu(\lim_n F_n)=\lim_n \mu(F_n)  $$
by continuity of the measures. But 
$$\lim_n F_n=\bigcup_{n=1}^\infty F_n=\lim\inf E_n $$
This is an useful result. Since $\bigcap_{i=n}^\infty E_n\subset E_k$ for $k\geq n$ which implies that $\mu(\bigcap_{i=n}^\infty E_n)\leq \mu(E_k)$ then $\mu(\bigcap_{i=n}^\infty E_n)\leq \inf\{\mu(E_k): k\geq n\}$ taking $n\rightarrow \infty$ and using the result given above we get $\mu(\lim\inf E_n)\leq \lim\inf \mu(E_n)$. 
In analoguos way using the continuity from above measures we can get the other result. 
A: The second statement is false.
Consider the sequence $E_n = [n-1,n)$ in $\mathbb{R}$ with the Lebesgue measure. Every non-negative real number is in one and only one of the $E_n$, so $\limsup E_n = \emptyset$. However, we have $\mu(E_n) = 1$ for all $n$, so $\limsup \mu(E_n) =1$. Therefore $\mu ( \limsup E_n) < \limsup \mu(E_n)$.
