TITLE: Measure and limsup and liminf
QUESTION [6 upvotes]: 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!

REPLY [9 votes]: 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 $\mu ( \limsup E_n) < \limsup \mu(E_n)$.