Explanation of this group of equalities of lim inf and lim sup Let $(X, A, u)$ be a finite measure space ($u(X) < \infty$, e.g. $u$ is a probability measure) and $(E_n)$ a sequence of sets in A. Then
$u$(lim inf $E_n$) $\le$ lim inf $u(E_n) \le$ lim sup $u(E_n) \le u$(lim sup $E_n$)
So $u$(lim inf $E_n$) is measure of the lim inf of the sequence of sets $(E_n)$.
But I don't get what meant by: lim inf $u(E_n)$
How do you take the measure of a sequence of sets $E_n$....Is it possible to have such a thing?
Or ios it is supposed  to means $u(E_n)$ is a sequence of real numbers which we are taking the lim inf of. Is that correct?
 A: There are two different notions: the limes inferior of a real-valued sequence and the limes inferior of a sequence of sets.
Let $(u_n)_{n \in \mathbb{N}}$ a sequence in $\mathbb{R}$. Then
$$\liminf_{n \to \infty} u_n := \lim_{n \to \infty} \bigg( \inf \{u_k; k \geq n\} \bigg)$$
This means that $\liminf_{n \to \infty} u_n$ equals the smallest accumulation point of the sequence.
On the other hand, for a sequence of sets $(E_n)_{n \in \mathbb{N}}$ we have
$$\liminf_{n \to \infty} E_n := \bigcup_{n \in \mathbb{N}} \bigcap_{k \geq n} E_k$$
Therefore, $\liminf_{n \to \infty} E_n$ consists of all points which are contained in all but finitely many sets $E_n$, $n \in \mathbb{N}$.
Using these definition, it's not difficult to show that
$$1_{\liminf_{n \to \infty} E_n}(x) = \liminf_{n \to \infty} 1_{E_n}(x)$$
where
$$1_E(x) = \begin{cases} 1 & x \in E \\ 0 & x \notin E \end{cases}$$
denotes the indicator function of a set $E$. This turns out to be quite useful for the first inequality you have to prove.
A: If $\{E_n\}$ is a sequence of sets then $$\liminf E_n = \bigcup_{n=1}^\infty \bigcap_{k \ge n} E_k.$$ If each $E_n \in A$ then $\liminf E_n \in A$ too, so that $u(\liminf E_n)$ makes sense. On the other hand, $\liminf u(E_n)$ is just the usual numerical liminf of the sequence $\{u(E_n)\}$. 
