liminf inequality in measure spaces 
Let $(X;\mathscr{M},\mu)$ be a measure space and $\{E_j\}_{j=1}^\infty\subset \mathscr{M}$. Show that $$\mu(\liminf E_j)\leq \liminf \mu(E_j)$$
  and, if $\mu\left(\bigcup_{j=1}^\infty E_j\right)<\infty$, that
  $$\mu(\limsup E_j)\geq \limsup \mu(E_j).$$

I'm trying to parse what's going on.  On the left, we're taking the measure of $\liminf E_j$, which is $\cup_{i=1}^\infty\cap_{j=i}^\infty E_i$.  This is the union of the tails... okay.
On the right, we've got $\lim_{n\to\infty}\inf\{\mu(E_j):n\leq j\}$.  The smallest $\mu$ for everything after $n$ (or the greatest lower bound, anyway).
I can't make any progress, I've been stuck here for quite a while. I just don't know where to make the comparison. Can I get a nudge?
 A: $\left(\bigcap_{j=i}^\infty E_j\right)_{i=1}^\infty$ is an increasing sequence of sets, so you may have a theorem that states that $$\mu\left(\bigcup_{i=1}^\infty \bigcap_{j=i}^\infty E_j\right) = \lim_{i \to \infty} \mu\left(\bigcap_{j=i}^\infty E_j\right).$$ Then, note that $\mu\left(\bigcap_{j=i}^\infty E_j\right) \le \mu(E_j)$.
A: Apply Fatou's Lemma to the characteristic functions of these sets.
More precisely, use the notation $\chi_A$ to denote the characteristic function of the set $A$.
Then we have that $\chi_{\liminf E_j} = \liminf \chi_{E_j}$. You should check this (the proof is short).
Thus we have by Fatou's Lemma,
$\mu(\liminf E_j) = \int \chi_{\liminf E_j} d \mu = \int \liminf \chi_{E_j} d \mu \leq \liminf \int \chi_{E_j} d \mu = \liminf \mu(E_j)$
For the second inequality, the idea is similar, but first prove and apply the reverse of Fatou's Lemma: http://en.wikipedia.org/wiki/Fatou%27s_lemma#Reverse_Fatou_lemma
A: Assume that for any sequences $a_n$, $b_n$ in $[0,\infty]$, the following are known:

*

*$0 \leq \lim\inf_{n \to \infty}  a_n \leq \infty,$

*$\lim\inf_{n \to \infty}  a_n \,\,\,+\,\,\, \lim\inf_{n \to \infty}  b_n \,\,\,\leq\,\,\,\lim\inf_{n \to \infty}  (a_n + b_n).$
Because $\cap_{j \geq 1} E_j \,\,\subset\,\, \cap_{j \geq 2} E_j \,\,\subset\,\, \cap_{j \geq 3} E_j \,\,\subset\,\, \dots \,\,$ is a growing sequence of sets, use the well-known result:
$$\mu\left( {\lim\inf} \, E_i\right) \,\,=\,\,\mu\left(\bigcup_{i \geq 1} \, \bigcap_{j \geq i} E_j\right)  \,\,=\,\, \lim_{i \to \infty} \mu\left(\bigcap_{j \geq i} E_j\right).$$
Now our goal is to show:
$$\lim_{i \to \infty} \mu\left(\bigcap_{j \geq i} E_j\right) \,\,\leq\,\,  {\lim\inf} \,\, \mu\left(E_i\right).$$
This we can show as follows. The last equality is by the additivity of measure.
$$\lim_{i \to \infty} \mu\left(\bigcap_{j \geq i} E_j\right) \,\,=\,\,  \lim\inf \mu\left(\bigcap_{j \geq i} E_j\right)$$
$$\leq \,\, \lim\inf \mu\left(\bigcap_{j \geq i} E_j\right) \,\,+\,\,  \lim\inf \mu\left(E_i \setminus \bigcap_{j \geq i} E_j\right)$$
$$\leq \,\, \lim\inf \left( \, \mu\left(\bigcap_{j \geq i} E_j\right) \,\,+\,\, \mu\left(E_i \setminus \bigcap_{j \geq i} E_j\right) \, \right)$$
$$= \,\, {\lim\inf} \,\, \mu\left(E_i\right).$$
For the other part we can reason similarly, but relying also on the following.
for any sequences $a_n$, $b_n$ in $(-\infty,\infty)$:

*

*$\lim\sup_{n \to \infty}  (-a_n) \,\,=\,\, -\lim\inf_{n \to \infty} a_n$

*$\lim\sup_{n \to \infty}  a_n \,\,\,+\,\,\, \lim\sup_{n \to \infty}  b_n \,\,\,\geq\,\,\,\lim\sup_{n \to \infty}  (a_n + b_n).$
