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?

$\left(\bigcap_{j=i}^\infty E_j\right)_{i=1}^\infty$ 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 $\mu\left(\bigcap_{j=i}^\infty E_j\right) \le \mu(E_j)$.
