liminf inequality in measure spaces

Question

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?

Answer 1

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

Answer 2

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