Supremum and Infimum of measures of collection of sets edit: I am using |*| as notation for measure, and all sets are assumed to be measurable.
Define $\liminf_{k \rightarrow \infty}E_k= \cup_{k=1} \cap_{j=k} E_j$ and $\limsup_{k \rightarrow \infty}E_k= \cap_{k=1} \cup_{j=k} E_j$. 
I'm wondering if there is an example where $\liminf_{k \rightarrow \infty}E_j$ is a proper subset of $\limsup_{k \rightarrow \infty}E_j$.
Also, I'm trying to prove the inequality $|\liminf_{k \rightarrow \infty}E_k| \leq \liminf_{k \rightarrow \infty}|E_k|$
I think this is an interesting inequality, because the right hand is asking for the infimum of a sequence of positive numbers, where the left is the measure of a definition of infimum defined as a set operation.
Any advice is greatly appreciated! Thanks everyone!!
 A: What if$$E_n=\begin{cases}\{\text{odd natural numbers}\}&\text{ if $n$ if odd}\\\{\text{even natural numbers}\}&\text{ if $n$ if even?}\end{cases}$$Then $\liminf_{n\to\infty}E_n=\emptyset$ and $\limsup_{n\to\infty}E_n=\mathbb N$.
A: The inequality is trivial from the definition of $\liminf|E_j|$. Say $$I_k=\bigcap_{j\ge k}E_j.$$ Since $I_k\subset E_j$ for $j\ge k$  we have $$|I_k|\le\inf_{j\ge k}|E_j|.$$And now since $I_k\subset I_{k+1}$ it follows that $$|\bigcup_k I_k|=\lim_k|I_k|\le\lim_k\inf_{j\ge k}|E_j|=\liminf|E_j|.$$
A: The lemma of Fatou tells us that for integrals wrt any measure $\mu$:$$\int\liminf f_n\leq\liminf\int f_n$$ where the $f_n$ are non-negative measurable functions. 
Taking for $f_n$ the indicator function of set $E_n$ and for $\mu$ the the measure in your question this results in:$$|\liminf E_n|\leq\liminf|E_n|$$
A: Let $E_n=(n,n+1)$ when $n$ is odd and $E_n=(0,1)$ when $n$ is even. Then $\lim \inf E_n=\phi$ and $\lim \sup E_n=(0,1).$ And $|\lim \inf E_n|=0$ but $|E_n|=1$ for all $n$. 
If you want all the $E_n$ to be distinct, amend this by letting $E_n=(-1/n,1)$ when $n$ is even.
Remark: $x\in \lim \inf E_n$ iff $\{n\in \Bbb N: x\not \in E_n\}$ is finite, i.e. $x$ belongs to $E_n$ for "almost all" $n.$ And  $x\in \lim \sup E_n$ iff $\{n\in \Bbb N: x\in E_n\}$ is infinite, i.e. $x$ belongs to $ E_n$ for infinitely many $n$. 
