Lebesgue-measurable subsets of $[0, 1]$ Let $A_1, A_2,\cdots$ be Lebesgue-measurable subsets of $[0, 1]$ of measure $1/2$, and let $A$ denote the set of points $x$ such that $x$ belongs to infinitely many of the sets $A_n$. How we can show that the Lebesgue measure of $A$ is at least $1/2$. 
 A: Take $B=\cap _k \cup_{n\geq k} A_k $
Any x is present in an infinite number of $A_n$s, if and only if they are present in any $B_k=\cup_{n\geq k} A_k $ for all $n$. (You should be able to show both directions, or, see lim sup and lim inf of sequence of sets..)
Therefore, $\mu(B)=\mu(\cap_k B_k)=\mu (\limsup A_n)\geq \limsup \mu ( A_n)\geq .5$
The second last step following from Fatou's Lemma using characteristic functions of the sets $A_n$ involved. It can also be shown using the continuity of measures. The last step follows from the fact that all $A_n$s have measure .5.
A: First, $\bigcap_k\bigcup_{n>k} A_n$ is exactly the set of points which appear in infinitely many $A_k$s.
$\Rightarrow$: If $\exists i: x\in A_i$, then $\forall k < i:x\in \bigcup_{n>k}A_n$. If there are infinitely many such $i$, a choice would be available for any $k$. So $x$ is in every such union, hence in the infinite intersection.
$\Leftarrow$: If $x$ is in the infinite intersection, it is in every union $\bigcup_{n>k} A_n, \forall k$. So $\forall k:\exists i > k:x\in A_i$. The set of such $i$ has to be infinite. Otherwise it would have an upperbound $M$, and for $k > M$, no such $i$ exists that $x\in A_i$, since $M$ is the largest one possible.
Note that all $\bigcup_{n>k} A_n$ are measurable, and every measure is at least $1/2$, and that they form a non-increasing sequence of real numbers. Then we can use the infinite intersection property:
$$\mu\left(\bigcap_{k}^\infty \left(\bigcup_{n>k} A_n\right) \right) = \lim_{k\to\infty} \mu\left(\bigcup_{n>k} A_n\right)$$
which directly results from the $\sigma$-additivity. Note that the right limit exists, since they form a non-increasing, bounded sequence.
