Let $m$ be a lebesgue measure and $\{A_n\}_{n\in \mathbb{N}}$ a family of measurable sets on $[0,1]$,and $F=\{x:\forall n\in\mathbb{N} \exists k\geq n \text{ s.t } x\in A_k \}$
Prove the following:
$F=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$
if $m(A_n)>\delta>0$ for all $n$ then $m(F)>\delta$
if $\sum_{n=1}^{\infty}m(A_n)<\infty$ then $m(f)=0$
there is a familly $\{A_n\}_{n\in \mathbb{N}}$ s.t $\sum_{n=1}^{\infty}m(A_n)=\infty$ and $m(F)=0$
1) I started with taking and element from the RHS and LHS and show it bidirectional contained to prove the equality.
Let $x\in \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$ that mean the for all $n\in \mathbb{N}$ the element $x\in\bigcup_{k=n}^{\infty}A_k$ which means that for all $n\in \mathbb{N}$ there are some $k\geq n$ such that $x\in A_k$ but this is the definition of $F$ so $x\in F$.
Now to take $x\in F$ and show that $x\in \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$ seems to be the same claims as before
2) I have thought about way to start using the definition that $E$ is lebesgue measurable then for all $A$ we have $m^*(A)=m^*(A\cap E)+^*(A\cap E^C)$ but I do not have much to say on the compliment of the set $A_k$
3) I tried using the subadditive of the measure and the fact the a bounded set a measure $0$ but did not manage
4) I do not have a clear idea how to approach this
2)Νote that $F=\limsup_nA_n$
Use the inequality $$m(\limsup_nA_n) \geq \limsup_nm(A_n)$$
3) It is the Borel-Cantelli lemma.
4)Take the sets $A_n=[0,\frac{1}{n}]$