Countable covering and disjoint covering I am curious about the following problems:
Let a bounded set $A\subseteq \mathbb{R}$ be a Lebesgue null set (i.e., $\lambda(A)=0$, $\lambda$ being the Lebesgue measure on the real line). 
Question 1: if $\{I_n\}_{n\ge 0}$ is a sequence of closed intervals such that $A\subseteq\bigcup_{n\ge 0}I_n$. Do there exist finite many intervals in $\{I_n\}$ whose union also covers $A$ (i.e., can we reduce a countable cover into a finite cover?)
Question 2: Is it true that $A$ can always be covered by at most countably many pairwise disjoint intervals (closed or open)?
Can anyone give me some hint or suggestion on these questions? Many thanks!

Question 1 : 
Counter-example: $A=\{\frac 1{n+1} / n \in {\mathbb N} \}$ and $I_n=\left]-\frac{1}{2(n+1)},\frac{3}{2(n+1)}\right[$
Question 2 : Hint:
$\bullet$ if $x \in A$ then $x \in ]r_x-1,r_x+1[$, wher $r_x$ is a rational number such that :$$r_x-1 < x < r_x+1$$
($r_x$ exists as a rationnal number st : $x-1 < r_x <x+1$ by density of $\mathbb Q$  in $\mathbb R$.)
Since $\mathbb Q$  is  a countable set we have a countable cover of $A$ by intervals.
$\bullet$ if $A \subset \bigcup_{n \geq 0} I_n$, we can define $J_0=I_0$ and  $\forall n \geq 1 \quad J_n=I_n \backslash (\mathop \cup_{k=0}^{n-1} I_k) $.
Since $J_n$ is a finit union of intervals and the sets $J_n$  are  disjoints and $\mathop \cup_{n \geq 0} J_n=A$,  we  have a cover by disjoint intervals...
NB:We can't say that these intervals are all open or closed so this solution holds if you need a disjoint interval cover only.
