a subset of a zero Lebesgue measure set is measurable? This question is related to What's the quickest way to see that the subset of a set of measure zero has measure zero?  But I'm specifically concerned about Lebesgue measure, $m$, on a real interval, $X=[a,b]$, and specifically about why a subset, $A$, of zero measure set, $E$, is measurable?  Is it by the construction of the Lebesgue measure?
I can understand that the outer measure of $A$ is zero, i.e. $m^*(A)=0$, since any open cover of $E$ also covers $A$.  But to claim that $A$ is measurable, I think (by definition) we need to show that $A$ is a countable union of finitely $m$-measurable sets.  It's not clear to me how this may be done?  I'd appreciate some help.
BTW, according to Rudin's Principles of Mathematical Analysis, a set $B$ is finitely $m$-measurable, if there is a sequence $\{B_n\}$ of elementary sets such that $B_n \to B$.
 A: I will use the following well known result: If $ N \subseteq \mathbb{R}^n$ and $m^*(N)=0$ then $N$ is Lebesgue measurable.
Also, remember that the Lebesgue measure $m$ is the restriction of $m^*$ to the Lebesgue measurable set $\mathscr{L}(\mathbb{R}^n)$: $m=m^*\big|_{\mathscr{L}(\mathbb{R}^n)}$. It means $m(M)=m^*(M),\forall M \in  \mathscr{L}(\mathbb{R}^n)$.
Now, turning to the problem, suppose $M \in  \mathscr{L}(\mathbb{R}^n):m(M)=0$ and let $E\subseteq M $. Then $0 \leq m^*(E)\leq m^*(M)$ by monotonicity of outer measures. Since $m=m^*\big|_{\mathscr{L}(\mathbb{R}^n)}$, we have that $0=m(M)=m^*(M)$. Therefore, using the result of the first line above, we conclude that $E \in \mathscr{L}(\mathbb{R}^n)$. It means that the set $E$ is Lebesgue measurable.
So, this is an easy way to see the completeness of $(\mathbb{R}^n,\mathscr{L}(\mathbb{R}^n),m)$ if you know those basic concepts of the Lebesgue measure.
A: Suppose $\lambda(E)=0$, and that $A\subseteq E$. Then as you said, $\lambda^*(A) = \lambda(E)$. In order for $A$ to be measurable, it must be the case that for each $B\subseteq [a,b]$, we have that
$$\lambda^*(A) = \lambda^*(A\cap B)+\lambda^*(A\cap B')$$
where $B' = [a,b]\backslash B$. Indeed, since $A\cap B, A\cap B' \subseteq E$, both of these sets have outer measure zero, and so the equality is satisfied. Thus, $A$ is measurable.
A: Let us use the Caratheodory criterion.
If E is measurable, then for any B, $\mu(B)=\mu(B\cap E)+\mu(B\cap E^c)$
Now, as E is measure zero $\mu(B\cap A)\leq\mu(B\cap E)\leq \mu( E)=0$, and as measure is non-negative all measures must be zero.
Therefore, $\mu(B)=\mu(B\cap A)+\mu(B\cap E^c)\leq \mu(B\cap A)+\mu(B\cap A^c)$
But outer measure is sub additive, hence 
$\mu(B)=\mu(B\cap A)+\mu(B\cap A^c)$
