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 $B_n \to B$.

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 
$\mu(B)=\mu(B\cap A)+\mu(B\cap A^c)$
