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 $B_n \to B$.

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.
