The Lebesgue sigma algebra is complete with respect to Lebesgue measure, which means that if $A$ is a Lebesgue measurable set with Lebesgue measure $0$ and $B$ is a subset of $A$, then $B$ is Lebesgue measurable as well. But I'd like to know if something stronger is true.
Suppose that $A$ is a subset of $B$ which is a subset of $C$, where $A$ and $C$ are Lebesgue measurable sets and the Lebesgue measure of $A$ is equal to the Lebesgue measure of $C$. Then my question is, does $B$ have to be Lebesgue measurable as well?
If $A$ and $C$ are each Lebesgue measurable, then so is $C\setminus A$ and so by finite additivity it has measure zero.
But then $B\setminus A$ also has measure zero and hence is measurable, so $$B=A\cup(B\setminus A)$$ is the union of two measurable sets, hence is measurable.