# Equivalence of condition for Lebesgue-measurability

We had a proof in lecture, that showed a bunch of equivalences for Lebesgue-measurability. I have a problem understanding the following implication, where the professor said the reasoning was "trivial".

Let $A\subseteq\mathbb{R^n}$. If there exists $B\in F_\sigma$ such that $\lambda_n^*((A\setminus B)\cup(B\setminus A))=0$ then $A$ is Lebesgue-measurable. $F_\sigma$ denotes sets that are a countable union of closed sets, $\lambda_n^*$ is the $n$-dimensional Lebesgue outer measure.

Can anyone jump start me as to why this is "trivial"?

$$A=(B\setminus(B\setminus A))\cup(A\setminus B).$$ The rest should be more trivial.
• $B$, $B\setminus A$ and $A\setminus B$ are measurable. – Julián Aguirre Nov 13 '13 at 14:22