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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"?

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$$ A=(B\setminus(B\setminus A))\cup(A\setminus B). $$ The rest should be more trivial.

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  • $\begingroup$ I'm still stuck. How do I apply this? $\endgroup$ – blst Nov 13 '13 at 14:03
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    $\begingroup$ $B$, $B\setminus A$ and $A\setminus B$ are measurable. $\endgroup$ – Julián Aguirre Nov 13 '13 at 14:22

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