Note: $m$ is an outer measure on the power set of $\mathbb{R}$.
Attempt:
Suppose the condition holds,choose closed sets $F_n\subset{E}$ such that $$m(E\setminus F_n)<1/n$$ for $n=1,2,\dots$ Let $A=\bigcup{F_n}$. Then $$E=A \cup (E\setminus A)$$ Now I think I just need to show $A$ and $E\setminus A$ are $m$-measurable, which I am unsure how to do. I do not have any idea how to prove the contrary.
[Math] Subset $E$ of $\mathbb{R} $ is $m$-measurable iff for every $\epsilon>0$ $\exists$ a closed set $F\subset{E}$ such that $m(E\setminus F)<\epsilon$
lebesgue-measuremeasure-theory
Best Answer
$m(E \backslash A) \leq m(E \backslash F_n) < 1/n$ for all $n$ because $E \backslash A \subseteq E \backslash F_n$ for all $n$; hence $E \backslash A$ is measurable with measure $0$. Since the $F_n$ are in our $\sigma$-algebra, so is $\bigcup F_n = A$? Also note that $E = A \sqcup(E \backslash A)$, where $\sqcup$ means disjoint union, so what can we say about $m(E)$?