I've been working through an exercise which deals with showing that two definitions of a set being Lebesgue measurable are equivalent. The definition that I was thinking about was the following:
$E \subseteq \mathbb{R}^{d}$ is measurable if for every $\epsilon$ there is a closed set $F$ contained in $E$ with $m_{*}(E-F)<\epsilon$.
The $m_{*}$ is the outer measure obtained from taking the infimum of the total volume of rectangular covers.
I am having trouble showing that the compliment $E^{c}$ is also measurable when $E$ is measurable
My hunch is to show that $E^{c}$ is the intersection of such measurable sets. One observation I made is that for every $n \in \mathbb{N}$, we have a closed set $F_{n}$ such that
$m_{*}(E-F_{n}) < \epsilon$
If we set $O_{n}=F_{n}^{c}$ and $S=\bigcap_{n=1} O_{n}$, I think we should be might be able to show that $E^{c}=S$ or use $S$ in some way to show that $E^{c}$ is measurable.
I'm happy or accept to hear any alternative methods, but some help along the above lines would be very much appreciated.
Best Answer
You need to use the definition of "measurable". As you refer to the Lebesgue measure in your question, the definition of "measurable" is likely:
Let $\mathcal A$ denote the class generated by complements and finite unions of half open (on the left) intervals. Let $m$ be the length function on $\mathcal A$. Define $m^*$ to be the outer measure associated to $m$. Then a set $E$ is said to be $m^*$ measurable iff $$m^*(A) = m^*(A\cap E) + m^*(A\setminus E)$$ for every $A\in \mathcal A$.
It is now immediate, because $E\setminus A = E\cap A^c$, and $E\cap A = E\setminus A^c$.