How to show that every set with Lebesgue outer measure zero is Lebesgue measurable?

718 Views Asked by At

Definition of Lebesgue measurable if for each $ε>0$, there exist a closed set $F$ and an open set $G$ with $F⊂E⊂G$ such that $m$ * $(G-F)<ε$. About this problem, $F$ can be a empty set that is closed. However, I've no idea how to find a open cover such that $m$*$(G-F)<ε$ satisfied due to the fact that $E$ may be uncountable like cantor set.

Does anyone have some idea? Thanks.

1

There are 1 best solutions below

1
On BEST ANSWER

I will assume you're working in $\mathbb{R}$. The definition of outer-measure is here: $$m^{*}(E)=\inf \{\sum_{n=1}^{\infty}l(I_n)| E \subset \bigcup_{n=1}^{\infty}I_n, I_n \text{being disjoint sequence of open intervals}\}$$

If $E$ has outer measure zero, that means that for every positive $\epsilon$, there exists a sequence of disjoint open intervals $(I_n)\subset \mathbb{R}$ such that $$ \sum_{n=1}^{\infty}l(I_n)<\epsilon $$

I guess these $(I_n)$ can be your open cover, then you use $\emptyset =F\subset E\subset \cup I_n = G$, with $m^*(G-F)=m^*(G)<\epsilon$.