# Doubt in Understanding Lebesgue measure.

I am studying Measure theory form Stein and Shakarachi:Real Analysis.

I come across observation regarding the outer measure.

For any $$E\in R^d$$ $$m_*(E)=\inf m_*(O)$$ where $$O$$ is the open set containg E.

i.e $$\forall \epsilon>0 \exists O\in R^d$$ such that $$m_*(O) and $$E\subset O$$.

Also By Lebesgue measurable set definition

E is said to be Lebesgue measurable iff $$\forall \epsilon>0,\exists O\in R^d$$ such that $$m_*(O\setminus E)<\epsilon$$ and $$E\subset O$$

From this, I thought both definitions are the same. then every set in $$R^d$$ become Lebesgue measurable which is of course not true .where is my interpretation fails?

When it is possible that $$m_*(O \setminus E)\neq m_*(O)-m_*(E)$$

• You should rather use \setminus=$\setminus$ for set difference. – Berci Feb 16 at 9:37
• getting $\mu^\ast(O-E)$ to be very small should lead you to a compact subset of $E$ with positive measure... – Tim kinsella Feb 16 at 11:02
Let $$E$$ be a vitali set. One feature of $$E$$ that's easy to prove is that every compact subset has zero measure (thus the same is true of closed subsets of $$E$$). Also $$E$$ has positive outer measure. Then let $$\epsilon>0$$ and suppose we can find open $$O$$ containing $$E$$ such that $$m^\ast(O-E)<\epsilon$$. Then there exists open $$U$$ such that $$O-E\subset U$$ and $$m(U)< 2\epsilon.$$ Now $$O$$ is the disjoint union:$$O= (O\cap U) \cup(O-U) ,$$ and so $$m(O) = m(O\cap U)+m(O-U).$$ But $$O-U$$ is a closed subset of $$E$$, so has zero measure. So $$m(O) = m(O\cap U)\leq m(U)\leq 2\epsilon.$$ But $$\epsilon$$ was arbitrary, and $$E$$ supposedly has positive outer measure. Contradiction.