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 $m_*(O)<m_*(E)+\epsilon$ 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 $m_*(O \setminus E)\neq m_*(O)-m_*(E)$

Please help me out to solve this misinterpretation.

Any help will be appreciated

  • 1
    $\begingroup$ You should rather use \setminus=$\setminus$ for set difference. $\endgroup$
    – Berci
    Feb 16 '19 at 9:37
  • $\begingroup$ aren't you just asking for a proof that there's a nonmeasurable set? $\endgroup$ Feb 16 '19 at 10:56
  • $\begingroup$ @Timkinsella tried to prove for Vitali set, But am not able to do so. Sir Can you give me hint to do so? $\endgroup$ Feb 16 '19 at 10:58
  • $\begingroup$ i think you want to use the fact that vitali set has zero inner measure. $\endgroup$ Feb 16 '19 at 11:00
  • $\begingroup$ getting $\mu^\ast(O-E)$ to be very small should lead you to a compact subset of $E$ with positive measure... $\endgroup$ Feb 16 '19 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.


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