Let $E\subset [a,b]$. Show that $E$ is Lebesgue measurable if and only if the Lebesgue outer measure of $E$ is equal to the Lebesgue inner measure of $E$.

I have seen the proof for this above statement for the Caratheodory definition of Lebesgue measurable, but I was wondering if someone could help me prove it for a different (but equivalent) definition of Lebesgue measurable set.

The definition my book is using:

$E\subset \mathbb{R}$ is said to be Lebesgue measurable if $E$ can be squeezed between an open set $G$ and a closed set $F$ where we have that $m^*(G\setminus F)<\varepsilon$

I need some help to start the problem please. Thanks!

Just for completeness I provide the definitions of inner and outer lebesgue measure below:

Lebesgue outer measure: For a subset $E$ of $\mathbb{R}$, we have that $m^*(E)=\inf\{\sum_{n=1}^{\infty}\ell(I_n):E\subset \cup_{n=1}^{\infty}I_n\}$

where $\ell(\cdot)$ denotes the length

Lebesgue inner measure: For a subset $E$ of a bounded interval $[a,b]$, we have that $m_*(E)=b-a-m^*([a,b]\setminus E)$


Basically you are asked to prove the equivallence of to definitions of Lebesgue measurable set.

(1) $\implies$(2) Suppose $\mu^*(E)=\mu_*(E)$. Than for any $\varepsilon>0$ there are two sets of open intervals $\{I_n\}_{n\in\mathbb{N}}$ and $\{J_n\}_{n\in\mathbb{N}}$ such that

(i) $\quad E\subset I=\bigcup_{n\in\mathbb{N}} I_n$

(ii) $\ \ \ [a,b]\setminus E\subset J=\bigcup_{n\in\mathbb{N}} J_n$

(iii) $\ \ \sum\ell(I_n)+\sum\ell(J_n)-\varepsilon < b-a$

So we have an open set $I$ that contains $E$, closed set $H=[a,b]\setminus J$ that's contained in $E$. Note that $I\cup J=[a,b]$ and so $m(I\setminus J)+ m(I\cap J)+m(J\setminus I)=b-a$, where $m$ is measure on intervals and their countable unions. In (iii) we have $m(I)+m(J)-\varepsilon<b-a$ which implies $$ m(I\setminus J)+2\cdot m(I\cap J) + m(J\setminus I)-\varepsilon<b-a $$ $$ \underbrace{m(I\setminus J)+ m(I\cap J) + m(J\setminus I)}_{=b-a}+m(I\cap J)-\varepsilon<b-a $$ and that's why $m(I\cap J)<\varepsilon$. Finally $I\cap J=I\setminus H$ and $m^*=m|_{\mathrm{Dom}(m)}$ so we have $m^*(I\setminus H)<\varepsilon$ as desired.

(2) $\implies$(1) Suppose we have open set $G\supset E$ and closed set $F\subset E$, and $m^*(G\setminus F) < \varepsilon$. As an open set $G$ can be expressed as a countable union of pairwise disjoint open intervals. The same works for $K=[a,b]\setminus F$. We see that $G\setminus F=G\cup K$ and $m(G\cup K)=m^*(G\setminus F) < \varepsilon$, and clearly $m^*(E)\leq m(G)$ and $m_*(E)\geq m(F)$. As $G\supset F$ we have $m(G\setminus F)=m(G)-m(F)$. Summing this all up we get:

$$ m^*(E)-m_*(E)\leq m(G)-m(F)=m(G\setminus F)= m^*(G\setminus F) < \varepsilon, $$ and your result stands.

  • $\begingroup$ Thanks for your answer. Would you mind explaining how you know that $m^*(I\cap J)<\epsilon$? $\endgroup$ – Tomas Jun 24 '16 at 16:53
  • $\begingroup$ And actually, why does $m^*(E)-m_*(E)<2\epsilon$ in the last line? $\endgroup$ – Tomas Jun 24 '16 at 16:57
  • $\begingroup$ @Tomas, I've edited the answer, added details. Please take a look and ask if something is unclear $\endgroup$ – Glinka Jun 24 '16 at 18:51

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