2
$\begingroup$

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 $m_*(E)=b-a-m^*([a,b]\setminus E)$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.