1
$\begingroup$

Let $A$ be a set in $E=[0,1]\times[0,1]$. Then, there exists a Lebesgue measurable set $A'$ such that $A\subseteq A' \subseteq E$ and $\mu^*(A)=\mu^*(A')$.

$\mu^*$ is Lebesgue outer measure.

I was searching for a proof for bigger theorem on Lebesgue measure, and I encountered this claim. But, I think it is too hard for me to prove it. I tried to find the proof in some books such as Real Analysis by Royden and Introductory Real Analysis by Kolmogorov, and some websites, but to no success.

I do hope that anyone could provide suitable reference for the proof of this claim. I just start to learn this Lebesgue measure, so maybe I do not have sufficient tools to prove it.

Additional info: The definition of set $S$ to be Lebesgue measurable is: for all $\epsilon > 0$, there exists an elementary set $B$ in $E$ such that $\mu^*(S\Delta B)<\epsilon$.

$\endgroup$
  • 1
    $\begingroup$ Couldn't you just add a single point to the set? $\endgroup$ – Matt Samuel Nov 2 '14 at 22:29
  • $\begingroup$ But is that make it Lebesgue measurable? $\endgroup$ – Haley13 Nov 2 '14 at 23:07
  • $\begingroup$ It's the union of two Lebesgue measurable sets. $\sigma$-algebras are closed under finite union. So yes. $\endgroup$ – Matt Samuel Nov 2 '14 at 23:08
  • 1
    $\begingroup$ We do not know whether $A$ is Lebesgue measurable or not. $\endgroup$ – Haley13 Nov 2 '14 at 23:10
  • $\begingroup$ Oh, sorry I misunderstood. $\endgroup$ – Matt Samuel Nov 2 '14 at 23:11
1
$\begingroup$

Just use the definition of the outer measure. For each $n\ge 1$ let $A=\bigcup_{k\ge1} I_{n,k}$ be a union of open rectangles such that $\mu(\bigcup_{k\ge1}I_{n,k})\le\mu^*(A)+\frac1n$ and set $$A'=\bigcap_{n\ge1}\bigcup_{k\ge1} I_{n,k}$$ Surely $A'\supseteq A$ and $\mu(A')=\mu^*(A)$.

$\endgroup$
  • $\begingroup$ I am sorry, but what do you mean by intervals here? Because $A$ is in a plane. $\endgroup$ – Haley13 Nov 2 '14 at 23:18
  • $\begingroup$ @Haley13 Oh, right. Rectangles then. $\endgroup$ – user2345215 Nov 2 '14 at 23:19
  • $\begingroup$ Oh, I think I get what you mean here. Thank you! $\endgroup$ – Haley13 Nov 2 '14 at 23:26

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.