I met with a multiple choice question,

Q. Let $E\subset M \subset \Bbb R^n$, where $M$ is a measurable (Lebesgue) set with $m(M)<\infty, $ then $E$ is measurable if

($m^*$ denotes the Lebesgue outer measure)

  1. $m(M)=m^*(E)+m^*(M\backslash E)$.

  2. $m(M)=m^*(E)+m^*(M\cap E)$.

  3. $m(M)=m^*(E)+m^*(M \triangle E)$, $\triangle$ denotes the symmetric difference.

  4. $m(M)=m^*(E)+m^*(M\cap E^c)$.

Caratheodory criterion says '$E$ is measurable if and only if $m^*(A)=m^*(A \cap E)+m^*(A \cap E^c)$ for all $A \subseteq \Bbb R^n$'. Thus, here I can see the measurability of $E$ implies the statements of First, Third and Fourth options (which are exactly the same). But how can we make a necessary and sufficient statement using the condition 'being a subset of a measurable set $M$'?


1 Answer 1


I'll do the proof for $1$.

I'll denote set difference by "$-$" for easier typing.

There exists a $G_{\delta}$ set $G$ such that $M-E\subset G $ and $m^{*}(M-E)=m(G)$ (I am assuming you know that Borel sets are measurable).

Then $M-E\subset M\cap G\subset G$ and hence $m^{*}(M-E)\leq m(M\cap G)\leq m(G) = m^{*}(M-E)\implies m^{*}(M-E)=m(M\cap G)$ .

As $M\cap G$ is measurable we have by Caratheodory cut condition on $M\cup G$ with $M$

$M\cap(M\cap G)= M\cup G$ and $M\cap(M\cap G)^{c}=M-G$

So $m(M)=m(M\cup G)+m(M-G)=m^{*}(M-E)+m(M-G)$ .

Now $m^*(M-E)=m(G)$ and we have the following claim that if $E$ is measurable and $F$ is any set then $m^{*}(E\cup F)+m^{*}(E\cap F)=m(E)+m^{*}(F)$ (You can prove this by using Cut condition on $E\cup F$ and $ F$ and is a standard result if you already know it. It is widely used in probability) . Now using this result for $E\subset F$ and $F-E$ to get $m^{*}(E\cup(F-E))+m^{*}(E\cap(F-E))=m^{*}(F)+m^{*}(\emptyset)=m(E)+m^{*}(F-E)$

Hence we have for $E$ measurable and $E\subset F$ that $m^{*}(F)=m(E)+m^{*}(F-E)$

Thus we can apply this to have $m(M)-m^{*}(M-E)=m^{*}(E)$ and this would mean $m^{*}(E)=m(M-G)$

Thus we have a measurable set $M-G$ whose measure equals the outer measure of $E$ and $m^{*}(E)<\infty$ .

This means that $m^{*}(E-(M-G))=m^{*}(E)-m^{*}(M-G)=0$ and hence $E-(M-G)$ is of measure $0$ and hence measurable.

This means $E=(E-(M-G))\cup (M-G)$ is measurable.

  • $\begingroup$ The equation $m^{*}(E-(M-G))=m^{*}(E)-m^{*}(M-G)=0$ is not well clear for me. How can we justify the additivity of $m^*$ on the sets $E-(M-G)$ and $M-G$ using the measurability of $M-G$? $\endgroup$
    – Messi Lio
    Jun 27 at 17:42
  • 1
    $\begingroup$ We are merely using the assumptions. We have $m^{*}(M-E)=m(G)$ . And $m(M)-m^{*}(M-E)=m(E)$. Hence So $m(M)-m(G)=m^{*}(E)$ and hence $m(M-G)=m^{*}(E)$ $\endgroup$ Jun 27 at 17:57
  • 1
    $\begingroup$ Now $M-G$ is a measurable set and as I said in the claim that if $E$ is measurable(and of finite measure) and $F$ is ANY set then $m^*(F-E)+m(E)=m^{*}(F)$. $\endgroup$ Jun 27 at 18:00
  • $\begingroup$ Thanks a lot Sir $\endgroup$
    – Messi Lio
    Jun 28 at 1:34

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