Effect of Caratheodory criterion on a subset of a measurable set

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 $$M$$'?

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.

• 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 $M-G$? Jun 27 at 17:42
• 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 $m(M-G)=m^{*}(E)$ Jun 27 at 17:57
• 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 $m^*(F-E)+m(E)=m^{*}(F)$. Jun 27 at 18:00
• Thanks a lot Sir Jun 28 at 1:34