A Question on Sets of Full Outer Measure

Question

I came across this problem whilst studying for a comprehensive exam in real analysis; for reference, see Exercise 1.24(A) in Folland's Real Analysis; it's a modification of that.

Consider the unit interval $I:=\left[0,1\right]$, and let $\mathcal{M}$ be the $\sigma$-algebra of all Lebesgue measurable subsets of $I$. Denote by $m^*$ the Lebesgue outer measure on $\mathcal{M}$. Suppose that $E$ is a subset of $I$ such that $m^{*}(E)=1$. Prove that, if $A,B\in\mathcal{M}$ and $A\cap E=B\cap E$, then $m(A)=m(B)$.

This result is trivial if $E$ is Lebesgue measurable; the sticky wicket occurs if $E$ fails to be so. I and my colleagues have attempted to solve this using the Carathéodory characterization of Lebesgue measurability, and failed as yet. We are aware that it need not be the case that $E$ be measurable. Any hints would be appreciated.

Answer 1

Let $D$ be borel such that $E\subseteq D$ and $m(D)=1$. Let $F \subseteq D-E$ be measurable, then $D-F$ is measurable and $m(D)=m(D-F)+m(F)$ now $E \subseteq D-F$ so $m(D-F)=1$ and we have $m(F)=0$. Also $m(I-D)=0$ is obvious. Thus we have if $F\subseteq I-E$ is measurable then $m(F)=0$.

Answer 2

Here is one solution written by Xiudi Tang in his course notes:

It suffices to show that $m^*(A \cap E) = m^*(A)$ for any $A \in \mathcal{M}$.

By the measurablity of $A$, we have that $$m^*(I)=m^*(A)+m^*(I-A)$$ $$m^*(E)=m^*(A \cap E)+m^*(E-A).$$

Since $m^*(E)=m^*(I)$ and $E \subset I$, we have that $$m^*(I-A) \geq m^*(E-A),$$ $$m^*(A) \geq m^*(A \cap E)$$ and $$m^*(A)+m^*(I-A)=m^*(A \cap E)+m^*(E-A).$$

Therefore $m^*(A \cap E) = m^*(A)$.