Proof of contrapositive of Lemma 1.5.3 of book Measure Theory, by Donald L. Cohn

Question

I came across the following Lemma in Cohn's book and would like to prove the contrapositive of it.

The Lemma : Let $A$ be a Lebesgue measurable subset of $\Bbb{R}^d$. Then there exist Borel subsets $E$ and $F$ of $\Bbb{R}^d$ such that $E \subseteq A \subseteq F$ and $\lambda(F-E) = 0$.

I want to prove that if Borel sets $E, F \subseteq \Bbb{R}^d$ satisfy $E \subseteq F$ and $\lambda_d(F-E)=0$, then every set $A$ with $E \subseteq A \subseteq F$ is Lebesgue measurable.

Thanks for the help !

Answer 1

Notice that, $A=E\cup(A-E)$ and $A-E\subseteq F-E$, so $\lambda(A-E)\leq \lambda(F-E)=0$ implies $A-E$ is lebesgue measurable with measure zero.

Since $E$ is measurable and $A$ union of $E$ with a set of zero measure, $A$ is measurable. To prove this, choose any $C\subseteq \Bbb R^n$, then we have to show, $$\lambda^*(C)\geq\lambda^*(A\cap C)+\lambda^*(A\cap C'),C'=\Bbb R^n-C.$$ But, $$A\cap C=(E\cap C)\cup\big((A-E)\cap C\big)$$$$\implies\lambda^*(A\cap C)\leq\lambda^*(E\cap C)+\lambda^*\big((A-E)\cap C\big)\leq\lambda^*(E\cap C)+\lambda^*(A-E)$$$$=\lambda^*(E\cap C)+0=\lambda^*(E\cap C)$$Also, $E\cap C\subseteq A\cap C\implies \lambda^*(E\cap C)\leq \lambda^*(A\cap C)$. So using above $\lambda^*(E\cap C)=\lambda^*(A\cap C)$.

Similarly, $\lambda^*(A\cap C')=\lambda^*(E\cap C')$.

So combining we have, $$\lambda^*(A\cap C)+\lambda^*(A\cap C')=\lambda^*(E\cap C)+\lambda^*(E\cap C')\leq \lambda^*(C),$$ the last inequality due to the fact $E$ is lebesgue measurable.