Exercise 1.2.19 (Null set approximation of Lebesgue measurable sets)

We use the chain of equivalences.

• $(i)⟹(\mathit{ii})$
  Let $E$ be Lebesgue measurable. By outer regularity and the axiom of choice we can find a sequence of open sets ${(U_n)}_{n\in \mathbb{N}}$ such that for all $n\in \mathbb{N}$ we have $E\subseteq U_n$ and $m(U_n)-m(E)\le 1∕n$. Since both $E$ and $U_n$ are Lebesgue measurable, we can use the additivity property and see that

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:m(U_n)-m(E)=m(U_n∕E)\le 𝜖$$

  Set $U:={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{U}_n$. We then have $E\subseteq U$ or, to match the theorem assertion, $U=E\cup (U∕E)$. Regarding the last term we have $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:$ $m(U∕E)\le m(U_n∕E)\le 𝜖$, in other words $m(U∕E)=0$, as desired.

• $(\mathit{ii})⟹(\mathit{iii})$
  Suppose that $E$ is an intersection of countably many open sets minus the null set. Since a countable intersection of measurable set is again measurable, and since null sets are also Lebesgue measurable, $E$ must be Lebesgue measurable too. By inner regularity and by the axiom of choice, find a collection ${(F_n)}_{n\in \mathbb{N}}$ of closed sets such that $F_n\subset E$ and $m(E∕F_n)=m(E)-m(F_n)\le 1∕n$. Set $F={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{F}_n$. We then have $F\subseteq E$ and $E=F\cup (E∕F)$. Looking at the latter term we see that it is a null set $m(E∕F)=m(E)-m(F)\le m(E)-m(F_n)\le 1∕n$. Since this is true for any $n\in \mathbb{N}$ we have $m(E∕F)=0$ as desired.
• $(\mathit{iii})⟹(i)$
  Since a countable union of Lebesgue measurable sets is again Lebesgue measurable, and since set difference of Lebesgue measurable sets is also Lebesgue measurable, $E$ must be Lebesgue measurable.