Let $A$ be a Lebesgue measurable subset of $\Bbb R$.
1) Show that there exists a Borel measurable subset $B$ of $\Bbb R$ such that $A\subseteq B$ and such that $l^*(B\setminus A)=0$.
2) Show that every Lebesgue measurable set is the union of a Borel measurable set (with same measure) and a set of Lebesgue measure zero.
(Note that $l^*$ denote the outer measure. )
I already showed that if $C$ is a Lebesgue measurable subset of $\Bbb R$ and $\epsilon \gt 0$, then there exists an open set $G_{\epsilon}\supseteq C$ such that $l^*(C)\le l^*(G_{\epsilon})\le l^*(C)+\epsilon$.
Moreover, if $D$ is a Lebesgue measurable subset of $\Bbb R$, if $\epsilon \gt 0$, and if $D\subseteq I_n=(n,n+1]$, then there exists a compact set $K_{\epsilon} \subseteq D$ such that $l^*(K_{\epsilon})\le l^*(D)\le l^*(K_{\epsilon})+\epsilon$.
My thought is to somehow manipulate the inequality and have $\epsilon$ shrinking, but I'm stuck to show this and come up with the result.
Could someone help to provide a proof please? Any help is appreciated. Thanks.
Best Answer
You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).
Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that $$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$ And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)
Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.
If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$
This shows the first corollary.
For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set. Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$
$G^C$ is again a borel set (As a completement of borel set).
As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:
It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).