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.


2 Answers 2


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:

  1. It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
  2. It is defined on borel sets, bounded sets have finite exterior-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).


The set $L$ of Lebesgue sets is usually defined as the smallest $\sigma$-algebra that has the set $B$ of Borel sets as a subset, and contains every set with outer measure $0$.

Let $L'=\{b$ \ $a: b\in B\land l^*(a)=0\}.$ Obviously $L'\supset B$ and $L'\supset \{a:l^*(a)=0\}.$

(i). If $A\subset \cup \{a_f:f\in F\}$ where $F$ is countable and $l^*(a_f)=0$ for each $f\in F$ then $l^*(A)=0.$

So if $F=\{b_f$ \ $ a_f:f\in F\}\subset L'$ where $F$ is countable, with $b_f\in B$ and $l^*(f)=0$ for each $f\in F,$ then $\cup F= (\cup_{f\in F} b_f)$ \ $A$ where $A\subset \cup_{f\in F}a_f.$ So $l^*(A)=0$. Therefore $\cup F \in L'.$

(ii). For $b\in B$ and $l^*(a)=0,$ let $c$ be a $G_{\delta }$ set with $l(c)=0$ and $c\supset a.$ We have $$\mathbb R \backslash (b \backslash a)=b' \backslash a'$$ where $b'= (\mathbb R$ \ $b)\cup c$ and $a'= (c\cap b)$ \ $a.$ Obviously $b'\in B.$ And $l^*(a')\leq l^*(c)=l(c)=0.$ Therefore $\mathbb R $ \ $(b$ \ $a)\in L'.$

(iii). So $L'$ is closed under countable unions, and complements, and is not the empty set. Therefore $L'$ is a $\sigma$-algebra. So by the def'n of $L$ we must have $L'\supset L.$ Since obviously $L'\subset L$, we have $L'=L.$


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