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This is a problem from Elias Stein's real analysis book. I'll use $m(E)$ to denote the Lebesgue measure of a measurable set $E$. And $m^{*}(A)$ to denote the exterior measure (outer measure) of not-assumed-measurable set $A$. The problem is: Suppose $E$ is a measurable set and $E=A\cup B$, where $A$ and $B$ are not-assumed-measurable sets. The intersection of $A$ and $B$ is empty.


  1. $m(E)=m^{*}(A)+m^{*}(B)$
  2. $m(E)$ is finite.

then prove both $A$ and $B$ are measurable.

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The basic facts you have to use are the following: (as there are references almost everywhere I won't give you a proof. I advice you to try them yourself. If you have questions feel free to ask)

Fact 1. An element $C\in \mathcal P(\mathbb R^n)$ is Lebesgue measurable if and only if

  • $m^*(C)<+\infty$
  • $$m^*(C)=\sup_{\overset{K\Subset C}{K \text{compact}}} m(K)$$ i.e. the measure of $C$ can be approximated by the measure of the compact subsets inside $C$.

Fact 2. Given $A,B$ disjoint subsets of $\mathbb R^n$, and a compact $K\Subset A\cup B$, it is possible to find two compact subsets $A_1\Subset A$, $B_1\Subset B$ respectively, such that $K\subset A_1\cup B_1$.

Immediately notice that, as $m(E)<+\infty$, it follows that $m^*(A)<+\infty$, $m^*(B)<+\infty$.

Now, for your problem first let's see that if we prove that $A$ is measurable, then it will immediately follows that also $B$ is measurable, since $B=E\cap A^c$, and both $E$ and $A^c$ are measurable.

Then fix any $\varepsilon >0$. By the fact that the Lebesgue measure of a set $C$ can be approximated from below by the measure of compact subsets inside $C$, it follows that $\exists K\Subset E=A\cup B$ compact such that $$m(K)+\varepsilon \geq m(E)=m^*(A)+m^*(B).\tag{1}$$

Now we use fact $2$ to say that there exists $A_1\Subset A$, $B_1\Subset B$ compact such that $K\subset A_1\cup B_1$. By monotonicity of the Lebesgue measure, from $(1)$ it follows that $$m(A_1)+m^*(B)+\varepsilon \geq m(A_1)+m(B_1)+\varepsilon \geq m(K)+\varepsilon\geq m^*(A)+m^*(B)\tag{2}$$ i.e. we have shown that for any $\epsilon>0$ we can find a compact $A_1$ inside $A$ such that $$m(A_1)+\varepsilon\geq m^*(A)\tag{3}$$ and this says precisely $$m^*(A)=\sup_{\overset{K\Subset A}{K \text{compact}}} m(K).\tag{4}$$ By fact $1$ we conclude that $A$ is indeed measurable, and so is $B$.

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thank you. But what do you mean by K⋐E? – Li Xinghe Feb 1 '13 at 7:32
That K is Well contained in E, That is $\bar K \subset E $ – uforoboa Feb 1 '13 at 8:34
But if K is compact, the closure of K is K itself? – Li Xinghe Feb 2 '13 at 7:02
right.. You do not have to worry then in this case that something might go wrong. The fact holds in much more generality, in your case it's a special case let's say. – uforoboa Feb 2 '13 at 11:11

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