Outer measure of a union of 2 subsets of disjoint measurable sets of real numbers.

Every set mentioned is a subset of the real numbers.

Let $m^*(C)$ denote the outer measure of a set $C$. Let $E$ be $any$ set and $A,B$ be measurable, disjoint sets. I'm trying to show that $$m^*(E\cap (A\cup B))=m^*(E\cap A)+m^*(E\cap B).$$

Proof: ($\le$) follows by the countable subadditivity of the outer measure since $$E\cap (A\cup B)= (E\cap A)\cup (E\cap B).$$

Here's where I get stuck:

($\ge$) My attempts have reduced to something of the form:

1. There are bounded open sets $G_1, G_2$ containing $E\cap A, E\cap B$, respectively, such that $$m(G_1)\ge m^*(E\cap A),\quad m(G_2)\ge m^*(E\cap B).$$ Hence $$m^*(E\cap A)+m^*(E\cap B)\le m(G_1)+m(G_2).$$ And I would like to extend this inequality to $m(G_1\cup G_2)$ but I know that's not even true, especially since the sets $G_1, G_2$ may not even be disjoint.
2. I also tried de la Vallée-Poussin Criterion: Let $\epsilon>0$. Since $A, B$ are measurable, there are closed subsets $F_1, F_2$ of $A,B$ respectively, such that $m^*(A\cap E - F_1)+ m^*(B\cap E-F_2)<\epsilon$. Even if I could show, $$|m^*(E\cap (A\cup B)-[(F_1\cup F_2))+ m^*(A\cap E - F_1)+ m^*(B\cap E-F_2)]|<\epsilon.$$

I'm not sure what that would mean.

What I know:

• Measure has only been defined for bounded sets.

• A bounded set $A$ is $measurable$ if its outer and inner measures are equal; if so, the measure of $A$ is the common value of these measures.

• Differences, countable unions, countable intersections of measurable sets are measurable.

• The union of a set of pairwise disjoint measurable sets is measurable, with the measure of the union equal to the sum of the measures of the sets in the union.

• Outer and inner measures are monotone increasing functions.

• Countable subadditivity for outer measure, which states that if $A$ is a countable or finite union of sets $A_i$ then $m^*(A)\le \sum m^*(A_i)$.

• De la Vallée-Poussin Criterion, which states that a bounded set $A$ is measurable iff for every $\epsilon >0$ there is a closed set $B\subset A$ such that $m^*(A-B)< \epsilon$.

• For any bounded set $B$, I can always find a set $C$ that is a countable intersection of open sets for which $B \subset C$ and $m^*(B)=m^*(C)$.

• If $A$ and $B$ are measurable sets, then $m(A\cup B) + m(A\cap B) = m(A) + m(B)$.

• If $A$ is bounded and $I$ is an open interval containing $E$, then $m^*(E) + m_*(I-E) = m(I)$.

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Why are subsets of measurable sets bounded? Consider for example $]0,\infty[\subset \mathbb{R}$, where $\mathbb{R}$ is measurable and $]0,\infty[$ is unbounded. –  Thomas E. Aug 13 '12 at 4:04
So far my notes have only defined measurability for bounded sets. Haven't gotten there yet, but thanks for the insight. –  The Substitute Aug 13 '12 at 4:10
Perhaps useful –  leo Aug 17 '12 at 0:58
Notice that Caratheodory Criterion is equivalent to the problem you pose. You can get a proof of the Caratheodory Criterion for example from the Wheeden & Zygmund's Measure and Integral book, to get some inspiration. –  leo Aug 17 '12 at 1:08
@leo Thanks for the link you sent, but the equation in that question has both inner and outer measure. Even though my set $A$ is measurable, we cannot conclude that $A\cap E$ is. I proved the Caratheodory Criterion by using the result I am trying to prove. My goal is to create/find a solution to the reverse inequality that I posed, rather than using the proof of Caratheodory (which is a later problem in my notes). –  The Substitute Aug 17 '12 at 2:14

Lemma. Let $E\subseteq \Bbb R$. If $H\supseteq E$ is a $G_\delta$ set (countable intersection of open sets) such that $$m(H)=m^\ast(E),$$ then for every $C\subseteq\Bbb R$ $$m^\ast(H\cap C)=m^\ast(E\cap C).$$

Proof. Let $C\subseteq\Bbb R$. In the following the superscript $^c$ means complement. \begin{align*} m^\ast(H\cap C) &\leq m^\ast(H\cap C\cap E\cap C)+m^\ast((H\cap C)\setminus (E\cap C))\\ &= m^\ast(E\cap C) + m^\ast((H\cap C)\cap (E\cap C)^c)\\ &= m^\ast(E\cap C) + m^\ast(C\cap (H\setminus E))\\ &\leq m^\ast(E\cap C) + m^\ast(H\setminus E)\\ &= m^\ast(E\cap C) \end{align*} The inequality $m^\ast(H\cap C)\geq m^\ast(E\cap C)$ comes free by the monotony of the outer measure since $H\supseteq E$.

Proof of $m^\ast(E\cap (A\cup B))\geq m^\ast(E\cap A)+m^\ast(E\cap B)$.

Pick $H\supseteq E$ a $G_\delta$ set so that $m(H)=m^\ast(E)$. Then \begin{align*} m^\ast(E\cap (A\cup B)) &= m^\ast(H\cap (A\cup B)) &&\text{by the Lemma}\\ &= m(H\cap A) + m(H\cap B) &&\text{(^\ast)}\\ &\geq m^\ast(E\cap A) + m^\ast(E\cap B) &&\text{by the monotony of the outer measure.} \end{align*} ($^\ast$) because here we are dealing with measurable sets (of finite measure).

Observation. Notice that such a $G_\delta$ set $H$ always exist even if $E$ is unbounded.

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I can see that such a set $H$ with the same outer measure of $E$ will exists. But how do you know that the countable intersection $H$ will be always be measurable? –  The Substitute Aug 19 '12 at 18:30
If you can prove that open sets are measurable then that's a corolary. Do you can? –  leo Aug 21 '12 at 2:10
Yes, but the intersection of open sets isn't always open. Take $A_n = (-1/n,1/n)$ in the real line. Then $\cap A_n$ is a singleton, which won't be open in the real line. –  The Substitute Aug 21 '12 at 3:09
You're right, it's a singleton, a singleton have outer measure zero so it is measurable. In the other hand, if you can proof that an open set is measurable by the third bullet of What you know it follows that a countable intersection of measurable sets is always measurable. –  leo Aug 21 '12 at 3:34
Thank you, and thanks to everyone who contributed. –  The Substitute Aug 21 '12 at 4:21
Put $F=E\cap (A\cup B)$ then, since $A$ is $m^*$-measurable, we have that $$m^*(F)= m^*(F\cap A) + m^*(F\cap A^c) = m^*(E\cap A)+m^*(E\cap B)$$ since $A\cap B =\emptyset$