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

Question

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 $m^*(E) + m_*(I-E) = m(I)$.

Answer 1

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 $A\cap B =\emptyset$

Answer 2

We start with a little Lemma:

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.

Answer 3

It can be shown that there exists nonvoid disjoint subsets A,B of R such that m*(A U B) is strictly less than m*(A) + m*(B) using the Axiom of Choice.