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

  • $\begingroup$ 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. $\endgroup$
    – T. Eskin
    Commented Aug 13, 2012 at 4:04
  • $\begingroup$ So far my notes have only defined measurability for bounded sets. Haven't gotten there yet, but thanks for the insight. $\endgroup$ Commented Aug 13, 2012 at 4:10
  • $\begingroup$ Perhaps useful $\endgroup$
    – leo
    Commented Aug 17, 2012 at 0:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – leo
    Commented Aug 17, 2012 at 1:08
  • $\begingroup$ @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). $\endgroup$ Commented Aug 17, 2012 at 2:14

3 Answers 3


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 $A\cap B =\emptyset$

  • $\begingroup$ I want to use the equation in my problem to prove the Caratheodory Crtieria / the equation you just used, so I don't want to assume this equation is true yet. $\endgroup$ Commented Aug 14, 2012 at 0:31
  • 2
    $\begingroup$ @Broseph: I suggest making explicit in the question your definition of measurability, as well as the results that are available so that a more appropiate answer may be given. $\endgroup$
    – Jose27
    Commented Aug 14, 2012 at 5:09

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.

  • $\begingroup$ 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? $\endgroup$ Commented Aug 19, 2012 at 18:30
  • $\begingroup$ If you can prove that open sets are measurable then that's a corolary. Do you can? $\endgroup$
    – leo
    Commented Aug 21, 2012 at 2:10
  • $\begingroup$ 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. $\endgroup$ Commented Aug 21, 2012 at 3:09
  • $\begingroup$ 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. $\endgroup$
    – leo
    Commented Aug 21, 2012 at 3:34
  • $\begingroup$ Thank you, and thanks to everyone who contributed. $\endgroup$ Commented Aug 21, 2012 at 4:21

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.

  • 1
    $\begingroup$ You still need to elaborate $\endgroup$
    – Shailesh
    Commented May 6, 2016 at 23:54

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