Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The problem statement, all variables and given/known data

The question is from Stein and Shakarchi, Real Analysis 2, Chapter 1, Problem 5:

Suppose $E$ is measurable with $m(E) < \infty$, and $E=E_1\cup E_2$, $E_1\cap E_2=\emptyset$.


a) If $m(E) = m^{*}(E_1) + m^{*}(E_2)$, then $E_1$ and $E_2$ are measurable.

b) In particular, if $E \subset Q$, where $Q$ is a finite cube, then $E$ is measurable if and only if $m(Q) = m^{*}(E) + m^{*}(Q − E)$.

The definition of a 'measurable set' given in the book is that for any $\epsilon > 0$ there exists an open set $O$ with $E \subset O$ and $m^{*}(O − E) \leq \epsilon$, so I'm looking for a set of implications that lead me back to this definition.

all i could prove is that if $E$ measurable from my definition up, iff $ m(A) = m( A \cap E) + m(A \cap E^{c}) $

Thanks in advance for any help you can give me - it's very much appreciated.

share|cite|improve this question
up vote 6 down vote accepted

We define the inner measure $m_*$ of a set $X$ as $$m_*(X)=\sup_{F\in\mathcal{C}}\ m(F),$$ where $\mathcal{C}$ is the family of closed subsets of $X$.

Then you can prove the following lemmas:

Lemma 1 For all $E$:

$i)$ $m_{\star}(E)\leq m^{\star}(E)$

$ii)$ If $E$ is measurable then $m_*(E)=m^*(E)$. If $m_*(E)=m^*(E)\lt \infty$ then $E$ is measurable.

Lemma 2 If $E$ is measurable and $A$ is any subset of $E$, then $$m(E)=m_*(A)+m^*(E\setminus A).$$

Now, note that if $E_1\cap E_2=\emptyset$ and $E=E_1\cup E_2$ then $$\begin{align*} E\setminus E_2&= (E_1\cup E_2)\setminus E_2\\ &= E_1\setminus E_2\\ &= E_1\setminus (E_1\cap E_2)\\ &= E_1. \end{align*}$$

Also note that is enough to show that $E_1$ is measurable. Since $E_2\subseteq E$, $m^*(E_2)\lt \infty$. By your hypothesis and the lemma 2 you have $$m(E)=m^*(E_1)+m^*(E_2)$$ and $$m(E)=m_*(E_1)+m^*(E_2).$$

I think you can conclude the proof from this point.

share|cite|improve this answer
Thanks!! got it :) – alice Oct 15 '11 at 15:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.