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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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