# Proving sets are measurable

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 $E=E_1\cup E_2$, $E_1\cap E_2=\emptyset$.

Prove:

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 $m(A) = m( A \cap E) + m(A \cap E^{c})$

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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.

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Thanks!! got it :) –  alice Oct 15 '11 at 15:32