Stein 1.26

$\textbf{Exercise 1.26}$ Given the conditions that $A \subseteq E \subseteq B$, $A,B \in{\mathcal{M}}$ where $\mathcal{M}$ is the $\sigma$-algebra of Lebesgue measurable sets, $m(A)=m(B) < \infty$ then $E\in{\mathcal{M}}$

Stategy: We wish to write $E$ as the union or intersection of some measurable sets.

$\textit{Proof}$ : Consider $B-A=B \cap A^c$. Since $A,B\in{\mathcal{M}}$, then by the $\sigma$-algebra structure of $\mathcal{M}$, $A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under compliments) and $B \cap A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under countable intersections). We may write $B=A \cup (B-A)$ and so $A \cap (B-A) = \emptyset$. The measure of a union of disjoint measurable sets is the sum of their measures (this is theorem 3.2) so, $m(A)+m(B-A)=m(B)$. Because of our assumption that $m(A)=m(B) < \infty$, then $m(B-A)=0$. Since $A \subseteq E \subseteq B$ then $E-A \subseteq B-A$. By the monotinicity of $m_*$, $m_*(E-A) \leq m_*(B-A)=0$. So, $m_*(E-A)=0 \Longrightarrow E-A \in{\mathcal{M}}$ since sets of outer measure 0 are measurable. Again using the $\sigma$-algebra structure of $\mathcal{M}$, $E= (E-A) \cup A$...a union of 2 measurable sets. $\therefore$ $E \in{\mathcal{M}} \quad \blacksquare$