MHB Measure Theory - Existence of Fsigma set contained in measurable set

[joypav]

Problem:
Let $E$ have finite outer measure. Show that $E$ is measurable if and only if there is a $F_\sigma$ set $F \subset E$ with $m^*\left(F\right)=m^*\left(E\right)$.

Proof:
"$\leftarrow$"
To Show: $E=K\cup N$ where $K$ is $F_\sigma$ and $m^*(N)=m(N)=0$.

By assumption, $\exists F$, and $F_\sigma$ set, $F\subset E$, and $m^*(F)=m^*(E)$
Write,
$E=F\cup (E-F)$

To Show: $m^*(E-F)=0$

By a theorem given in class,
$\exists G$, a $G_\delta$ set, such that $E\subset G$ and $m^*(E)=m(G)$.
Then,
$m(E-F) \leq$ (G is "bigger" than E) $m(G-F) = m(G)-m(F) = m(E) - m(E) = 0$
$\implies m^*(E-F)=m(E-F)=0 \implies E$ is measurable.

Could someone get me started in the other direction??

 

[Euge]

If $E$ is measurable, use the representation you wrote above: $E = K\cup N$ where $K$ is $F_\sigma$ and $m^*(N) = 0$. By countable subadditivity of the outer measure $m^*(E) \le m^*(K) + m^*(N) = m^*(K)$. On the other hand, since $K \subset E$, $m^*(K) \le m^*(E)$ by monotonicity of $m^*$. Therefore, $m^*(E) = m^*(K)$.