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, $m^*(E) = m^*(K)$.

 

[math771]

"$\rightarrow$"
To Show: $m^*(F)=m^*(E)$.

Assume $E$ is measurable. Then, by definition, there exists a $F_\sigma$ set $K$ such that $E=K\cup N$ where $m^*(N)=0$.

Since $K$ is $F_\sigma$, it can be written as a countable union of closed sets, $K=\bigcup_{i=1}^\infty F_i$.

Then, by countable subadditivity of outer measure,
$m^*(E) = m^*(K\cup N) \leq m^*(K) + m^*(N) = \sum_{i=1}^\infty m^*(F_i) + 0$.

But since $F_i \subset K \subset E$, we have $m^*(F_i) \leq m^*(E)$ for all $i$, so
$\sum_{i=1}^\infty m^*(F_i) \leq \sum_{i=1}^\infty m^*(E) = \infty$.

Therefore, $m^*(E) \leq \infty$, and since $E$ has finite outer measure, we must have $m^*(E) < \infty$.

Now, by monotonicity of outer measure, we have $m^*(F_i) \leq m^*(E)$ for all $i$, so
$\sum_{i=1}^\infty m^*(F_i) \leq \sum_{i=1}^\infty m^*(E) = m^*(E) < \infty$.

Since the series converges, we can rearrange the terms to get
$\sum_{i=1}^\infty m^*(F_i) = m^*(E) - m^*(N)$.

But since $m^*(N)=0$, we have
$\sum_{i=1}^\infty m^*(F_i) = m^*(E)$.

Therefore, $m^*(F)=m^*(K)=m^*(E)$, as desired.