- #1

joypav

- 151

- 0

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??