Definition of measurability
By James Williams
The alternative definition of measurability is:
A subset $E \subset \mathbb{R}^n$ is measurable if for every $\epsilon > 0$ there is a closed set $F \subset E$ such that $m^*(E\setminus F)<\epsilon$.
I am asked to show that this is equivalent to the usual definition of measurability:
A subset $E \subset \mathbb{R}^n$ is measurable if for every $\epsilon > 0$ there is an open set $O \supset E$ such that $m^*(O \setminus E)<\epsilon$.
My approach is:
If $E$ is measurable, then $E^c$ is measurable. So for all $\epsilon > 0$, there exists a closed $F \subset E^c$ such that $m^*(E^c-F)<\epsilon$. Let $O=F^c$
- Since $F$ is closed, $O=F^c$ is open.
- If $F \subset E^c$, then $O=F^c \supset E$.
- Finally, $E^c-F=F^c-(E^c)^c=F^c-E=O-E$, which implies $m^*(O-E)=m^*(E^c-F)<\epsilon$.
Other direction:
Suppose for all $\epsilon > 0$ there exists an open $O \supset E^c$ such that $m^*(O-E)<\epsilon$. This means $E$ is measurable, which in turn means $E^c$ is measurable. So there exists an open $O \supset E^c$ such that $m^*(O-E^c)<\epsilon$. Let $F=O^c$. If $O$ is open, then $F=O^c$ is closed. If $O \supset E^c$, then $F=O^c \subset E$. Finally, $$O-E^c=(E^c)^c-O^c=E-O^c=E-F,$$ which means $m^*(E-F)=m^*(O-E^c)<\epsilon$.
$\endgroup$1 Answer
$\begingroup$Your idea is on the right track.
To prove this equivalence, first suppose $E$ is measurable using the closed set definition.
We need to show the open set definition holds. Let $\epsilon > 0$. Since $E$ is measurable, so is $E^{c}$. But $E^{c}$ measurable means there is some closed $F \subseteq E^{c}$ such that $m(E^{c} - F) < \epsilon$.
But if $F \subseteq E^{c}$, then $E \subseteq F^{c}$, and since $F$ is closed, $F^{c}$ is open. Finally, $m(E^{c} - F) = m(E^{c} \cap F^{c}) = m(F^{c} - E)$, and so since $m(E^{c} - F) < \epsilon$, we have $m(F^{c} - E) < \epsilon$. So, given an arbitrary $\epsilon > 0$, we found an open set $O = F^{c}$ containing $E$ such that $m(O - E) < \epsilon$, and this was exactly the definition of measurability using the open sets.
You should try proving the other direction (assume the definition for open sets, and prove it for closed sets), which is very similar, on your own.
$\endgroup$ 4
