How can one understand Pugh about outer measures?

Question

I'm reading Pugh now and I feel a bit inconvenient about what he says about the outer measure.

He says on page 400 that $m^\ast(A)$ is the infinum of the measure of open sets that contain A. Yes, this is definition.

But then he writes: This infinum is achived by a $G_\delta$-set that contains A.

How can we get that set he denotes as $H_A$.

i don't think I am right if I say that $H_A=\bigcap G$, where $G$ contains $A$.

Answer 1

Assume $m^*(A)<\infty$ first. Use the definition of the outer measure, there are open sets $A\subset \cdots \subset U_{n+1}\subset U_n$ such that $$ m^*(A)>m (U_n)-1/n $$ for $n\in \mathbb N$. By monotone class theorem in measure theory, $U_n\subset \cdots \subset U_1$ implies that $$ m^*\left (\bigcap_{n\in \mathbb N} U_n\right)=\lim_{n\to \infty }m(U_n) $$ Thus $$ m^*(A)\geqslant m\left (\bigcap_{n\in \mathbb N} U_n \right)\tag1 $$ Clearly for any $n\in \mathbb N$, there is $$ A\subset\bigcap_{n\in \mathbb N} U_n \quad\text{and}\quad m^*(A)\leqslant m\left(\bigcap_{n\in \mathbb N} U_n\right)\tag2 $$ With $(1)$ this implies that $$ m^*(A)= m\left (\bigcap_{n\in \mathbb N} U_n \right) $$ It is proved if we let $$ H_A=G_{\delta}=\bigcap _{n\in \mathbb N}U_n\tag3 $$ If $m^*(A)=\infty$, then for each $U_n$, $m(U_n)=\infty$. By $(2)$ we have $$ m^*(A)= m\left (\bigcap_{n\in \mathbb N} U_n \right)=\infty $$ and $(3)$ still holds.

Answer 2

If $m^*(A)=\infty$ let $H_A=\mathbb R.$

If $m^*(A)\lt\infty$ then, for each $n\in\mathbb N,$ choose an open set $G_n$ such that $A\subseteq G_n$ and $m(G_n)\lt m^*(A)+\frac1n.$ (I guess the axiom of choice is needed here.) Let $H_A=\bigcap_{n\in\mathbb N}G_n.$