How can one understand Pugh about outer measures? 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 $A$.
 A: 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 $H_A=\bigcap_{n\in\mathbb N}G_n.$
A: 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.
