Jtdylktuy

I am reading Bogachev's book "Measure Theory" which is in my opinion is very good book on measure theory. Let me ask you the following question:

Definition: Let $mu$ - non-negative set function with domain $mathcal{A}subset 2^X$. The set $A$ is called $mu$-measurable if
for any $varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$
such that $mu^*(Atriangle A_{varepsilon})<varepsilon,$ where
$mu^*$-outer measure.

The set of all $mu$-measurable sets is denoted by
$mathcal{A}_{mu}$.

One thing began to confuse me during reading. When author takes some set $Ain mathcal{A}_{mu}$ and then he considers $mu(A)$. This confuses me a lot.

We know that $mathcal{A}subset mathcal{A}_{mu}$ and when $Ain mathcal{A}_{mu}$ it means that $A$ is $mu$-measurable set. But $mu$ is defined only on $mathcal{A}$. Why he does not consider $mu^*(A)$?

More precisely, if $A$ is $mu$-measurable ($Ain mathcal{A}_{mu}$) then why $mu(A)$ makes sense? and does $mu(A)=mu^*(A)$?

P.S. I know the fact that $mathcal{A}_{mu}$ is $sigma$-algebra and $mu$ can be uniquely extended from algebra $mathcal{A}$ to $sigma$-algebra $mathcal{A}_{mu}$. And this extension is given by $mu^*$.

I would be very grateful if somebody can detailed answer. Because it is important two understand such things.

EDIT: I will divide my questions into 3 parts.

1) Corollary 1.5.8. (page 21).

I was not able to understand this part. He just uses the definition of outer measure $mu^*$ but not the definition of $mu$-measurability.

My approach: Since $A$ is $mu$-measurable then $forall varepsilon>0$ there exists $A_{varepsilon}in mathcal{A}$ such that $mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using monotonicity of $mu^*$ we have $mu^*(A-A_{varepsilon})leq mu^*(Atriangle A_{varepsilon})<varepsilon$. Then using subadditivity of $mu^*$ we have $mu^*(A)-mu^*(A_{varepsilon})leq mu^*(A-A_{varepsilon})<varepsilon.$ But since $A_{varepsilon}in mathcal{A}$ then $mu^*(A_{varepsilon})=mu(A_{varepsilon})$ so $mu^*(A)-mu(A_{varepsilon})<varepsilon$. We have that $mu^*(A)$ is finite so we can apply the definition of outer measure. Is it correct? But I am sure that it is OK.

2)

He takes $Bin mathcal{A}_{mu}$ but why $mu(B)$ makes sense? $mu(B)$ is defined only on $malcal{A}$ but $B$ may not be in $mathcal{A}$. This also confusing me a lot.

3)

The same question $B$ and $C$ are $mu$-measurable sets but why $mu(B)$ and $mu(C)$ makes sense?

I guess that all these questions are related to each. So would be very grateful for detailed answer and help!

EDIT 2: Also one moment which I have forgot to ask in the previous questions:

In the above proof note two moments which I have underlined with red line. Why Bogachev writes $mu^*(A)=mu^*(A'')$ but in he writes just $mu(B)=mu^*(X-A)$. The first is OK and I am agree with that but in the second he omits $mu^*$ and just write $mu(B)$. Could you explain it, please?

In 1) he does indeed not use the definition of $mu$-measurability, he refers to the definition of the outer measure.

2) He just started to denote the unique nonnegative countable additive extension $mu^*$ of $mu$ by $mu$ itself. Indeed, he does so already before definition 1.5.10 when he calls $(X,mathcal{A}_mu,mu)$ the Lebesgue completion of $(X,mathcal{A},mu)$. Using the same name for a function and an extension of it is a widespread harmless abuse of notation in much of mathematics.

3) Same as 2)