Carathéodory's theorem and $\sigma$-finite measures

Question

In class I was given this version of Carathéodory's theorem:

If $(\Omega,R,\mu)$ is a pre-measure space, the there is $\mu'$ that extends $\mu$ to $\sigma(R)$ such that $(\Omega,\sigma(R),\mu')$ is a measure space.

Now, there is an additional statement regarding this theorem that says:

if $\mu$ is $\sigma$-finite then $\mu'$ is unique and $\sigma$-finite too

I'm having touble with the "$\mu'$ is $\sigma$-finite too" part.

So far this is what I've got:

First I prove this lemma

Let $v$ be another measure in $\sigma(R)$ that extends $\mu$, then $v(E)\leq \mu'(E)$ for every $E\subseteq\sigma(R)$, and $v(E)= \mu'(E)$ only when $\mu'(E)<\infty$

Let $E\in\sigma(R)$ such that $E\subseteq\bigcup_{j=1}^{\infty}A_j|A_j\in R$, then

$v(E)\leq v(\bigcup_{j=1}^{\infty}A_j)=\sum_{j=1}^{\infty}v(A_j)=\sum_{j=1}^{\infty}\mu(A_j)=\sum_{j=1}^{\infty}\mu'(A_j)=\mu'(\bigcup_{j=1}^{\infty}A_j)$

which, in turn, means that $v(E)\leq\mu'(E)$.

Now, suppose $\mu'(E)<\infty$ and define $A:=\bigcup_{j=1}^\infty A_j$, then

$v(A)=\lim_{n\to\infty} v(\bigcup_{j=1}^n A_j)=\lim_{n\to\infty} \mu'(\bigcup_{j=1}^n A_j)=\mu'(A)$

If one chooses $A_j$'s such that $\mu'(A)<\mu'(E)+\epsilon$ then one would have that $\mu'(A-E)<\epsilon$ and

$\mu'(E)\leq\mu'(A)=v(A)=v(E)+v(A-E)\leq v(E)+\mu(A-E)\leq v(E)+\epsilon$

So $\mu'(E)=v(E)$ since $\epsilon$ is arbitary.

$\Box$

Now that the lemma is proven one can take $\Omega=\bigcup_{j=1}^{\infty} A_j$ with $\mu(A_j)<\infty$ and disjoint $A_j$'s, and

$\mu'(E)=\sum_{j=1}^\infty \mu'(E∩A_j)=\sum_{j=1}^\infty v(E∩A_j)=v(E)$

$\Box$

Now I need to use what I have proven (or maybe not?) to prove that $\mu'$ is $\sigma$-finite, this is, there are $\{A_n\}_{n=1}^\infty\subseteq\sigma(R)$ such that $\Omega=\bigcup_{n=1}^\infty A_n|\mu'(A_n)<\infty$

I've been trying but I'm stuck, any ideas or help?

Thanks in advance.

Answer 1

$\mu$ being $\sigma$-finite means that there are $A_1, A_2, \ldots \in R$ s.t. $\cup_n A_n = \Omega$ and $\mu(A_n)< \infty$ for all $n \in \mathbb{N}$. That $\mu'$ is $\sigma$-finite follows simply, since $A_1, A_2, \ldots \in R \subset \sigma(R)$ and, since $\mu'$ is an extension of $\mu$ it agrees on R. This means $\mu(A_n) = \mu'(A_n) < \infty$.

So all in all: there are $A_1,A_2, \ldots \in \sigma(R)$ s.t. $\cup_n A_n = \Omega$ and $\mu'(A_n) < \infty$ for all $n \in \mathbb{N}$ $\Longrightarrow \mu'$ is $\sigma$-finite

Edit: Typo