Example of $\sigma$-finite extension to non-$\sigma$-finite functional I asked this question this morning. The question there is pretty much answered, since I was missing a little detail that answered. But @Ramiro, the commenter who pointed that detail out, just told me that:

The extension being $\sigma$-finite does NOT ensure the original $\mu$ is $\sigma$-finite.

Here is the context. We have a ring of sets $\mathcal{A}\subseteq\mathcal{P}(X)$, that is a collection of subsets of a set $X$ which is stable under finite intersections, finite unions, and relative complement, and I think must also contain $\varnothing$. In fact, $\varnothing\in\mathcal{A}$ as a consequence of relative complements, since $A\smallsetminus A=\varnothing$. Anyways, we are talking of extending a $\sigma$-additive positive functional $\mu:\mathcal{A}\to\mathbb{R}^+$ to a measure on $\sigma(\mathcal{A})$, the minimal $\sigma$-algebra on $X$ containing $\mathcal{A}$. So if an extension $\tilde\mu$ is $\sigma$-finite, $\mu$ can apparently still be non-$\sigma$-finite. Can you give me an example of a situation where $\mu$ is not $\sigma$-finite though $\tilde\mu$ is?
 A: Here is an exemple. Let $X$ be $\mathbb{N} \cup \{\infty\}$.  Let $\mathcal{A}$ be defined as 
$$
\mathcal{A}=\{C\subseteq \mathbb{N} \:|\: C \textrm { is finite } \} \cup \{D\cup \{\infty\} \:|\: \mathbb{N} \setminus D \textrm { is finite}  \} 
$$
It is easy to prove that $\mathcal{A}$ is a ring (in fact, an algebra). Consider $\mu$ the counting measure defined on $\mathcal{A}$. Then, $\mu$ is not $\sigma$-finite.
On the other hand, we know that $\sigma(\mathcal{A})=P(X)$. For each $r$ positive (finite) real number, let $\nu_r$ be the measure defined on $P(X)$ by $\nu_r(\{a\})=1$, if $a\neq \infty$ and $\nu_r(\{\infty\})=r$. All those $\nu_r$ measures (there are uncountable many of them) are $\sigma$-finite extensions of $\mu$.
Remark 1: 


*

*$\mathcal{A}$ is a ring (in fact, an algebra).


Just check directly the definition of ring and algebra. $\mathcal{A}$ is closed under set difference and (finite) union. Also $X \in \mathcal{A}$. 


*$\mu$ is not $\sigma$-finite. 


Let $\{E_n \in  \mathcal{A} \:|\: n \in \mathbb{N}\}$ be a countable family of sets in $\mathcal{A}$ such that $\bigcup_n E_n = X$. Then, there is $k \in \mathbb{N}$ such that $\infty \in E_k$ and so  $\mu(E_k)=+\infty$. So no countable union of set of finite measure $\mu$ can     cover $X$.


*$\sigma(\mathcal{A})=P(X)$.


Clearly, we have that  $\sigma(\mathcal{A}) \subseteq P(X)$. So, all we need to prove is that $P(X) \subseteq \sigma(\mathcal{A})$. 
Firstly note that, for all $a \in \mathbb{N}$, $\{a \} \in  \mathcal{A} \subseteq \sigma(\mathcal{A})$. Now, for all  $n \in \mathbb{N}$, let $E_n= \{a\in \mathbb{N} \:|\: a > n\}\cup \{\infty\}$. It is easy to see that $E_n \in \mathcal{A}$ and $\{\infty \}= \bigcap_n E_n$. So $\{\infty \} \in \sigma(\mathcal{A})$.
So, for all $a \in X$, $\{a \} \in \sigma(\mathcal{A})$. But, since $X$ is countable, any set $E \in P(X)$ is a countable union of singletons, so $P(X) \subseteq \sigma(\mathcal{A})$. 
Remark 2:
If we consider $\mu^*$, the outer measure induced be $\mu$, we have that  $\mu^*$ is, in fact, a measure on $P(X)$ and $\mu^*(\{a\})=1$, if $a\neq \infty$ and $\mu^*(\{\infty\})=+\infty$. 
Note that $\mu^*$ extends $\mu$ and note also that $\mu^*$ is NOT $\sigma$-finite. 
It is easy to see that, for all $r$ positive (finite) real number, $\mu^* \neq \nu_r$. 
