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I'm wondering if there's an example of a measure $\mu$ on a $\sigma$-algebra $\mathcal{F}$ that is $\sigma$-finite on $\mathcal{F}$ but not on $\mathcal{F}_0$, the field that generates $\mathcal{F} := \sigma\langle\mathcal{F}_0\rangle$?

Also, a related question: An example when $\mu$ is $\sigma$-finite on $\mathcal{F}_0 \cup \mathcal{N}$ where $\mathcal{N} = \{F \in \mathcal{F}: \mu[F] = 0\}$, but not on $\mathcal{F}_0$? And an example where $\mu$ is $\sigma$-finite on $\mathcal{F}$ but not on $\mathcal{F}_0 \cup \mathcal{N}$?

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Yes, consider the case where $\mathcal{F}$ is the Borel $\sigma$-algebra on the reals, $\mathcal{F}_0$ is the collection of finite unions of half-open intervals $(a,b]$ ($a,b\in\mathbb{R}\cup\{\pm\infty\}$) and $\mu(S)$ just counts the number of rationals in a set $S$.

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How is $\mu$ $\sigma$-finite on $\mathcal{F}$? I'm thinking that you mean to divide $\mathbb{R}$ into countable union of sets, each with only 1 rational point? But how to do that? –  user18115 Nov 6 '11 at 23:51
    
@user18115: The rationals are countable, $\mathbb{Q}=\{q_1,q_2,\ldots\}$. Let $A_n$ be the union of the irrational numbers together with $\{q_1,\ldots,q_n\}$. Then, $\mu(A_n)=n < \infty$ and $\bigcup_nA_n=\mathbb{R}$. –  George Lowther Nov 6 '11 at 23:54
    
That helps a lot. Thanks! I'm thinking that this answer can also be applied to my followed up question: an example where $\mu$ is $\sigma$-finite on $\mathcal{F}$ but not on $\mathcal{F}_0 \cup \mathcal{N}$? How about the other one, an example when $\mu$ is $\sigma$-finite on $\mathcal{F}_0 \cup \mathcal{N}$ but not on $\mathcal{F}_0$? –  user18115 Nov 7 '11 at 0:12
    
...now your question has changed, and I'm not sure why you would want to consider $\mathcal{F}_0\cup\mathcal{N}$ which is not even an algebra of sets. –  George Lowther Nov 7 '11 at 1:20
    
I guess it's because of this result: $\alpha[B] := \inf\{\mu[B \triangle A]: A \in \mathcal{F}_0\} = 0$ with all $B \in \mathcal{F}$ such that $\mu[B] < \infty$ if and only if $\mu$ is $\sigma$-finite on $\mathcal{F}_0 \cup \mathcal{N}$. That makes me wonder the difference between being $\sigma$-finite on $\mathcal{F}_0$ and being $\sigma$-finite on $\mathcal{F}_0 \cup \mathcal{N}$. –  user18115 Nov 7 '11 at 5:48

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