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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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    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 $\mathcal{F}_0 \cup \mathcal{N}$.2011-11-07