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This is a question from Erhan Cinlar's Probability and Stochastics book:

If $\mu$ is absolutely continuous with respect to a finite measure $\nu$, then $\mu$ is $\Sigma$-finite.

Can i have a hint on how i should construct the sequence of finite measure that sum to $\mu$?

green onion
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2 Answers2

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I will assume $\mu$ to also be a non-negative measure (the case for a signed, or a complex measure can be easily adapted). Now by the Radon Nikodym Theorem (for the existence of the Radon Nikodym derivative it suffices to only assume that $\nu$ is $\sigma$-finite, see: https://math.stackexchange.com/a/3342557/485385) we may find $f\geq 0$ measurable such that $$\mu(A)= \int_A f d\nu$$ for all $A$, so with some fancy notation this becomes $\mu = f \odot \nu$. Now define the sequence of measurable functions $\{f_n\}$ by $f_n=\chi_{\{f \in [n, n+1)\}} f$ and consider the measures $\mu_n = f_n \odot \nu$. Can you deduce the rest?

h3fr43nd
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  • Doesn't Radon Nikodym Theorem require sigma finiteness of $\mu$? – Kavi Rama Murthy Oct 31 '20 at 12:08
  • No, RN Theorem does not require the $\sigma$-finiteness of $\mu$; only $\sigma$-finiteness of $\nu$ is required. See, for example, https://math.stackexchange.com/questions/1760977/is-sigma-finiteness-unnecessary-for-radon-nikodym-theorem – user5280911 Nov 01 '20 at 04:31
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This is false. Let $X=\mathbb N$, $\mu (\emptyset)=0$ and $\mu (E)=\infty$ for all non-empty subsets of $X$. Let $\nu (E)=\sum_{n \in E} \frac 1 {2^{n}}$. then $\nu$ is a finite measure and $\mu <<\nu$. But $\mu$ is not sigma finite.

Kavi Rama Murthy
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  • $\mu$ is indeed not $\sigma$-finite, but it is $\Sigma$-finite. The construction of such the $\Sigma$-finiteness is to use countable number of Dirac measures at each integer. Countable union of countable sets is still countable. So, Erhan Cinlar's claim in the original question may be correct, only h3fr43nd's proof is incorrect in that it assumes $\sigma$-finiteness of $\mu$. Another proof which is not based on this assumption is needed. – user5280911 Nov 01 '20 at 04:28