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Let $(\Omega,\mathcal{A},\mu)$ be a measure space and $\mu$ $\sigma$-finite. Let $h$ be a probability density on $(\Omega,\mathcal{A},\mu)$. Consider the measure $h\mu(A):=\int_{\Omega}h1_A\, d\mu$ on $(\Omega,\mathcal{A})$. Is $h\mu$ a $\sigma$-finite measure?

Hello, do not know if one needs it here but we had a criterion for that in our lecture:

A measure $\mu$ on a measurable space $(\Omega,\mathcal{A})$ is $\sigma$-finite exactly then if there is a strictly positive function $f\in\mathcal{L}_{\mu}^1$.

Approved to this situation if I want to show that $h\mu$ is $\sigma$-finite, I have to find a striclty positive function $f\in\mathcal{L}_{h\mu}^1$.

That $h$ is a probability density means that $$ \int_{\Omega}h\, d\mu=1, $$ right? Then my idea is to use a constant function $f(x)\equiv c$ for $c>0$. Then $f$ is strictly positive, it is measurable and furtermore $$ \int_{\Omega}\lvert f\rvert\, d(h\mu)=c\cdot\int_{\Omega}h\, d\mu=c<\infty $$

Is that already the proof that $h\mu$ is a $\sigma$-finite measure?

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  • $\begingroup$ You just need the definition of $\sigma$-finiteness for this proof. $\endgroup$
    – mez
    Jan 18, 2014 at 12:41
  • $\begingroup$ @mezhang I added my idea. PS: I do not understand your first objection. The measure $h\mu$ is defined for $A\in\mathcal{A}$. $\endgroup$
    – user34632
    Jan 18, 2014 at 12:42

2 Answers 2

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Recall the definition of $\sigma$-finiteness. Since $(\Omega, \mathcal{A},\mu)$ is $\sigma$-finite, there exist $\{\Omega_i\}$ countable family such that $\Omega = \bigcup_i \Omega_i$ and $\mu(\Omega_i)<\infty$. Now clearly $$h\mu(\Omega_i) = \int_\Omega h\chi_{\Omega_i} d\mu = \int_{\Omega_i} h d\mu\le 1 <\infty$$ Thus $h\mu$ is $\sigma$-finite since $\Omega$ can be covered by a countable family of $h\mu$ measure finite sets.

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  • $\begingroup$ Ok, thanks. You used the definition of $\sigma$-finite. In my idea that I added I use the characterization of $\sigma$-finite we recently had in our lecture. Can you say mé if my added proof is right? $\endgroup$
    – user34632
    Jan 18, 2014 at 12:50
  • $\begingroup$ Are you sure the criterion you have is necessary and sufficient? If it is, then your proof is correct. $\endgroup$
    – mez
    Jan 18, 2014 at 12:55
  • $\begingroup$ Yes, anyhow this is in our script. $\endgroup$
    – user34632
    Jan 18, 2014 at 12:55
  • $\begingroup$ As more or less indicated in my answer, I am not sure these steps are the best way to go (anyway they are not necessary). $\endgroup$
    – Did
    Jan 18, 2014 at 13:43
  • $\begingroup$ @Did You are right. $\endgroup$
    – mez
    Jan 19, 2014 at 3:08
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Better to go straight to the heart of the matter, I would say... By hypothesis, $h$ is a probability density hence $(h\mu)(\Omega)=1$, which shows that $h\mu$ is actually a finite (not only $\sigma$-finite) measure.

The hypothesis that $\mu$ is $\sigma$-finite is not needed. Nor is the criterion you recall in your post.

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