0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You just need the definition of $\sigma$-finiteness for this proof. $\endgroup$ – mez Jan 18 '14 at 12:41
  • $\begingroup$ @mezhang I added my idea. PS: I do not understand your first objection. The measure $h\mu$ is defined for $A\in\mathcal{A}$. $\endgroup$ – math12 Jan 18 '14 at 12:42
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – math12 Jan 18 '14 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 '14 at 12:55
  • $\begingroup$ Yes, anyhow this is in our script. $\endgroup$ – math12 Jan 18 '14 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 '14 at 13:43
  • $\begingroup$ @Did You are right. $\endgroup$ – mez Jan 19 '14 at 3:08
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.