proof verification - Is this a $sigma$-finite measure?

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?

Answer

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.