On conditions for the existence of $h\in L^1$ such that $h>0$ a.e

Question

I've been reading the chapter of Uniform Integrability on Probability Theory by Achim Klenke which has the following proposition.

If $\mu$ is $\sigma$-finite, then there is a function $h \in L^{1}(\mu)$ such that $h > 0$ almost everywhere.
In the proof, he constructed an increasing sequence of sets $A_{n}$ with limit being the whole space $\Omega$ and each element possessing finite measure, i.e. $\mu(A_{n}) < \infty$.

I am not sure whether this is only for $\mu$ being $\sigma$-finite, since in the following theorem he showed result similar to the Vitali convergence theorem but without the condition that restrict set measure.

Answer 1

It is false without assuming that the measure is $\sigma$-finite. Let $\mu$ be the counting measure on $\mathbb{R}$. Then $h>0$ a.e. means $h>0$ everywhere and clearly $$ \int_{\mathbb{R}}hd\mu=\sum_{x\in\mathbb{R}} h(x)=+\infty. $$

Answer 2

This is a characterization of $\sigma$-finiteness: If $h$ is an $L^1$-function with values in $(0,\infty)$ then $A_n=\{h\ge 1/n\}$ are measurable sets with $\bigcup_{n\in\mathbb N} A_n=\Omega$ and $\mu(A_n)=\int I_{A_n}d\mu \le \int nh d\mu <\infty$.

If $h$ is only almost everywhere strictly positive you may replace it by $\tilde h= hI_{\{h>0\}} + I_{\{h\le 0\}}$.