how to prove $\int{f}d\mu=\sum_{x\in\Omega}f(x)$

Question

Prove that $\int{f}d\mu=\sum_{x\in\Omega}f(x)$ when $f$ is absolutely summable, where $\mu$ is a counting measure on the measure space $(\Omega,\mathscr{F})$. Can someone give me hints?

Answer 1

Firstly, we see that if there are uncountably many points of $\Omega$ for which $|f(x)|>0$, then there are infinitely many so that $|f(x)|>{1\over n}$ for some $n\in\Bbb N$, i.e. $f$ is not integrable. So we may, WLOG, assume $\Omega$ is countable--since only points where $f(x)\ne 0$ are counted in integration. But then we may write $\Omega =\{x_n\}_{n\in\Bbb N}$. We have that $\Omega$ is a countable, disjoint union

$$\Omega=\coprod_{n=1}^\infty \{x_n\}$$

so that

$$\int_{\Omega}f\,d\mu=\sum_{n=1}^\infty\int_{\{x_n\}}f\,d\mu=\sum_{n=1}^\infty f(x_n)=\sum_{x\in\Omega}f(x).$$