Let $(\Omega, 2^{\Omega}, \mu)$ be the measure space with $\mu$ the counting measure on a set $\Omega \neq \varnothing$ and $f : \Omega \to [0,\infty]$ a function. (All such functions are automatically measurable.)
Prove the following statement: If $f(x) > 0$ for uncountably many $x \in \Omega$, then $\int_\Omega f \: d \mu = \infty$.
It's an exercise I thought of myself and will answer it, too. Any comments / improvements are of course welcome.
Suppose $\int_\Omega f \: d\mu < \infty $. We prove $f^{-1}((0,\infty])$ is a countable set. First, remark that $E_1 = f^{-1} ([1,\infty])$ is finite. Indeed, if this set were infinite, we'd have $$\int_\Omega f \: d\mu = \sum_{x \in \Omega} f(x) \geq \sum_{x \in f^{-1} ([1,\infty])} f(x) \geq \sum_{x \in f^{-1} ([1,\infty])} 1 = \infty,$$ contrary to the assumption $\int_\Omega f \: d\mu < \infty $. In similar fashion one can show that $E_n = f^{-1}\left(\left[\frac{1}{n}, \frac{1}{n-1} \right) \right)$ cannot be infinite for each $n \geq 2$. Hence: \begin{align*} f^{-1} \left( \left( 0 , \infty \right] \right) &= f^{-1} \left( \bigcup_{n \geq 1} E_n \right) \\ &= \bigcup_{n \geq 1} f^{-1} \left( E_n \right), \end{align*} is a countable union of finite sets. We conclude that $f^{-1}((0,\infty])$ is a countable set.