Proof. WLOG, we assume that $f$ is real valued and is nonnegative. Let $S_0$ denote the preimage of the interval $[1,\infty)$ and, for every positive integer $n$ , let $S_n$ denote the preimage of the interval $\left[\frac{1}{n+1},\frac{1}{n}\right)$ . Since the integral of $f$ is bounded, each $S_n$ can be at most finite. Taking the union of all the $S_n$ , we get the support $\displaystyle \supp f = \bigcup_{n=0}^\infty S_n$ . Thus, $\supp f$ is a union of countably many finite sets and hence is countable.