TITLE: $A:=\left\{y\in\mathbb{R}:\mu\left(f^{-1}(\left\{y\right\})\right)>0\right\}$ is countable, if $\mu$ is a finite measure and $f$ has compact support
QUESTION [8 upvotes]: Let

$E$ be a metric space
$\mathcal{E}:=\mathcal{B}(E)$ be the Borel algebra on $E$
$\mu:\mathcal{E}\to [0,1]$ be a measure
$f:E\to\mathbb{R}$ be measurable and have compact support

Assume that $\mu(D_f)=0$, where $$D_f:=\left\{x\in E:f\text{ is discontinuous at }x\right\}$$ How can we prove that $$A:=\left\{y\in\mathbb{R}:\mu\left(f^{-1}\left(\left\{y\right\}\right)\right)>0\right\}$$ is countable?

REPLY [7 votes]: Put $F_y=f^{-1}(\{y\})$, and $a_y=\mu(F_y)$. Let $n\geq 1$, and $A_n=\{y; a_y\geq 1/n\}$. It is clear that we have $A=\cup_{n\geq 1} A_n$. Fix $n$, and suppose that $A_n$ is infinite, and let $y_k$, $k\geq 1$ be a sequence of distincts elements in $A_n$. Then the $F_{y_k}$ are disjoints, and $\mu(F_{y_k})\geq 1/n$. Hence if $F=\cup F_{y_k}$, we get $\mu(F)=\sum\mu(F_{y_k})=+\infty$, a contradiction, as $\mu$ is finite. Hence all $A_n$ are finite, and $A$ is (at most) countable.