Measure space and measurable function

Question

Let $f :\mathbb R\rightarrow \mathbb R$ is a continuous function then the set $\{x \in \mathbb R : \mu ((f^{-1}(x)) >0 \}$ has a zero measure. I think in the case, if f is a step function this will hold. In general measurable functions I couldn't get any idea how to begin. Hints are appreciated !!!

Answer 1

Your set looks countable to me. If I parsed it correctly, the hint is that any family of pairwire disjoint sets of positive measure is countable.

Answer 2

Denote $$ X_k=\left\{x\in\mathbb{R}\colon\ \mu(f^{-1}(x)\cap[-k,k])\ge\frac1k\right\}. $$ Then $$ X=\{x\in\mathbb{R}\colon\ \mu(f^{-1}(x))>0\}=\bigcup_{k=1}^{\infty} X_k. $$ Fix $k$ and let $x_1, \ldots,x_m$ be different points from $X_k$. Since $f^{-1}(x_j)$ are disjoint we have $$ 2k=\mu([-k,k])\ge\sum_{j=1}^m\mu(f^{-1}(x_j)\cap[-k,k])\ge\frac{m}{k}, $$ that is $m\le 2k^2$. It means that the set $X_k$ contains at most $[2k^2]$ points, i.e. is finite, and $\mu(X_k)=0$ for every $k$. Therefore, $\mu(X)=0$.