Measure theory question involving a $\sigma$ finite measure space

Question

Q/ Let $f: X \rightarrow \bar{\mathbb{R}}$ be measurable on a $\sigma$-finite measure space $(X,\mathscr{A},\mu)$. Show that the set $\{x\in \mathbb{R} :\mu(f^{-1}(x))>0\}$ is countable.

So since we know we can write $X=\cup_1^{\infty} A_i$ with $A_i \in \mathscr{A}$ and $\mu(A_i)<\infty$ for all i.

I spent a while looking at it without really getting anywhere, I can't really see the connection between the measurability of f and the sigma finiteness of the measure space, I am assuming that there is more to the measurability of f than just the fact it means that set makes sense. I thought possibly about writing f as the limit of simple measurable functions as they only have a finite number of values they each put out but couldn't really make it go anywhere. It was only a brief thought. I don't really have time to spend ages looking at it no matter how much I want to so a small to medium push in the right direction would be appreciated.

Answer 1

Whenever you are asked to show a property holds for a $\sigma$-finite space, it is a good idea to show that the property holds for finite measure spaces first, then use the $\sigma$-finiteness.

So let's assume for now that $\mu(X) < \infty$. We can write $$X = \bigcup_{y\in\overline{\mathbb{R}}} f^{-1}(y)~;$$ this is just basic set theory. Moreover, the union is disjoint.

Suppose now that $\{y\in \overline{\mathbb{R}}: \mu(f^{-1}(y)) > 0\}$ is uncountable. Then there are uncountably many preimages $f^{-1}(y)$ in the above union with nonzero measure. A union of uncountably many disjoint sets of nonzero measure cannot have finite measure. (See Set of Finite Measure: Uncountable disjoint subsets of non-zero measure if you are unfamiliar with this fact.) But this contradicts our assumption that $\mu(X)<\infty$. Therefore $\{y\in\overline{\mathbb{R}}: \mu(f^{-1}(y))>0\}$ must be countable.

Now that we have this for $X$ of finite measure, the generalization to $\sigma$-finite $X$ is not difficult, using the familiar Cantor diagonalization rule: "a countable union of countable sets is countable."