Is the function measurable with respect to Lebesgue measure

Question

Let $h: [a,b]\rightarrow \mathbb{R}$ be a continuous function. For every $y\in \mathbb{R}$, denote $$ S(y)=\begin{cases} 0 & \text{, if } h^{-1}(y) = \emptyset \\ \# h^{-1}(y) &\text{, if } h^{-1}(y) \text{ is finite}\\ \infty, &\text{, if } h^{-1}(y) \text{ is infinite} \end{cases} $$ Is the function measurable (as a function from $\mathbb{R} \mapsto \mathbb{R}$) with respect to the Lebesgue measure in $\mathbb{R}$?

Answer 1

Here's a neat proof that $f$ is Borel-measurable:

It suffices to show that $S^{-1}(\{n\})$ is Borel for all $n \geq 0$, as the image of $S$ is contained in $\{0,1,2,\ldots\}\cup\{\infty\}$. For $n = 0$, $S^{-1}(\{0\}) = \mathbb{R} \setminus f([a,b])$, which is open. For $n \geq 1$, let $g = f\times \cdots \times f: [a,b]^n \to \mathbb{R}^n$. For $\Delta \subseteq [a,b]^n$ the set of points where at least two coordinates are equal (the "big diagonal") and $D: \mathbb{R} \to \mathbb{R}^n$ the map $D: t \mapsto (t,\ldots, t)$ (the diagonal map), I claim $$ S^{-1}(\{n\}) = D^{-1}(g([a,b]^n \setminus \Delta)).$$ As $g$ is continuous and $[a,b]^n \setminus \Delta$ is a countable union of compact sets, $g([a,b]^n \setminus \Delta)$ is Borel, and hence $S^{-1}(\{n\})$ is.