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This is a problem from the S.-T YAU COLLEGE MATH CONTEST in 2012. The original problem is

Let $f(x)$ be a real measurable function defined on $[a,b]$. Let $n(y)$ be the number of solutions of the equation $f(x) = y$. Prove that $n(y)$ is a measurable function on $\mathbb{R}$.

Later it was proven that the condition $f$ is measurable is not strong enough, and counterexamples could be easily constructed using the Cantor-Lebesgue function. However, I still find it difficult to tackle if $f$ is assumed continuous, in which condition this proposition holds.

By the continuity of $f$ we know that the set of roots of $f(x)=y$ for any given $y$ should be closed, but I don't know how to get around from this to the measurability of $n(y)$. Any help will be appreciated.

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Suppose that $f : [a,b] \to \mathbf R$ is continuous.

For each $k \ge 1$ let $P_k$ be a partition of $[a,b]$ into intervals with the property that $P_{k+1}$ refines $P_k$ and $\max\{ \mathrm{diam\ } E \mid E \in P_k\} \to 0$.

For example, if $[a,b] = [0,1]$ you could take \begin{align*}P_1 &= \{[0,1]\} \\ P_2 &= \{[0,1/2),[1/2,1]\} \\ P_3 &= \{[0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1]\} \end{align*} et cetera.

For each $y \in \mathbf R$ the sum $$f_k(y) = \sum_{E \in P_k} \chi_{f(E)}(y)$$ counts the number of intervals in the partition $P_k$ containing a solution $x \in [a,b]$ to $f(x) = y$.

Since $P_{k+1}$ refines $P_k$ it is clear that $f_k \le f_{k+1}$.

If $n(y)$ is finite, the fact that the diameters of the intervals in $P_k$ are tending to $0$ implies that $n(y) = f_k(y)$ for all sufficiently large $k$. Likewise if $n(y) = \infty$, then $f_k(y) \nearrow \infty$. In both cases you obtain $f_k(y) \nearrow n(y)$.

Finally, the fact that each $E \in P_k$ is an interval implies that each set $f(E)$ is an interval, and therefore each $f_k$ is Lebesgue--and in fact Borel--measurable. (This argument can be done in greater generality using Borel partitions).

It follows that $n$ is a limit of measurable functions, hence measurable.

Note: the increasing nature of the convergence means that $$\int_{f([a,b])} n(y) \, dy = \lim_{k \to \infty} \sum_{E \in P_k} m(f(E))$$ which leads to some nice formulas if you make some assumptions on the variation of $f$.

Umberto P.
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  • Thank you so much for this detailed explanation! I thought about using mesurable functions to approximate $n(y)$ but failed to construct one like this. – Apocalypse Mar 27 '19 at 00:34