Question: If f : [a, b] → R is continuous, let M(y) be the
number of points x in [a, b] such that f(x) = y. M(y) may be
finite or infinite. Prove that M is Borel measurable
Attempted Proof: Notice we can decompose M(y). The graph of f is closed because its continuous and so M(y)= n($\pi(\cup (y, f(x)) \in Gr f$)), where n is the counting measure and $\pi$ is the projection. Since, the projection is a closed map and the graph is closed and the union of closed sets* is closed, the inside is closed and hence, borel measurable. I'm stuck on how to proceed.
Notes: (*) is only true if M(y) is finite which is not always the case. I'm stuck on how to use the counting measure. Any hints would be appreciated.
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Frog2341
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@Ramiro I know the proof doesn't work, but since [a, b] is compact, doesn't it become closed because https://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map – Frog2341 Jul 23 '24 at 14:50
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Ok. You are right. – Ramiro Jul 23 '24 at 17:27
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Hint: since $f$ is continuous, $M(y) \ge m$ if and only if there is a $\delta > 0$ such that for every $\varepsilon > 0$ there are $m$ points $x_i \in [a,b]$ with $|x_i - x_j| > \delta$ for $i \ne j$ and $|f(x_i) - y| < \varepsilon$.
Olius
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@Sm465 Are you stuck proving the hint or using it to show that $M$ is measurable? – Olius Jul 21 '24 at 22:42