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Suppose $B$ is a Borel set and $f$ mapping from $B$ to $R$ is a non decreasing function.Prove that $f(B)$ is a Borel set.

f is a Borel measurable function because the inverse image of $(a, \infty)$ will be of the form $(t, \infty)$ or $[t,\infty)\cap B$ which is a Borel set. I cannot define inverse function because it may not be strictly increasing. How do I proceed?

mathgirl
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    This is surely overkill, but there is a theorem in Descriptive Set Theory asserting that the image of a Borel set under an injective function is also Borel. A reference for this is Kechri's Classical Descriptive Set Theory – Reveillark Aug 11 '20 at 22:32
  • So if its not injective, f(B) might not be a borel set? – mathgirl Aug 11 '20 at 22:35
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    No, this can fail even for continuous $f$. This was a famous mistake done by Lebesgue in the early 20th century. – Reveillark Aug 11 '20 at 22:37
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    @pooja - That's correct, but notice monotone maps are always injective. $x \neq y$ implies $x < y$ (without loss of generality) so $f(x) < f(y)$. In particular, $f(x) \neq f(y)$ – HallaSurvivor Aug 11 '20 at 22:38
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    Monotone doesnt mean strictly increasing function.even constant functions come under increasing functions – mathgirl Aug 11 '20 at 22:40
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    I should clarify that when I said "injective function" in my first comment, I meant injective Borel function. Also, this is Theorem 15.1 in Kechris – Reveillark Aug 11 '20 at 22:44
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    I looked around to see if I could find a proof that doesn't rely on the descriptive set theoretic fact that @Reveillark mentioned, and I was unsuccessful. Here's a MSE post that does it in detail, though. – HallaSurvivor Aug 11 '20 at 23:26
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    A proof of the result mentioned by Reveillark can be found in Cohn's Measure Theory. – Kavi Rama Murthy Aug 11 '20 at 23:32

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