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Suppose f and g are two Borel function on [0, 1]. The push-forward of the Lebesgue measure on [0,1] by f and by g are the same. Then are there some Borel measurable function from [0,1] to [0,1], such that

f = g(h)?

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  • $\begingroup$ I know If f and g are simple functions, the answer is yes. $\endgroup$ – Banan.SUN Jul 25 '18 at 2:00
  • $\begingroup$ Do you mean $f = g\circ h$ almost everywhere? I can give a simple counterexample if you demand $f(t) = g(h(t))$ for all $t$. $\endgroup$ – Nik Weaver Jul 25 '18 at 3:45
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Assuming push-forward means this ...

Hint... Consider the two maps $f(x) = \{2x\}$ and $g(x) = \{3x\}$, where the brackets are the fractional part. What would $h$ be then?

2x

3x

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  • $\begingroup$ I thought of this too, but then I tried to show no $h$ could exist and discovered that it's easy to find an $h$ that works. $\endgroup$ – Nik Weaver Jul 25 '18 at 14:04
  • $\begingroup$ Perhaps we have to distinguish between the problem in the title (measure preserving) and the problem in the text that does not say $h$ is measure preserving. $\endgroup$ – Gerald Edgar Jul 25 '18 at 14:46
  • $\begingroup$ Ah, I didn't notice that. Yes, you certainly have a counterexample for $h$ measure preserving. The problem in the text seems more subtle --- here I feel the existence of $h$ should follow from some measurable selection theorem which could only fail for really pathological $g$, if it fails at all. $\endgroup$ – Nik Weaver Jul 25 '18 at 15:56
  • $\begingroup$ For the case where $h$ need not be measure preserving, the hypothesis that $f, g$ have the same distribution is irrelevant: it is enough that $f$ and $g$ have the same range. But, as Nik notes, the selection $h$ may fail to be Borel measurable: but it may be chosen universally measurable. $\endgroup$ – Gerald Edgar Jul 25 '18 at 16:53

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