Let $f$ be a measurable function on real numbers and $g$ is a monotonic continuous function on real numbers. Is the function composition $f \circ g$ Lebesgue measurable? Thanks.


In general, the answer is no. But if you have an extra condition that $g^{-1}$ is Lipschitz, the answer is yes.

Recall that, if $h$ is Lipschitz, then $\mu(A) = 0 \Rightarrow \mu(h(A)) = 0$ (you can try proving this). Now we can express $f^{-1}(A)$ as a disjoint union of $B$ and $C$ where $B$ is borel measurable and $C$ has measure zero. So we will have,

$$f \circ g \, (A) = g^{-1}(f^{-1}(A)) = g^{-1}(B \cup C) = g^{-1}(B) \cup g^{-1}(C)$$

$g^{-1}(B)$ is borel and $g^{-1}(C)$ has measure zero since $g^{-1}$ is Lipschitz, hence proving that $f \circ g$ is measurable.

  • $\begingroup$ is condition of Lipschitz necessary? $\endgroup$ – nim Mar 23 '13 at 17:53
  • $\begingroup$ @safa: Yes - if you want the composition to be measurable. $\endgroup$ – user62089 Mar 23 '13 at 18:21
  • $\begingroup$ let for any set measure zero example C ,function g inverse C be Lebesgue measurable. is composition fog Lebesgue measurable? thanks $\endgroup$ – nim Mar 23 '13 at 18:26
  • $\begingroup$ @safa: What has this got to do with the composition being lebesgue measurable. Can you clarify as to which part of the solution are you referring to? $\endgroup$ – user62089 Mar 23 '13 at 20:15
  • $\begingroup$ @safa: That's exactly my point. The statement in the current form that you have stated is not true. Only with the lipschitz assumption is it true. And $g^{-1}$ taking null sets to null sets is an off shoot of the assumption. In general why would $g^{-1}$ take map null sets to null sets. I have not come across any property that would make it do so. If there is any, do let me know. $\endgroup$ – user62089 Mar 24 '13 at 3:35

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