6
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.