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Let $h:[0,\infty)\to\mathbb{R}$ be a monotone function, with $\int_0^\infty |h(x)|x^2\,dx<\infty$.

And let $f:\mathbb{R}^3\to\mathbb{R}$ with $f(x)=h(|x|)$ for all $x$.

Prove that $f$ is (Lebesgue) measurable on $\mathbb{R}^3$.


I tried several techniques but did not manage to prove.

Try 1) $h$ is monotone function, thus continuous almost everywhere. $g(x)=|x|$ is a continuous function. However $f=hg$ is not necessarily continuous almost everywhere.

Try 2) I know that if $g$ is continuous and $h$ measurable, then $gh$ is measurable. Unfortunately, the order is wrong, we need $hg$ measurable.

Thanks for any help!

Another thing is that $g(x)=|x|$ is Lipschitz, but again that doesn't seem to help as we need $g^{-1}$ Lipschitz instead.

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    $\begingroup$ Monotone functions are Borel measurable, but is it true that they are Lebesgue measurable as well? The example that I am suspecting is the right-inverse $f$ of the Cantor-Lebesgue function $f:[0,1]\to[0,1]$. It is strictly increasing and has measure-zero image. So the image of a non-measurable subset $A$ of $[0,1]$ under $f$ is a Lebesgue null-set and thus Lebesgue measurable. $\endgroup$ – Sangchul Lee Dec 9 '16 at 8:56
  • $\begingroup$ I thought it is true that monotone functions are Lebesgue measurable: math.stackexchange.com/questions/662099/… $\endgroup$ – yoyostein Dec 9 '16 at 9:02
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    $\begingroup$ I forgot that Lebesgue measurable function actually means 'Lebesgue-to-Borel' measurable function and mistakenly thought that the target space should also be endowed with Lebesgue $\sigma$-algebra. It is well-explained in this Wikipedia article. I must have been working with only Borel-measurability too long to forget this caveat... $\endgroup$ – Sangchul Lee Dec 9 '16 at 9:07
  • $\begingroup$ For this question, the definition that I would take is that $f$ is measurable iff $f^{-1}(G)$ is (Lebesgue) measurable for every open $G\subseteq\mathbb{R}$. $\endgroup$ – yoyostein Dec 9 '16 at 9:09
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This can be proved "by hand" in a couple of ways: One would be to notice that since $f$ is monotone, $f^{-1}((\alpha,\infty))$ is an interval. Once we have that, it's simple to finish. Another way would be to recall that $f$ has at most a countable number of discontinuities. It follows that $f\circ a$ is continuous on $\mathbb R^n$ minus a countable number of spheres. It's easy to finish from there.

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With $g(x)=|x|$ , we have that $g$ is continuous, hence measurable. $h$ is monotone, therefore $h$ is measurable. It follows that $f = h \circ g$ is measurable.

In this case, what saves this argument is that both $h$ and $g$ are Borel measurable, so the composition is also Borel measurable. – Sangchul Lee

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  • $\begingroup$ I thought the composition of measurable functions may not be measurable? math.stackexchange.com/questions/283443/… $\endgroup$ – yoyostein Dec 9 '16 at 8:47
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    $\begingroup$ @yoyostein, True, and it is a mistake that Lebesgue himself committed. $\endgroup$ – Gary Moore Dec 9 '16 at 8:48
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    $\begingroup$ In this case, what saves this argument is that both $h$ and $g$ are Borel measurable, so the composition is also Borel measurable. $\endgroup$ – Sangchul Lee Dec 9 '16 at 9:12
  • $\begingroup$ @SangchulLee Yes indeed, thanks! $\endgroup$ – yoyostein Dec 9 '16 at 9:14
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    $\begingroup$ @SangchulLee Nothing saves the argument in the first paragraph. $\endgroup$ – zhw. Dec 9 '16 at 22:16

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