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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.

yoyostein
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    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. – Sangchul Lee Dec 09 '16 at 08:56
  • I thought it is true that monotone functions are Lebesgue measurable: http://math.stackexchange.com/questions/662099/show-that-if-e-subset-mathbbr-is-a-measurable-set-so-fe-rightarrow-mathb – yoyostein Dec 09 '16 at 09:02
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    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... – Sangchul Lee Dec 09 '16 at 09:07
  • For this question, the definition that I would take is that $f$ is measurable iff $f^{-1}(G)$ is (Lebesgue) measurable for every open $G\subseteq\mathbb{R}$. – yoyostein Dec 09 '16 at 09:09

2 Answers2

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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

yoyostein
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Fred
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  • I thought the composition of measurable functions may not be measurable? http://math.stackexchange.com/questions/283443/is-composition-of-measurable-functions-measurable – yoyostein Dec 09 '16 at 08:47
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    @yoyostein, True, and it is a mistake that Lebesgue himself committed. – Yes Dec 09 '16 at 08:48
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    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 Dec 09 '16 at 09:12
  • @SangchulLee Yes indeed, thanks! – yoyostein Dec 09 '16 at 09:14
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    @SangchulLee Nothing saves the argument in the first paragraph. – zhw. Dec 09 '16 at 22:16
  • @zhw. What kind of issue persists, knowing that the composition of two Borel measurable functions is again Borel measurable? – Sangchul Lee Dec 09 '16 at 23:36
  • @SangchulLee None, just saying the argument in the first paragraph, as it continues to be stated, is not correct. – zhw. Dec 10 '16 at 00:15
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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.

zhw.
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