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Let $(X, \Sigma )$ be a measurable space and $f:X \rightarrow \mathbb R$ a measurable function. If $h: \mathbb R \rightarrow \mathbb R $ is continuous, then prove that $h \circ f$ is a measurable function.

I unfortunately lost my lecture notes so I really need some help with this module (linear analysis).

Please can someone guide me on how to do this question.

Thanks

janmarqz
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snowman
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2 Answers2

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Since $h$ is continuous, $h^{-1}((-\infty ,a))$ is open and thus borelien. Since $f$ is mesurable and $h^{-1}((-\infty ,a))$ is borelien, $f^{-1}(h^{-1}((-\infty ,a)))$ is measurable. Therefore $h\circ f$ is measurable.

Rick
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On an abstract level, the composition of two measurable maps $f:X\to Y$ and $g:Y\to Z$ is measurable; this is very easy to verify from the definition of a measurable map.

The only remaining step is to prove, as Rick did, that continuous real maps are measurable with respect to the Borel sigma-algebra.

Justpassingby
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  • this is not true that a composition of measurable is measurable. – Rick Nov 26 '15 at 22:14
  • Are we using the same definition? A map between measurable spaces is measurable if the inverse image of every measurable set on the right hand side is measurable on the left hand side? – Justpassingby Nov 26 '15 at 22:16
  • This is not the definition of a measurable function. $f:\mathbb R\to\mathbb R$ is measurable if $f((-\infty,a))$ is measurable for all $a$. And you have a theorem that say that $f$ is measurable if for all borelien $B$, $f^{-1}(B)$ is measurable, but it's not necessarily a borelien ! Therefore if $f,g$ are measurable, $f\circ g$ is measurable if $g^{-1}(f^{-1}(B))$ is measurable when $B$ is borelien. Of course $f^{-1}(B)$ is measurable, but there is no reason that it's a borelien. Therefore, $g^{-1}(f^{-1}(B))$ is not necessarily measurable. – Rick Nov 26 '15 at 22:23
  • What is the meaning of a borelien set if not a member of the Borel sigma-algebra? – Justpassingby Nov 27 '15 at 06:23
  • A borelien is of course a member of the sigma algebra. – Rick Nov 27 '15 at 07:53
  • Then I do not uderstand your statement "of course $f^{-1}(B)$ is measurable, but there is no reason that it's a borelien." Measurable set = member of a sigma-algebra, in this case the Borel sigma-algebra. – Justpassingby Nov 27 '15 at 07:58
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    Because a measurable is not necessarily borelien. The best is an example. Let $f$ continuous and $g$ measurable (continuous is much stronger that mesurability). Let $\mathcal C_1$ and $\mathcal C_2$ two cantor set generalized where $m(\mathcal C_1)>0=m(\mathcal C_2)$. Let $F:[0,1]\longrightarrow [0,1]$ bijective, increasing and such that $F(\mathcal C_1)=\mathcal C_2$ (such function exist). Let $\mathcal N\subset \mathcal C_1$ a non mesurable set (we take $\mathcal N$ the vitali set, i.e. $\mathcal N\subset [0,1]$. Take $g=F$ and $f=1_{g(\mathcal N)}$. Both are measurable but not $f\circ g$ – Rick Nov 27 '15 at 08:07
  • @Rick : Very nice example. – idm Nov 27 '15 at 08:18
  • I am beginning to suspect that we are using different definitions of a measurable set (or measurable function) because I never interpreted the question to involve Lebesgue measure. What is your standard definition of (1) a measurable set and (2) a measurable function? – Justpassingby Nov 27 '15 at 08:26
  • If you want me to see your comment, you have to put a @Justpassingby . You know what, I sent a mail to my measure teacher, and I will have more precise information. But I know that I already have this discussion with him, and I thought as you (that $f$ measurable if $f^{-1}(M)$ is measurable when $M$ is measurable), and he told me precisely that it was not correct. But he will write me a mail soon, and I will let you know. But as my example say, the composition of mesurable are not mesurable. And even if one of them is continuous. – Rick Nov 27 '15 at 08:48
  • @Rick eagerly looking forward to the reply, even worth the risk of a few more vandals downvoting my reply :-) In my course as a student and any book I have seen since then, your definition between brackets is used. – Justpassingby Nov 27 '15 at 08:52
  • @Justpassingby: I have the answer which is going to clarify everything, we consider function $f:(\mathbb R,\mathcal M)\longrightarrow (\mathbb R,\mathcal B)$ where $\mathcal M$ is the measurable set $\sigma -$ algebra and $\mathcal B$ is the borelien $\sigma -$algebra$. So, I think it will clarify your doubt :-) – Rick Nov 27 '15 at 09:11
  • @Rick thanks for this clarification. In the original question there was no mention of measurable maps from the reals to the reals, only from an abstract measurable space to the reals without any specification of which sigma-algebra to use on the reals. I maintain that in the category of measurable spaces, the composition of two measurable maps is measurable. That does not diminish the fact that, with your definition of measureble maps of real numbers, the composition of two measurable maps need not be measurable. My original answer must be read with a single choice of sigma-algbra on the reals – Justpassingby Nov 27 '15 at 09:26