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I know that Cantor function $\varphi$ is not a measurable funtion. Also I understand that we can prove it by showing $g(x)=x+\varphi(x)$ is not a measurable function. But I don't know how can we prove it by the definition of measurable function.

pwj
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  • But if $\varphi$ is a measurable function,then $x+\varphi$ should be a measurable function while it's not.So is there something I misunderstand? – pwj Jul 08 '18 at 05:43
  • So far I'm not puzzled about that. If continuous function is measurable. Let's see the function $g^{-1}(x)$,it's continous but not measurable because the preimage of a measure zero is a non-meansurable set(Consider a nonmeasurable set in the $g(C)$,C is cantor set).So I don't think that a continous function must be measurable. – pwj Jul 08 '18 at 06:40
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    @qbert This comes up periodically on MSE; see, e.g., MSE 479441. – Benjamin Dickman Jul 08 '18 at 06:50
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    The key here is that "measurable" if otherwise unqualified generally means $(\mathbb R, \mathcal L), (\mathbb R,\mathcal B)$ measurable. Not $(\mathbb R, \mathcal L), (\mathbb R,\mathcal L)$-measurable. – spaceisdarkgreen Jul 08 '18 at 06:53
  • @BenjaminDickman I see, thank you for pointing this out! – operatorerror Jul 08 '18 at 06:56
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    @qbert you were right! If they actually mean $(\mathbb R,\mathcal L)-(\mathbb R,\mathcal L)$-measurable, they should say so. "It's continuous, so it's measurable" is a perfectly valid objection/answer to this question as currently written. – spaceisdarkgreen Jul 08 '18 at 07:24
  • All continuous function are Borel Measurable & Hence Measurable. Cantor function is continuous so it is Measurable too – Arghya Santra Mar 01 '24 at 11:32

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