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Let define $f:[0,1]\to [0,1]$ as follows:

$f(x) = \begin{cases} 0, & \text{if $$x$ \in \mathbb{C}$} \\ x^2, & \text{if $x$ $\notin \mathbb{C}$ } \end{cases},$ where $\mathbb{C}$ is the Cantor set. Show $f$ is Lebesgue measurable function.

I know a function $f : \mathbb{R} \to \mathbb{R}$ is called Lebesgue-measurable if preimages of Borel-measurable sets are Lebesgue-measurable. I think it is enough to show for any $a\geq 0$, $f^{-1}(a,\infty)=\{ f>a \}=\{x\in [0,1]\mid f(x)>a \}$ is Lebesgue-measurable set. But I stuck here $f^{-1}(a,\infty)= \begin{cases} \mathbb{C}^c & a< 0\\ [0,1] & a\geq 0 \end{cases}. $

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Well, $[0,1]=C\hspace{1mm}\cup C^{c} $ (I discourage you to use $\mathbb{C}$ for the cantor set) and you can write f as: $$f(x)=0\cdot\chi_{C}(x)+x^{2}\chi_{C^{c}}$$ where by $\chi$ I mean the characteristic function of the set in subscript. The sum of measurable functions is measurable as well as the product, and polynomial functions are measurable.

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    $\begingroup$ More importantly, the product of measurable functions is measurable. The first summand is anyway zero :) already (+1) for discouraging that notation $\endgroup$ Apr 1, 2021 at 17:24
  • $\begingroup$ correct, I should have wrote the same also for the product, thank you for the note. $\endgroup$ Apr 1, 2021 at 19:35

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