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I'm trying to prove the integral of the cantor function on [0,1] is equal to 1/2.

I'm thinking of using a symmetry argument by using the fact that the function is 1/2 on [1/3,2/3]. Then arguing that for every region where the function is less than 1/2, there is an equal size region where the function is above 1/2 by the same amount.

But I'm not sure how to formalise this argument using mathematics as there are an infinite number of intervals.

Is this the correct approach to use? I haven't covered topics in measure theory yet so am fully sure how to use those sorts of concepts.

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  • $\begingroup$ Perhaps you can apply your argument on appropriate intervals of the form $(x,1-x)$ and then let $x\to0+$ $\endgroup$ – saulspatz Nov 11 '18 at 15:10
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Let $f$ denote the Cantor function. Then $\int_0^1 fdx=\int_0^1\frac{f(x)+f(1-x)}{2}dx=\int_0^1\frac{1}{2}dx.$

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  • $\begingroup$ How do you know that the first 2 integrals are equivalent? $\endgroup$ – user601175 Nov 11 '18 at 20:17
  • $\begingroup$ @user601175 Look at the graph of $f(x)$. How does $f(1-x)$ compare? $\endgroup$ – J.G. Nov 11 '18 at 20:17
  • $\begingroup$ @user601175 Oh sorry, that's why the second & third match. Quite simply, any function satisfies $\int_0^1 h dx=\int_0^1 h(1-x) dx$. $\endgroup$ – J.G. Nov 11 '18 at 20:32
  • $\begingroup$ Doesn't that reverse the order of integration? As h(1-0)=h(1) and h(1-1)=h(0) so the limits switch? $\endgroup$ – user601175 Nov 11 '18 at 21:34
  • $\begingroup$ @user601175 It's just a $y=1-x$ substitution. $\endgroup$ – J.G. Nov 11 '18 at 21:48

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