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I understand how this proof works geometrically. Considering the unit square with corners $(0,0),(0,1),(1,1)$, and $(0,0)$, taking the integral of $f(x)$ and then adding its inverse would produce an area equivalent to this unit square. However, I am having a hard time trying to prove this algebraically or anyway that is not geometric. I've tried:

$\int_{0}^{1} f(x)dx + \int_{0}^{1}f^{-1}(x)dx = $

Manipulating the right side:

$ \int_{0}^{1}f^{-1}(x)dx = \int_{0}^{1} f^-1(f(y))f'(y)dy = \int_{0}^{1} yf'(y)dy = f(y)(y)|_{0}^{1} - \int_{0}^{1} f(y)dy = 1 -\int_{0}^{1} f(y)dy $

Plugging back into the original equation we get:

$\int_{0}^{1}f(x)dx + 1 - \int_{0}^{1}f(y)dy $ which equals 1, since changing the variable we are integrating with respect to does not change the result of integrating the function.

The problem I've found in this path is that I'm assuming that $x = f(y)$, which I do not believe is necessarily true in this case. Does anybody have any advice on how to approach this problem? Thank you

Haris
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Srivivasa
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  • What is not necessarily true? – Gabriella Chaos Feb 24 '24 at 05:05
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    You assumed that $f$ is differentiable in your proof but the problem doesn't assume that. – Bobby Ocean Feb 24 '24 at 05:07
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    The assignment of $x=f(y)$ is itself no problem, but there is a problem with $dx = f'(y),dy$ since $f$ may not actually be differentiable (in fact, it may have a nonzero singular part). If you want to continue this route: Do you know the Stieltjes integrals? – Brian Moehring Feb 24 '24 at 05:07
  • There are dozens of answers to this question on MSE. – geetha290krm Feb 24 '24 at 05:20
  • @BrianMoehring Is it ok to assign x = f(y) because we are changing the variable of integration? Because there is an argument running in my head that there is a function f here that transforms x to y. To relate y back to x, we’d use the inverse of that function. I think I’m overthinking, but it seems potentially troublesome to now assign f(y) = x and assume f^(-1)(x) = f^-1(f(x)) = y (the inverse of the function that maps x to y is also the inverse of the function that maps y to x). Thanks for the differentiable comment, I will look into Stieltjes integrals – Srivivasa Feb 24 '24 at 06:10
  • @geetha290krm thank you, I originally was having trouble finding them. – Srivivasa Feb 24 '24 at 06:12

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