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I posted it because someone mentioned measure theory and I remembered this example from a graduate course in real analysis. Cantor's function shows there are fu

by memexy 6y ago

I posted it because someone mentioned measure theory and I remembered this example from a graduate course in real analysis. Cantor's function shows there are functions that are not the integrals of their derivatives. The Cantor function goes from 0 to 1, it's increasing and continuous, but it's derivative is 0 almost everywhere. Integrating the 0 function gives us back the 0 function so it's a kind of counter-example to the fundamental theorem of calculus. > Georg Cantor (1884) introduced the Cantor function and mentioned that Scheeffer pointed out that it was a counterexample to an extension of the fundamental theorem of calculus claimed by Harnack. I don't know if it has any real world applications other than as a reminder to be careful about assumptions and definitions. Most people think that increasing/monotone functions must have positive derivatives but the Cantor function shows this is not the case.

yorwba 6y ago
> Cantor's function shows there are functions that are not the integrals of their derivatives. That's because Cantor's function does not have a derivative that's defined everywhere. Of course a function without derivative is not the integral of that non-existent derivative. If most people think that increasing/monotone functions must have positive derivatives, they're apparently forgetting about non-differentiable functions.
- memexy 6y ago
  The function is differentiable almost everywhere (this is a technical term from measure theory). Here's a pretty good explanation of its properties: https://epicmath.org/2012/11/27/5-the-cantor-set-and-the-cantor-function/ https://epicmath.org/2012/11/27/5-the-cantor-set-and-the-can.... > The Cantor function, also known as the Cantor Staircase, is a bizarre function that is continuous and has a derivative of 0 at every point where it is differentiable. In fact, it is differentiable at every point other than on the Cantor set, which is a set of measure zero. So from a measure theoretic perspective the derivative of the Cantor function is well defined and it is equal to 0 almost everywhere. Almost everywhere equality is an equivalence relation and the derivative of the Cantor function is in the same equivalence class as the 0 function.
  - yorwba 6y ago
    > Almost everywhere equality is an equivalence relation and the derivative of the Cantor function is in the same equivalence class as the 0 function. And a function in that equivalence class will have an integral that is equal to a constant function on all intervals where that function is defined. If you also integrate over intervals where the function is only defined almost everywhere, all bets are off.
BlackFly 6y ago
> Integrating the 0 function gives us back the 0 function so it's a kind of counter-example to the fundamental theorem of calculus. I guess that is one way to think about it, but if that is the case you can just differentiate a constant function and you the fundamental theorem of calculus won't return it to you (if you don't like that example just take the Heaviside function). In finite dimensional linear spaces, the derivative operator has a rank 1 kernel, but the integral does not. From that perspective, the derivative destroys information (specifically it destroys constants), the integral operator cannot restore it automatically and the whole + C is meant to identify an equivalence class for the indefinite anti-derivative.

by memexy 6y ago

yorwba 6y ago

> Cantor's function shows there are functions that are not the integrals of their derivatives. That's because Cantor's function does not have a derivative that's defined everywhere. Of course a function without derivative is not the integral of that non-existent derivative. If most people think that increasing/monotone functions must have positive derivatives, they're apparently forgetting about non-differentiable functions.

memexy 6y ago
The function is differentiable almost everywhere (this is a technical term from measure theory). Here's a pretty good explanation of its properties: https://epicmath.org/2012/11/27/5-the-cantor-set-and-the-cantor-function/ https://epicmath.org/2012/11/27/5-the-cantor-set-and-the-can.... > The Cantor function, also known as the Cantor Staircase, is a bizarre function that is continuous and has a derivative of 0 at every point where it is differentiable. In fact, it is differentiable at every point other than on the Cantor set, which is a set of measure zero. So from a measure theoretic perspective the derivative of the Cantor function is well defined and it is equal to 0 almost everywhere. Almost everywhere equality is an equivalence relation and the derivative of the Cantor function is in the same equivalence class as the 0 function.
- yorwba 6y ago
  > Almost everywhere equality is an equivalence relation and the derivative of the Cantor function is in the same equivalence class as the 0 function. And a function in that equivalence class will have an integral that is equal to a constant function on all intervals where that function is defined. If you also integrate over intervals where the function is only defined almost everywhere, all bets are off.

BlackFly 6y ago

> Integrating the 0 function gives us back the 0 function so it's a kind of counter-example to the fundamental theorem of calculus. I guess that is one way to think about it, but if that is the case you can just differentiate a constant function and you the fundamental theorem of calculus won't return it to you (if you don't like that example just take the Heaviside function). In finite dimensional linear spaces, the derivative operator has a rank 1 kernel, but the integral does not. From that perspective, the derivative destroys information (specifically it destroys constants), the integral operator cannot restore it automatically and the whole + C is meant to identify an equivalence class for the indefinite anti-derivative.