4 ms·

I'll throw in my funky function (which is probably similar to a well-known one, but I'm no mathematician so I know not what is well known): f(x) = a random num

by sfink 2y ago

I'll throw in my funky function (which is probably similar to a well-known one, but I'm no mathematician so I know not what is well known): f(x) = a random number uniformly selected from [-1, 1]. (So every real number maps to a distinct random number.) It's nowhere continuous, nowhere differentiable, yet its integral over any interval is exactly zero. Which is kind of weird if you think of integrals and derivatives as inverses, because as I understand it the derivative of a constant g(x) = c is g'(x) = 0, yet the integral of the above function is F(x) = 0 yet F'(x) = f(x) = the random nonsense monster. So the derivative of zero can be either zero or random noise? (And while f(x) is not differentiable, its integral surely is.)

eafer 2y ago
> yet its integral over any interval is exactly zero That function doesn't actually have an integral in any sense that I know of. The Riemann integral doesn't work because, for every interval in every possible partition, the function's upper bound will be 1 and the lower bound will be -1.
- sfink 2y ago
  Oh. Er... I didn't really consider whether it could be proven to have an integral, I just went off of intuition. After an additional lifetime of study, magically compressed into 15 minutes of skimming Wikipedia articles, I'm going to (tenuously) claim that the function is Lebesgue integrable. Heck, if it works for the {0 if irrational, 1 if rational} function, this should be much easier. I hope? My original intuition is that every value in [0, 1] will be cancelled by a value in [-1, 0]. My updated intuition to make it more Lebesgue-compatible would be to sort the (infinite) set of values in the interval. This will give the everywhere continuous, linear function from -1 at the minimum coordinate of the interval to 1 at the maximum coordinate. Hm... I notice that when applying Lebesgue to Dirichlet, it uses countability, and reals aren't countable. I have no clue whether that's a problem here or not.
  - eafer 2y ago
    It won't be Lebesgue integrable because it won't be measurable. Your function is actually a good (though not rigorous) argument for the existence of non-measurable sets.
TheRealNGenius 2y ago
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by sfink 2y ago

eafer 2y ago

> yet its integral over any interval is exactly zero That function doesn't actually have an integral in any sense that I know of. The Riemann integral doesn't work because, for every interval in every possible partition, the function's upper bound will be 1 and the lower bound will be -1.

sfink 2y ago
Oh. Er... I didn't really consider whether it could be proven to have an integral, I just went off of intuition. After an additional lifetime of study, magically compressed into 15 minutes of skimming Wikipedia articles, I'm going to (tenuously) claim that the function is Lebesgue integrable. Heck, if it works for the {0 if irrational, 1 if rational} function, this should be much easier. I hope? My original intuition is that every value in [0, 1] will be cancelled by a value in [-1, 0]. My updated intuition to make it more Lebesgue-compatible would be to sort the (infinite) set of values in the interval. This will give the everywhere continuous, linear function from -1 at the minimum coordinate of the interval to 1 at the maximum coordinate. Hm... I notice that when applying Lebesgue to Dirichlet, it uses countability, and reals aren't countable. I have no clue whether that's a problem here or not.
- eafer 2y ago
  It won't be Lebesgue integrable because it won't be measurable. Your function is actually a good (though not rigorous) argument for the existence of non-measurable sets.

TheRealNGenius 2y ago

[dead]