I know that the derivative of a differentiable function doesn't have to be continuous. How discontinuous can a derivative be?.
Inspired by Limits and continuity of a derivative, I was thinking of defining the notion of pseudo-continuous: $f:(a,b) \to \mathbb R$ is pseudo-continuous at $x \in (a,b)$ if $$ f(x) = \lim_{y\to x} \frac1{y-x} \int_x^y f(t) \, dt .$$ And then I wanted to show that a function is the derivative of a differentiable function if and only if it is pseudo-continuous.
But then I realized that the derivative doesn't have to be Lebesgue integrable, for example $$ f(x) = \frac x{\log|x|} \sin\left(\frac1x\right) , \quad x \in (-\tfrac12,\tfrac12) ,$$ or $$ f(x) = x^2 \sin\left(\frac1{x^2}\right) ,$$ Does there exist a differentiable function $f:(0,1) \to \mathbb R$ such that its derivative restricted to any subinterval of $(0,1)$ fails to be in $L^1$?
from Hot Weekly Questions - Mathematics Stack Exchange
Stephen Montgomery-Smith
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