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Question Can there exist a function $F$ which is differentiable on $[a,b]$ but $F'$ is non-riemann integrable function on $[a,b]$. There is a similar question here, but the construction done is not clear to me.

Context In Baby rudin, in the statement of $\textit{Integration by parts}$ they explicitly assume that let $F$ be a differentiable function such that $F'$ is Riemann integrable, I tried of thinking a counter example, can't come up with any.

Kindly help me develop a feel for this problem and the counter example if exists.

PS:- I know that a function which has uncountably many discontinuties are non riemann integrable, but how to construct such an $F'$.

Debu
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    Yes, these pathologies exist, but unfortunately I’m not sure how to simplify this any further, because in order to produce a non-Riemann-integrable function you need something which is very discontinuous (the set of points of discontinuity must have positive Lebesgue measure), or something which is unbounded. The situation is so bad that there are functions like Volterra’s for which $F$ is everywhere differentiable, $F’$ is bounded, yet $F’$ is not Riemann-integrable. But if you merely look for functions with unbounded derivative, then consider $F(x)=x^2\sin(1/x^2)$ for $x\neq 0$ and $F(0)=0$. – peek-a-boo Jun 25 '24 at 16:00
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    See https://math.stackexchange.com/a/3338788/123905 for graphs of $F$s (that look like random walks) whose derivatives (that look like random scatterings of dust) are nowhere continuous so are not Riemann integrable (but are Lebesgue integrable). – Eric Towers Jun 25 '24 at 16:12
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    For nice diagrams when the non-Riemann integrability is entirely due to $f'$ having too many discontinuities (rather than $f'$ simply being unbounded), see this MSE answer. Also, such functions exist in great perfusion in the sense that they are very strongly densely distributed throughout all bounded derivative functions -- see #3 in this MSE answer. (Note that #3 there is a very weak consequence of much stronger statements given there.) – Dave L. Renfro Jun 25 '24 at 19:34

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Counter-example:$$F(x)=\begin{cases}x^2\sin(\frac{1}{x^2}), &x\neq 0\\ 0, &x=0\end{cases}$$

$F(x)$ is integrable on $[0,1]$, but $$F'(x)=\begin{cases}2\sin\big(\frac{1}{x^2}\big)-\frac{2}{x}\cos\big(\frac{1}{x^2}\big), &x\neq 0 \\ 0, &x=0 \end{cases}$$

is not integrable on $[0,1]$ since it is not bounded (consider the case when $x$ is close to $0$).

Bowei Tang
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