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Define $F(x)=\int_{a}^{x} f(t)\,dt$ on $[a,b]$, then by fundamental theorem of calculus, we know that if $f(x)$ is continuous then $F'(x)=f(x)$.

Say we remove the condition that $f(x)$ is continuous then how would we construct an example such that $F'(x)\neq f(x)$ for finite amount of points? How to come up an example for infinite amount of point?

Thanks

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    $\begingroup$ Literally take a continuous function and then just change its values at finitely many points. $\endgroup$ – mathworker21 Jun 7 at 13:58
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Take$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x=\frac1n\text{ for some }n\in\mathbb N\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $F$ is the null function and therefore$$(\forall n\in\mathbb N):F'\left(\frac1n\right)\neq f\left(\frac1n\right).$$

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  • $\begingroup$ sorry to sound stupid but how can I rigorously show that F is null function? (I have covered up to riemann integration using minorant and majorant so I may not understand some proofs) $\endgroup$ – JustWandering Jun 7 at 16:23
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    $\begingroup$ You know that $F(x)=\int_0^xf(t)\,\mathrm dt$. Obviously, $F(0)=0$. Now, suppose that $x>0$. Clearly, every lower sum is equal to $0$. On the other hand, given $\varepsilon>0$, it is not hard to prove that there is a partition $P$ of $[0,x]$ such that $\overline\Sigma(f,P)<\varepsilon$. So, $F(x)=0$. $\endgroup$ – José Carlos Santos Jun 7 at 18:09
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Perhaps a simpler example:

$$f:[0,1]\to\Bbb R\,,\,\,f(x)=\begin{cases}0,&x\neq\frac12\\{}\\1,&x=\frac12\end{cases}\implies F(x):=\int_0^x f(t)\,dt=0\;\;\forall\,x\in[0,1]$$

and thus of course $\;0=F'\left(\frac12\right)\neq1=f\left(\frac12\right)\;$

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Start with $f(x)$, then say a new function $g(x)$ equals $f(x)$ for all irrational numbers in your domain, but $g(x) = f(x)+1$ for all rational numbers. Just one of many ways.

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