Counterexample of Second fundamental theorem of Calculus if f is not continuous

Question

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

Answer 1

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$$(\forall n\in\mathbb N):F'\left(\frac1n\right)\neq f\left(\frac1n\right).$$

Answer 2

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.

Answer 3

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 $\;0=F'\left(\frac12\right)\neq1=f\left(\frac12\right)\;$