Fundamental theorem of calculus necessary assumption

Question

The fundamental theorem of calculus is stated as follows:

Let $f$ and $F$ be real-valued functions defined on a closed interval $[a, b]$ such that the derivative of $F$ is $f$. That is, $f$ and $F$ are functions such that for all $x \in [a, b]$,

$F'(x) = f(x)$

If $f$ is Riemann integrable on $[a, b]$ then

$\int_a^b f(x)\,dx = F(b) - F(a)$

My question is: is it necessary to say ''if $f$ is Riemann integrable''? It seems to me that if $f$ is the derivative of a function $F$ then it must be Riemann integrable.

Is there an example of $f$ and $F$ such that $F'=f$ and $f$ is not Riemann integrable?

Answer 1

To answer the question at the end of your post:

Is there an example of $f$ and $F$ such that $F'=f$ and $f$ is not Riemann integrable?

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Quote from van Rooij, Schikhof: A Second Course on Real Functions, p.4:

(8) A function that has an antiderivative but is not Riemann integrable.
The function $h$ defined by $$h(x):= \begin{cases} 2x\sin x^{-2}-2x^{-1}\cos x^{-2} & \text{if }x\in(0,1], \\ 0 & \text{if }x=0. \end{cases} $$ is unbounded, hence certainly not Riemann integrable. But it is the derivative of $H$, where $$H(x):= \begin{cases} x^2\sin x^{-2} & \text{if }x\in(0,1], \\ 0 & \text{if }x=0. \end{cases} $$

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If you prefer an example which is bounded, this answer mentions Volterra's function. Its derivative is not Riemann integrable.