If $f$ derivable on $[a,b]$ does $\int_a^t f'(x)dx=f(t)-f(a)$ true? Let $f:[a,b]\longrightarrow \mathbb R$ a derivable function. Is it true that for all $t\in [a,b]$ we have that $$f(t)=f(a)+\int_a^t f'(x)dx \ \ ?$$
The thing is since $f'$ is not supposed continuous, there is no reason for me for $f'$ to be Riemann integrable. So my questions are the followings one : 


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*Q1) In Riemann sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?

*Q2) If we assume $f'$ Riemann integrable, is the formula correct (in Riemann sense). If no, do you have a counter example ?

*Q3) In Lebesgue sense, is the formula correct (if we don't have other hypothesis on $f'$). If no, do you have a counter example ?

*Q4) If we assume $f'$ Lebesgue integrable, is the formula correct (in Lebesgue sense). If no, do you have a counter example ?
 A: Q1: No,  because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.
Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral  to be defined exactly the way it is. Say $a=t_0<\dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=\sum(f(t_{j+1})-f(t_j))=\sum f'(\xi_j)(t_{j+1}-t_j),$$precisely a Riemann sum for $\int_a^b f'(t)\,dt$.
Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=\begin{cases}t^2\sin(1/t^{100}),&(t\ne0),
\\0,&(t=0).\end{cases}$$
Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21.
Or see here for a proof based on the Vitali-Caratheodory theorem.
Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'\in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.
