Can a function $f$ have an antiderivative even though its indefinite integral $F(x) = \int_{a}^{x} f(t)\ dt$ is not one? The fundamental theorem of calculus states that if $f:[a,b] \to \mathbb{R}$ is integrable and $F(x) = \int_{a}^{x} f(t)\ dt$, then $F'(x) = f(x)$ at every point $x$ at which $f$ is continuous. This means that if $f$ is integrable, $F'(x) = f(x)$ almost everywhere. If $f$ is continuous on $[a,b]$, then $F$ is an antiderivative of $f$, since $F'(x) = f(x)$ holds for all $x \in [a,b]$. 
But what if $f$ has a discontinuity at some $x \in [a,b]$? In this case, it is not necessarily true that $F'(x) = f(x)$, and so we cannot necessarily conclude that $F$ is an antiderivative of $f$. Does this mean that there is no antiderivative of $f$? Is it possible for $f$ to have an antiderivative but the indefinite integral $F$ is not an antiderivative of $f$?
I know that if $f$ has a jump discontinuity, then $f$ can have no antiderivative (since the derivative of a function must satisfy the intermediate value property), but what if we have some other type of discontinuity?
 A: Consider the map$$\begin{array}{rccc}F\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $F$ is differentiable and$$(\forall x\in\Bbb R):F'(x)=\begin{cases}-\cos\left(\frac1x\right)+2x\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}$$So, $F'$ is discontinuous at $0$. But $F$ is an antiderivative of $F'$.
A: Assume $f:[a, b] \to\mathbb {R} $ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $g:[a, b] \to\mathbb {R} $ on $[a, b] $ ie $g'(x) =f(x) \, \forall x\in[a, b] $. Then by Fundamental Theorem of Calculus we have $$g(x) =g(a) +\int_{a} ^{x} f(t) \, dt=g(a) +F(x) $$ It follows that $$F'(x) =g'(x) =f(x) $$ for all $x\in[a, b] $.
Therefore the indefinite integral $F:[a, b] \to\mathbb {R} $ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ also acts as an anti-derivative of $f$ on $[a, b] $.
Without symbols we can summarize as

If a Riemann integrable function possesses an anti-derivative over some closed interval then its indefinite integral also acts an anti-derivative over the same interval. 

In short the answer to your question in bold is NO if we assume Riemann integrability of $f$. However not every Riemann integrable function possesses an anti-derivative. Continuity is only a sufficient for existence of anti-derivative and jump discontinuity is only a sufficient condition for non-existence of anti-derivative. 
A: If there is a function $g$ which is differentiable at every point with $g'=f$ and if $\int_a^{x} f(t)dt$ exists for all $x$ then it is necessary that $g$ is absolutely continuous and $g(x)=g(a)+\int_0^{x} g'(t)dt=g(a)+\int_0^{x} f(t)dt$. This is proved in Rudin's RCA. Hence $\int_0^{x} f(t)dt$ is an anti-derivative. 
A: One can prove that if $f:J \rightarrow \Bbb R $, $J$ is an interval, is locally Riemann integrable (i.e. integrable in every compact interval of $J$)  and has an antiderivative then the function
$$
h(x)=\int^x_af(t)dt,\; a\in J
$$
is an antiderivative of $f(x)$. That is because if $f(x)$ is Riemann integrable and have an antiderivative $g(x)$ then 
$$
\int^b_af(t)dt=g(b)-g(a)
$$
A proof can be find in Bartle's book.
A: As I understand the question: If $F'=f$ on $[a,b],$ is it true that $F(x)=F(a) + \int_a^x f$ for $x\in [a,b]?$ Answer: No if we stay in the realm of the Riemann integral. One reason is that $f$ need not even be Riemann integrable (RI). A very nice example shows that $F$ can be differentiable everywhere and strictly increasing, and yet $f(x)=F'(x)=0$ for $x$ in a dense subset of $[a,b].$ Such an $f$ cannot be RI, so we can't get even get off the ground. See https://en.wikipedia.org/wiki/Pompeiu_derivative.
What if $F'=f$ on $[a,b]$ and $f$ is RI? Then the answer is yes. This is a well known result, following quickly from the MVT.
You might ask: Suppose $F'=f$ on $[a,b]$ and $f\in L^1[a,b].$ Do we have $F(x)=F(a) + \int_a^x f$ for $x\in [a,b]?$ Yes. You can find a proof of this in Rudin's Real and Complex Analysis.
Beyond RI and LI you could look at improper integrals I suppose, but I'm out of ammo at this point.
