Edit: I understood integrability in the Lebesgue sense. The longer first part is the original answer.
I'm rearranging my answer since I'm a bit confused about the actual question, which seems to be:
If $F(t) = \int_{a}^{t} f(x)\,dx$ is a $C^1$-function, is it then true that $f$ is continuous?
The answer to that question is no, since we can always modify $f$ on a null set without changing $F$. The characteristic function of $\mathbb{Q}$, for example, is nowhere continuous but its integral is zero over every interval. These are true discontinuities, not only removable ones.
The point is that $f$ is completely underdetermined in this question. Any modification of $f$ on a null set will yield the same $F$, thus we can't hope for more than $F' = f$ almost everywhere. Amazingly enough, this turns out to be true.
The right question to ask was answered by Lebesgue. The starting point is that for an integrable function $f$ its definite integral $F$ will always be absolutely continuous. The Lebesgue differentiation theorem even tells us that $F$ is almost everywhere differentiable and $F' = f$ almost everywhere.
Lebesgue proved even more:
Assume that $F$ is differentiable almost everywhere. In order for $F$ to be the definite integral of its derivative it is necessary and sufficient that $F$ be absolutely continuous.
This is proved in any decent text on measure theory e.g. Royden or Rudin.
On the other hand, Luzin's theorem implies that a measurable $f$ is continuous on the complement on a set of arbitrarily small measure.
One further point you should be aware of: If $F$ is continuous and differentiable almost everywhere and $F' = 0$ wherever it is defined then $F$ can be non-constant, even monotonic. the standard example for this is called the Cantor-Lebesgue function, or more colorfully the devil's staircase.
Finally, I'd like to point out that it is an awfully subtle business to characterize the functions which are derivatives of everywhere differentiable functions.
Added:
Robert Israel left the following comment to this answer (thank you very much for that!):
According to Lebesgue's criterion, a function is Riemann integrable on $[a,b]$ if [and only if] it is continuous almost everywhere and bounded. To maintain the continuity almost everywhere in the counterexample, the set on which you modify $f$ can be a closed set of measure $0$, e.g. a Cantor set.
The only thing I'd like to add to that is:
Lebesgue's criterion is actually Riemann's criterion
see Über die Darstellbarkeit einer Funktion durch eine trigonometrische Reihe (1854). More precisely, the relevant passage is Section 5 on pages 226 and 227 of Bernhard Riemann's Gesammelte mathematische Werke und wissenschaftlicher Nachlass.
(Hat tip to Roy Smith and Pete L. Clark)
You can't assume left or right continuity of $f$. Use the same approach as in proof of FTC and show the following.
Lemma: If $f$ is Riemann integrable on $[a, b] $ and if for some $c\in[a, b) $ the limit $\lim_{x\to c^{+}}f(x) $ exists then the function $F$ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ has right derivative at $c$ and we have $D^{+} F(c)=\lim_{x\to c^{+}} f(x) $.
Let $f(x) \to L$ as $x\to c^{+} $ and $\epsilon >0$ be arbitrary. Then there is a $\delta >0$ such that $$|f(t) - L|<\epsilon $$ whenever $0<t-c<\delta$. Thus we have $$L-\epsilon <f(t) <L+\epsilon$$ whenever $0<t-c<\delta$. Integrating the above inequality in interval $[c, c+h] $ where $0<h<\delta$ we get $$h(L-\epsilon) <\int_{c} ^{c+h} f(t) \, dt<h(L+\epsilon) $$ or $$L-\epsilon<\frac{F(c+h) - F(c)} {h} <L+\epsilon$$ whenever $0<h<\delta$. Thus $$D^{+} F(c) =\lim_{h\to 0^{+}}\frac {F(c+h) - F(c)} {h} =L=\lim_{x\to c^{+}} f(x) $$ A similar lemma can be proved in same manner for left limit and left derivative. Going further if $f(x) \to L$ as $x\to c$ then $F$ is differentiable at $c$ with $F'(c) =L$ and if $f$ is continuous at $c$ then $L=f(c) $ so that $F'(c) =f(c) $ which brings us to the traditional form of FTC.
It is now obvious that if $f$ has a jump discontinuity at some point then $F$ is not differentiable at that point. However if $f$ has essential discontinuity at some point then $F$ may or may not be differentiable at that point.
Best Answer
First, I would suggest you look at Darboux's theorem and the definition of a Darboux function in that webpage. If we did in fact have differentiability (possibly not $C^1$, but differentiable nonetheless), then you can easily construct an example of a right continuous Darboux function which is not continuous. I'll leave this construction up to you. Taking $f'$ as your Darboux function will yield your desired contradiction.
Thus, the question boils down to, is the condition you've given equivalent to a derivative?
The answer to that is yes.