The Fundamental Theorem of Calculus (for the lebesgue integral)

I am interested in the following statement.

Let $f:[a,b]\rightarrow \mathbb{R}$ be continuous and differentiable. Then, if $f':(a,b)\rightarrow \mathbb{R}$ is integrable, then $$f(x)-f(a)=\int _a^x\mathrm{d}t\, f'(t)$$ for all $x\in [a,b]$.

Clarifications: I only define the derivative on the interior, so there is no condition being imposed on limits of (one-sided) difference quotients at either $a$ or $b$. "Integrable" means "borel" (preimage of open sets are measurable) and "integral of absolute value is finite". It does not mean bounded and continuous almost-everywhere.

Actually, though I have yet to check it explicitly, I believe that the Volterra's function may be a counter-example to this. If so, I would then actually be interested in the statement

Let $f:[a,b]\rightarrow \mathbb{R}$ be continuous and differentiable. Then, if $f:(a,b)\rightarrow \mathbb{R}$ is integrable and continuous almost everywhere, then $$f(x)-f(a)=\int _a^x\mathrm{d}t\, f'(t)$$ for all $x\in [a,b]$.

I don't have any examples up my sleeve that are potential counter-examples, but I am still struggling to prove this. What I have been able to do is the following: If you look at the proof for $f$ \emph{riemann} integrable, you apply the Mean Value Theorem to obtain a riemann sum for $f'$---I have been able to show that this riemann sum converges pointwise almost-everywhere (at least at points of continuity) to the derivative itself. To finish the proof then, it would suffice to show that these sequence of riemann sums is bounded by an integrable function and apply the Dominated Convergence Theorem. I have been struggling to do this, however. Hopefully I'm just being really dense.

Is there a counter-example to this statement? If not, how do I complete my proof?

Clarifaction: I am not interested in the version of the theorem that involves absolute continuity---I want the derivative to exist everywhere, not just almost-everywhere. (Otherwise, the Devil's Staircase provides an 'easy' counter-example.)

• @Henry Yeah, that was a stupid mistake of mine. Sorry about that. It has been corrected. – Jonathan Gleason Aug 11 '15 at 18:36
• A question that I recall was on a qualifying exam, that I never did figure out how to answer: if a function is of bounded variation and everywhere differentiable, then it is absolutely continuous. With this in mind, I think you could use the continuous extension of $f(x)=x^2 \sin(1/x^\beta)$ for $\beta \gg 1$ to get an example of a function which is everywhere differentiable but has no FTC. – Ian Aug 11 '15 at 18:40
• My answer here A Fundamental Theorem of Calculus shows the more general statement that it even suffices if $f$ is differentiable from the right everywhere except for a countable set. – PhoemueX Aug 11 '15 at 18:42

The first statement is a theorem in Rudin Real and Complex Analysis. It's not quite trivial. (The theorem in Rudin assumes that $f$ is differentiable on $\Bbb R$. So it shows that $f(x)-f(a+\epsilon)=\int_{a+\epsilon}^x f'$. Now let $\epsilon\to0$.)