A Fundamental Theorem of Calculus Here is a problem I have been working on recently:
Let $f \colon[a,b] \to \mathbb{R}$ be continuous, differentiable on $[a,b]$ except at most for a countable number of points, and $f^{\prime}$ is Lebesgue integrable, then the fundamental theorem of calculus holds, i.e. $\forall x,y \in [a,b]$ we have $$f(y) = f(x) + \int_x^yf'(t)\,dt.$$
The proof I have at the moment is somewhat indirect: I can show that the FTC holds by proving that $f$ is $AC([a,b])$ (http://en.wikipedia.org/wiki/Absolute_continuity). The way I can prove this is showing that the Luzin's N-property (http://en.wikipedia.org/wiki/Luzin_N_property) is satisfied. I have spent quite a lot of time looking for a direct proof, but nothing seems to work! Can anyone help me?
Here is a summary of useful things (which I'll update if I figure out something else interesting!):


*

*Thm: If $u \colon [a,b] \to \mathbb{R}$, is continuous and differentiable everywhere on [a,b], with $u' \in L^1$, then the FTC holds.
(This is a well know result for the Riemann integral, maybe a little less know in the context of the Lebesgue integral.. At least I've never heard of this result before looking for it! A proof can be found in Rudin's Real & Complex Analysis)

*Applying 1. we can prove the result in the case the set of non differentiability is finite.

*(not really useful!) I can also prove it if the set is countable but with only a finite number of accumulation points. This can be done using 2. and adding and subtracting terms to work with telescopic series.


Thank you in advance for your help!
 A: The following is a combination of a proof in the book "Principles
of mathematical analysis" by Dieudonne of a version of a mean value
theorem and of the proof of the Theorem (Theorem 8.21) in Rudin's
book "Real and Functional Analysis" that you also cite.
The proof actually yields the stronger statement that it suffices
that $f$ is differentiable from the right on $\left[a,b\right]$
except for an (at most) countable set $\left\{ x_{n}\mid n\in\mathbb{N}\right\} \subset\left[a,b\right]$.
Let $\varepsilon>0$ be arbitrary. As in Rudin's proof, there is a
lower semicontinuous function $g:\left[a,b\right]\to\left(-\infty,\infty\right]$
such that $g>f'$ and
$$
\int_{a}^{b}g\left(t\right)\, dt<\int_{a}^{b}f'\left(t\right)\, dt+\varepsilon.
$$
Let $\eta>0$ be arbitrary. Define
\begin{eqnarray*}
F_{\eta}\left(x\right) & := & \int_{a}^{x}g\left(t\right)\, dt-f\left(x\right)+f\left(a\right)+\eta\left(x-a\right),\\
G_{\eta}\left(x\right) & := & F_{\eta}\left(x\right)+\varepsilon\cdot\sum_{\substack{n\in\mathbb{N}\\
x_{n}<x
}
}2^{-n}.
\end{eqnarray*}
With these definitions, $F_{\eta}$ is continuous with $F_{\eta}\left(a\right)=0=G_{\eta}\left(a\right)$.
Furthermore, if $z_{n}\uparrow z$, then $F_{\eta}\left(z_{n}\right)\to F_{\eta}\left(z\right)$
and
$$
\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\
x_{m}<z_{n}
}
}2^{-m}\leq\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\
x_{m}<z
}
}2^{-m},
$$
which yields 
$$
\limsup_{n\to\infty}G_{\eta}\left(z_{n}\right)\leq G_{\eta}\left(z\right).\qquad\left(\dagger\right)
$$
For $x\in\left[a,b\right)$ there are two cases:


*

*$x=x_{n}$ for some $n\in\mathbb{N}$. By continuity of $F_{\eta}$,
there is some $\delta_{x}>0$ such that $F_{\eta}\left(t\right)>F_{\eta}\left(x\right)-\varepsilon\cdot2^{-n}$
holds for all $t\in\left(x,x+\delta_{x}\right)$. For those $t$,
we derive
\begin{eqnarray*}
G_{\eta}\left(t\right)-G_{\eta}\left(x\right) & = & F_{\eta}\left(t\right)-F_{\eta}\left(x\right)+\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\
x\leq x_{m}<t
}
}2^{-m}\\
 & > & -\varepsilon\cdot2^{-n}+\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\
x\leq x_{m}<t
}
}2^{-m}\geq0.
\end{eqnarray*}

*$x\notin\left\{ x_{n}\mid n\in\mathbb{N}\right\} $. By assumption,
this implies that $f$ is differentiable from the right at $x$, which
means
$$
\frac{f\left(t\right)-f\left(x\right)}{t-x}\xrightarrow[t\downarrow x]{}f'\left(x\right)<f'\left(x\right)+\eta.
$$
Together with $g\left(x\right)>f'\left(x\right)$ and with the lower
semicontinuity of $g$, we see that there is some $\delta_{x}>0$
such that
$$
f\left(t\right)-f\left(x\right)<\left(f'\left(x\right)+\eta\right)\cdot\left(t-x\right)\text{ and }g\left(t\right)>f'\left(x\right)\qquad\forall t\in\left(x,x+\delta_{x}\right).
$$
Hence, for each $t\in\left(x,x+\delta_{x}\right)$, we get
\begin{eqnarray*}
G_{\eta}\left(t\right)-G_{\eta}\left(x\right) & = & \varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\
x\leq x_{m}<t
}
}2^{-m}+F_{\eta}\left(t\right)-F_{\eta}\left(x\right)\\
 & \geq & F_{\eta}\left(t\right)-F_{\eta}\left(x\right)\\
 & = & \int_{x}^{t}\underbrace{g\left(s\right)}_{>f'\left(x\right)}\, ds-\left[f\left(t\right)-f\left(x\right)\right]+\eta\left(t-x\right)\\
 & > & f'\left(x\right)\cdot\left(t-x\right)-\left[f'\left(x\right)+\eta\right]\left(t-x\right)+\eta\left(t-x\right)=0.
\end{eqnarray*}
In summary, for each $x\in\left[a,b\right)$ there is some $\delta_{x}>0$
such that $G_{\eta}\left(t\right)>G_{\eta}\left(x\right)$ for all
$x\in\left(x,x+\delta_{x}\right)$. Using $G_{\eta}\left(a\right)=0$,
we see $G_{\eta}\left(t\right)\geq0$ for $t\in\left[a,a+\delta_{a}\right)$.
Define
$$
\varrho:=\sup\left\{ t\in\left(a,b\right)\mid G_{\eta}|_{\left[a,t\right)}\geq0\right\} .
$$
It is easy to see that the supremum is actually attained. Using $\left(\dagger\right)$,
we also see $G_{\eta}|_{\left[a,\varrho\right]}\geq0$.
If we had $\varrho<b$, the above would yield $G_{\eta}\geq0$ on
$\left[a,\varrho\right]\cup\left(\varrho,\varrho+\delta_{\varrho}\right)=\left[a,\varrho+\delta_{\varrho}\right)$,
in contradiction to maximality of $\varrho$. Hence, $\varrho=b$
which yields
\begin{eqnarray*}
0 & \leq & G_{\eta}\left(b\right)\\
 & = & \varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\
x_{m}<b
}
}2^{-m}+\int_{a}^{b}g\left(t\right)\, dt-f\left(b\right)+f\left(a\right)+\eta\left(b-a\right)\\
 & \leq & \varepsilon+\int_{a}^{b}f'\left(t\right)\, dt+\varepsilon-f\left(b\right)+f\left(a\right)+\eta\left(b-a\right).
\end{eqnarray*}
Letting $\varepsilon\to0$ and then $\eta\to0$, we conclude
$$
f\left(b\right)-f\left(a\right)\leq\int_{a}^{b}f'\left(t\right)\, dt.
$$
Now apply the above argument to $-f$ instead of $f$ (note that $-f$
fulfills all assumptions). This yields
$$
f\left(b\right)-f\left(a\right)\geq\int_{a}^{b}f'\left(t\right)\, dt
$$
and hence
$$
\int_{a}^{b}f'\left(t\right)\, dt=f\left(b\right)-f\left(a\right).
$$
It is clear that the same argument yields
$$
f\left(y\right)-f\left(x\right)=\int_{x}^{y}f'\left(t\right)\, dt
$$
for all $x,y\in\left[a,b\right]$.
A: although not stated this way in books ,I believe it is true after reading page 98 of 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 ) by Russell A. Gordon ' 
LEMMA 6.14. Let F : [a, b] --> R be measurable and let E subset of [a, b] be measurable. If F is differentiable at each point of E, then µ* (F (E)) <=∫E
|F'|dx
