General versions of (Second part of) Fundamental Theorem of Calculus This question arose from a discussion here, and maybe it is of interest as such:
We know the (second part of the) Fundamental Theorem of Calculus (FTC), but what conditions are really neccessary?
Version 1: Let $g$ be defined on $[a,b]$, and let $g$ have an antiderivative $f$ on $[a,b]$
(i.e. $g(x)= f'(x) \forall x \in [a,b]$).
If $g$ is integrable on $[a,b]$, then 
$$    \int_a^b g(x)\,dx\, = f(b) - f(a) \text{.}$$
Version 2: Let $f$ be absolutely continuous, and be differentiable almost everywhere then
$$  \int_a^b g(x)  dx =f(b) - f(a) \text{.} $$
(Both from Wikipedia, does somebody know real sources for these?)

Case A: Let $f$ be differentiable in $R \setminus \{x_1, x_2, \dots \}$ with bounded derivate, $f$ continuous on whole $R$. Then $f'$ defined almost everywhere, and we extend it's definition to whole R by setting it to $0$ on $\{x_1, x_2, \dots \}$. And then I cannot directly apply the FTC to $f'$, because the condition with the anti-derivative doesn't hold at $\{x_1, x_2, \dots\}$.
I want to apply it in the case where $f$ is piecewise linear (so $f'$ piecewise constant, Case B),
and just want to have this result in order to show that $f$ is Lipschitz, so I search for a way to directly show this in Case A (or B) or search a reference for a version of the FTC that applies to Case A/B.
Q: Is there a version of the FTC that applies for only at countable points not differentiable functions (Case A)?
Version 2 doesn't seem to help for my case, if I show absolute continuity for $f$ then I could better just show uniform continuity instantly and I would be back at 1.

I hope I didn't mix something up, it's a bit confusing with the different versions.
Here there has been a discussion, but rather on different types of integrals, 
 A: If $f$ is continuous, is differentiable except at countably many points $x_i$, and $|f'(x)| \le B$  where it is defined, then I claim that $f$ is Lipschitz and therefore is absolutely continuous.
Suppose $c < d$.
For any $x \in [c,d]$, there is an interval $(x - \delta(x), x + \delta(x))$ around $x$ such that either $f$ is differentiable at $x$ and $|f(t) - f(x)| \le (B + 1) |t - x|$ for $t \in (x - \delta(x), x + \delta(x))$ or $x = x_i$ for some $i$ and $|f(t) - f(x)| <  2^{-i} (d-c)$ for $t \in (x - \delta(x), x + \delta(x))$.  By compactness, a finite collection of these
covers $[c,d]$.  Thus we have $t_0 = c < t_1 \ldots t_n = d$ such that for each $k =0, 1, \ldots n-1$, there is $x$ such that $x - \delta(x) < t_k \le x \le t_{k+1} < x + \delta(x)$.  Moreover, for each $i$ the case $x = x_i$ occurs for at most one $k$.  If $x = x_i$ we than have $|f(t_{k+1}) - f(t_k)| < 2^{1-i} (d- c)$, while if $f$ is differentiable at $x$ we have $|f(t_{k+1}) - f(t_k)| \le (B+1) (t_{k+1} - t_k)$  We 
conclude that $|f(d) - f(c)| \le \sum_{k=0}^{n-1} |f(t_{k+1}) - f(t_k)| \le \sum_i 2^{1-i} (d-c) + \sum_k (B+1) (t_{k+1} - t_k) = (B + 3)(d-c)$.
