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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,

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    What kind of integrals are you considering? See for example [this answer](http://math.stackexchange.com/questions/47285/fundamental-theorem-of-calculus/47328#47328)2011-09-07
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    Checking the wikipedia page and the validity of the statements: Version 1 holds where "integrable" means Riemann integrable and, as the Riemann and Lebesgue integrals agree in that case, you can use either for the integral. Version 2 only holds for the Lebesgue integral. You do not have to assume that f is differentiable almost everywhere, as this is automatic for absolutely continuous functions. However, g might not be Riemann integrable, so you should use the Lebesgue integral to be precise.2011-09-07
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    yes, I think it's the best to consider Lebesgue-integral as it also extends Riemann2011-09-08

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