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,

  • 1
    $\begingroup$ What kind of integrals are you considering? See for example this answer $\endgroup$ – Arturo Magidin Sep 7 '11 at 19:16
  • $\begingroup$ 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. $\endgroup$ – George Lowther Sep 7 '11 at 21:40
  • $\begingroup$ yes, I think it's the best to consider Lebesgue-integral as it also extends Riemann $\endgroup$ – Johannes L Sep 8 '11 at 6:16

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 $|f(d) - f(c)| \le \sum_{k=0}^{n-1} |f(t_{k+1}) - f(t_k)| \le \sum_i 2^{1-i} (d-c) + \sum_k (B+1) (t_{k+1} - t_k) = (B + 3)(d-c)$.

  • $\begingroup$ Thanks a lot for your answer! I think it will help. In the case $x=x_i$ Why is $|f(t)-f(x)| < 2^{-i} (d-c) $ ? Why does it depend on $i$? What I would see is that for each $i$ where $x_i$ belongs to $[c,d]$ we could fix $|f(t)-f(x_i)| \leq B_i$ because we choose $\delta(x)$ small enough so on booth sides there is no other $x_k$ inside $(x-\delta(x), x+\delta(x)$ and the derivates are bounded (on $(x-\delta(x), x+\delta(x) \setminus {x_i}$), and ... okay but this is useless, because we need to have an expression with $(d-c)$ to obtain Lipschitz $\endgroup$ – Johannes L Sep 8 '11 at 7:32
  • $\begingroup$ Now I think I might have got it: You just choose $\delta(x)$ such that $|f(t)-f(x_i)| < 2^{-i}(d-c)$ which is okay by continuity and the $2^{-i} (d-c)$ is just so that it sums up properly... But if $|f(t)-f(x_i)| < 2^{-i}(d-c)$ also at the other points couldn't one not just cover $[c,d]$ with this and choose a finite subcover.. I still haven't really got it $\endgroup$ – Johannes L Sep 8 '11 at 8:04
  • $\begingroup$ The point is that around those $x$ where you have differentiability, you use that to get an interval where the change in $f$ is not too big compared to the length of the interval. For the $x_i$ you don't have that, because your function might not be differentiable there, but the continuity gives you an interval where the change in $f$ is small. There can be infinitely many of those, but the $2^{-i}$ makes sure the sum of them is not too big. $\endgroup$ – Robert Israel Sep 8 '11 at 16:38

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