TITLE: Is every "almost everywhere derivative" Henstock–Kurzweil integrable?
QUESTION [10 upvotes]: It is well known that the Henstock–Kurzweil integral fixes a lot of issues with trying to integrate derivatives. The second fundamental theorem of calculus for this integral states:

Given that $f : [a,b] \rightarrow \mathbb{R}$ is a continuous function. If $f$ is differentiable co-countably everywhere (in other words: differentiable everywhere except for possibly a countable set of points), then:

$f'$ is Henstock-Kurzweil integrable
$\int_a^bf'(x) dx = f(b) - f(a)$


My question is what happens if you replace "co-countably everywhere" with "almost everywhere"? Clearly the second statement no longer holds (the Cantor function provides a counter-example), but what about the first statement?
If $f$ is continuous everywhere and differentiable almost everywhere, is $f'$ necessarily Henstock-Kurzweil integrable?

REPLY [3 votes]: Here are  three observations that should assist in thinking about this.
Suppose that  $f:[a,b]\to\mathbb R $ is a continuous function that is differentiable almost everywhere.
QUERY. What are the necessary and sufficient conditions in order that  $f'$  is
integrable on $ [a,b] $  in some sense and that $ f $ is an indefinite integral for  $f'$  in that sense, i.e.,
$$  \int_a^xf'(t)\,dt = f(x)-f(a)  \ \ \  (a<x\leq b).   \tag{*}$$

For the Lebesgue integral the NASC for (*)  is that  $f $ is absolutely continuous on  $[a,b]$.

For the Denjoy-Perron (aka Henstock, Kurzweil, Henstock-Kurzweil, etc.) integral the NASC is that  $f$  is generalized absolutely continuous
in the restricted sense on  $[a,b]$.   [This is popularly known as  ACG${}_*$.]

Theorem (Lusin).  Suppose that  $ g:[a,b]\to\mathbb R$  is any finite-valued measurable function.  Then there exists a continuous function
$ f:[a,b]\to\mathbb R$   that is differentiable almost everywhere and with  $f'(x)=g(x)$  almost everywhere.


The moral lesson of #3 is that being the derivative a.e. of a continuous function gives you no information whatsoever beyond the obvious fact that it must be measurable.
Short Answer:  The poster asks "If $f$ is continuous everywhere and differentiable almost everywhere, is $f′$ necessarily Henstock-Kurzweil integrable?   The answer is no, not at all.  This condition says nothing about the properties of $f'$ beyond the trivial observation that it does have to be measurable.
Source:  See S. Saks, Theory of the Integral (1937) for this stuff (and many more questions of this nature).