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?
Best Answer
The answer is no.
Here is a brief explanation: Consider a function $f$ of unbounded variation, where the increasing part comes from a differentiable function whereas the decreasing part comes from a "Cantor-staircase-type" function. Since $\int f'$ can only capture the increasing part, which itself is unbounded, the resulting integral diverges to $+\infty$ and hence $f'$ is not Henstock–Kurzweil integrable.
The following construction is a fleshed-out version of this idea.
Consider a sequence $(x_k)_{k\geq 0}$ such that $0 = x_0 < x_1 < x_2 < \cdots < 1$ and $\lim_k x_k = 1$. Also, for each $k\geq 1$, we pick a point $c_k \in (x_{k-1}, x_k)$. Then define the function $f : [0, 1] \to \mathbb{R}$ as follows:
On each $[x_{k-1}, c_k]$, the graph of $f$ is a line joining $(x_{k-1}, 0)$ to $(c_k, \frac{1}{k})$.
On each $[c_k, x_k]$, the graph of $f$ is a Cantor staircase joining $(c_k, \frac{1}{k})$ to $(x_k, 0)$.
$f(1) = 0$.
For instance, the figure below illustrates the graph of $f$ when $x_k = \frac{k}{k+1}$ and $c_k$'s are chosen as the middle points of the subintervals.
The first two condition guarantees that $f$ is continuous on $[0, 1)$. Also, since $0 \leq f(x) \leq \frac{1}{k}$ on each interval $[x_{k-1}, x_k]$, it follows that $f(x) \to 0$ as $x \to 1^-$ and hence $f$ is also continuous at $1$. It is also obvious that $f$ is differentiable almost everywhere with the a.e.-derivative
$$ f'(x) = \begin{cases} \frac{1}{k(c_k - x_{k-1})}, & \text{if $x \in [x_{k-1}, c_k]$}, \\ 0, & \text{elsewhere.} \end{cases} $$
Since
$$ \lim_{k\to\infty} \int_{0}^{x_k} f'(x) \, \mathrm{d}x = \lim_{k\to\infty} \sum_{j=1}^{k} \frac{1}{j} = \infty, $$
it follows that $f'$ is not Henstock–Kurzweil integrable.