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:

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

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?


3 Answers 3


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.

graph of the counter-example

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

$$ f'(x) = \begin{cases} \frac{1}{k(c_k - x_{k-1})}, & \text{if $x \in [x_{k-1}, c_k]$}, \\ 0, & \text{elsewhere.} \end{cases} $$


$$ \lim_{k\to\infty} \int_{0}^{x_k} f'(x) \, \mathrm{d}x = \lim_{k\to\infty} \sum_{j=1}^{k} \frac{1}{j} = \infty, $$

it follows that $f'$ is not Henstock–Kurzweil integrable.

  • 1
    $\begingroup$ Ahhh lovely, thank you for this example, it’s exactly what I was looking for. I can also see the possible potential for $f$ to be “fractalised” by adding suitably smaller copies of it, to construct an example which is not HK integrable in any interval. That’s as general as I could possibly ask for, thank you. $\endgroup$ Sep 9, 2021 at 18:31
  • 1
    $\begingroup$ It would be more informative to say that your construction produces a nonnegative a.e. derivative that is not Lebesgue integrable. Then, obviously, it cannot be HK integrable either since the two integrals agree for nonnegative functions. $\endgroup$ Sep 13, 2021 at 23:25
  • 1
    $\begingroup$ Note, too, that while $f'$ is not integrable on $[0,1]$, it is, however, integrable on every closed subinterval of $[0,1)$. That might suggest something. But it does not. There exist examples of a.e. derivatives of continuous functions that are not integrable on any interval at all. Use this. $\endgroup$ Sep 17, 2021 at 18:29
  • $\begingroup$ @B.S.Thomson Good reference. Alternatively check out my answer below for an explicit construction of an $f$ which is a.e. differentiable, but $f'$ is not integrable in any interval. $\endgroup$ Sep 18, 2021 at 23:45

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., $$ \int_a^xf'(t)\,dt = f(x)-f(a) \ \ \ (a<x\leq b). \tag{*}$$

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

  2. 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${}_*$.]

  3. 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).

  • $\begingroup$ The text by Saks is available online and is a must have for anyone serious about classical real analysis: kielich.amu.edu.pl/Stefan_Banach/e-saks.html $\endgroup$ Sep 13, 2021 at 17:51
  • 1
    $\begingroup$ This is a really great answer, thank you. I had never heard of Lusin's theorem, but that settles my question perfectly. Now I'm curious how small the "almost all" can be made in the theorem. For example, could the exceptional set be made to have Hausdorff dimension zero? I assume not in general, but it gives me something to think about. I'll check out the book recommendation. $\endgroup$ Sep 17, 2021 at 23:15

Sangchul Lee's answer does a fantastic job of answering my question. But I wanted to share my own answer that I have come up with since, to provide an even more extreme example. A function $f : \mathbb{R} \rightarrow \mathbb{R}$ with the following properties:

  • $f$ is continuous.
  • $f$ is differentiable almost everywhere, with $f' \geq 0$ almost everywhere.
  • $\int_a^b f' = +\infty$ for all $a, b \in \mathbb{R}$ with $a < b$.

Preliminary Constructions

Start by letting $C$ denote the Cantor function. Then define $D : \mathbb{R} \rightarrow \mathbb{R}$ as:

$$D(x) := \begin{cases} x & \text{if $0 \leq x < \frac{1}{2}$} \\ x - C(2x-1) & \text{if $\frac{1}{2} \leq x \leq 1$} \\ 0 & \text{otherwise} \end{cases} $$

This gives an example of a continuous non-negative function, with a maximum output of $\frac{1}{2}$, which is almost everywhere differentiable with $D'=1$ almost everywhere in $[0,1]$, and $D(0) = D(1) = 0$.

So given any interval $[a,b]$, we can lay down multiple copies of rescaled versions of $D$ back to back to construct a function $D_{[a,b]} : \mathbb{R} \rightarrow \mathbb{R}$ satisfying:

  • $D_{[a,b]}$ is non-negative, with $D_{[a,b]}(x) \leq b-a$ everywhere.
  • $D_{[a,b]}$ is continuous, with $D_{[a,b]}(x) = 0$ whenever $x \not\in (a,b)$.
  • $D_{[a,b]}$ is differentiable almost everywhere, with $D_{[a,b]}'(x) \geq \frac{1}{b-a}$ almost everywhere in $[a,b]$.

Main Construction

Let $\{ \frac{p_n}{q_n} : n \in \mathbb{N^{\geq 1}}\}$ be an enumeration of the rational numbers in their simplest form, with $q_n > 0$. Then for each $n \in \mathbb{N^{\geq1}}$ define the interval $I_n := [\frac{p_n}{q_n} - \frac{1}{2^{n+1}}\frac{1}{q_n^3}, \frac{p_n}{q_n} + \frac{1}{2^{n+1}}\frac{1}{q_n^3}]$.

Then finally define $f(x) := \sum_{n=1}^{\infty} D_{I_n}(x)$.

I claim that this $f$ satisfies the criteria stated at the beginning. Firstly, the convergence and continuity of $f$ is immediate by using the Weierstrass M-test with $\frac{1}{2^n}$.

Preliminaries for Differentiability

The set $A_1 := \{x \in \mathbb{R} : x \text{ does not have a Liouville-Roth irrationality measure of } 2\}$ is known to be a null set.

The set $A_2 := \bigcup_{n=1}^\infty \{x \in \mathbb{R} : D_{I_n} \text{ is not differentiable at } x\}$ is a countable union of null sets, and is therefore also null.

Finally, the set $B := A_1 \cup A_2$ is a null set.


Given that $r \not\in B$.

Since $r$ has an irrationality measure of $2$, there exists $N \in \mathbb{N}^{\geq 1}$ such that $|r - \frac{p_n}{q_n}| \geq \frac{1}{q_n^3}$ whenever $n \geq N$.

So if $n \geq N$, this means $D_{I_n}(r+h) = 0$ whenever $|h| \leq \frac{1}{q_n^3} - \frac{1}{2^{n+1}}\frac{1}{q_n^3}$, and hence:

$$\small{\frac{|D_{I_n}(r+h)|}{|h|} \leq \frac{\frac{1}{2^n} \frac{1}{q_n^3}}{\frac{1}{q_n^3} - \frac{1}{2^{n+1}} \frac{1}{q_n^3}}}$$

$$\small{\frac{|D_{I_n}(r+h)|}{|h|} \leq \frac{2}{2^{n+1} - 1}}$$

$$\small{\frac{|D_{I_n}(r+h)|}{|h|} \leq \frac{2}{2^n}}$$

Also clearly $r \not\in I_n$, meaning that $D_{I_n}(r) = 0$ and $D_{I_n}'(r) = 0$. Therefore:

$$\small{- \sum_{k=n+1}^{\infty} \frac{2}{2^k} \leq \frac{f(r+h) - f(r)}{h} - \frac{\sum_{k=1}^n D_{I_k}(r+h) - \sum_{k=1}^n D_{I_k}(r)}{h} \leq \sum_{k=n+1}^{\infty} \frac{2}{2^k}}$$

$$\small{-\frac{2}{2^{n}} \leq \frac{f(r+h) - f(r)}{h} - \sum_{k=1}^n \frac{D_{I_k}(r+h) - D_{I_k}(r)}{h} \leq \frac{2}{2^{n}}}$$

$$\small{- \frac{2}{2^{n}} \leq \liminf_{h \rightarrow 0}\frac{f(r+h) - f(r)}{h} - \sum_{k=1}^n D_{I_k}'(r) \leq \limsup_{h \rightarrow 0} \frac{f(r+h) - f(r)}{h} - \sum_{k=1}^n D_{I_k}'(r) \leq \frac{2}{2^{n}}}$$

$$\small{- \frac{2}{2^{n}} \leq \liminf_{h \rightarrow 0}\frac{f(r+h) - f(r)}{h} - \sum_{k=1}^{\infty} D_{I_k}'(r) \leq \limsup_{h \rightarrow 0} \frac{f(r+h) - f(r)}{h} - \sum_{k=1}^{\infty} D_{I_k}'(r) \leq \frac{2}{2^{n}}}$$

As $n$ is arbitrary, we conclude that $\lim_{h \rightarrow 0} \frac{f(r+h) - f(r)}{h} = \sum_{k=1}^{\infty} D_{I_k}'(r)$.

Hence $f$ is differentiable at $r$ with $f'(r) = \sum_{k=1}^{\infty} D_{I_k}'(r)$.


Due to the above, clearly $\int_{I_n} f' \geq \int_{I_n} D_{I_n}' \geq 1$ for all $n \in \mathbb{N^{\geq 1}}$. Since any non-empty open interval contains infinitely many disjoint $I_{n_1}, I_{n_2}, ...$ we must have $\int_a^b f' = +\infty$ for all $a,b \in \mathbb{R}$ with $a < b$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .