Henstock–Kurzweil integral

Builds on Partition of an interval Riemann sum

Definition

Let $a < b$ be real numbers and let $f \colon [a, b] \to \R$ be a function. A gauge on $[a, b]$ is a function $\delta \colon [a, b] \to (0, \infty)$ . A tagged partition $a = x_0 < x_1 < \cdots < x_n = b$ with tags $t_i \in [x_{i-1}, x_i]$ is $\delta$ -fine if $[x_{i-1}, x_i] \subseteq (t_i - \delta(t_i), t_i + \delta(t_i))$ for every $i$ . The function $f$ is Henstock–Kurzweil integrable on $[a, b]$ , with integral $A \in \R$ , if the following holds. For every $\varepsilon > 0$ there is a gauge $\delta$ such that $\Bigl\lvert \sum_{i=1}^{n} f(t_i)(x_i - x_{i-1}) - A \Bigr\rvert < \varepsilon$ for every $\delta$ -fine tagged partition of $[a, b]$ .

Remarks

Cousin's lemma says that for every gauge a $\delta$ -fine tagged partition exists, so the integral, when it exists, is unique. If the gauges can always be taken constant, the definition is that of the Riemann integral.

Example: let $F(x) = x^2 \sin(1/x^2)$ for $x \ne 0$ and $F(0) = 0$ , and $f = F'$ on $[0, 1]$ . The function $F$ is differentiable everywhere, so $f$ is Henstock–Kurzweil integrable with integral $F(1) - F(0) = \sin 1$ . Here $f(x) = 2x \sin(1/x^2) - (2/x)\cos(1/x^2)$ for $x \ne 0$ , and $\abs{f}$ is not Lebesgue integrable.

Non-example: $f(x) = 1/x$ for $0 < x \le 1$ and $f(0) = 0$ . A nonnegative Henstock–Kurzweil integrable function is Lebesgue integrable, and $\int_0^1 dx/x = \infty$ , so $f$ is not integrable.

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Tag	`Y3KYKTAMCT`
URI	`m2://Y3KYKTAMCT`
Kind	definition
Writing	W0 stub
MSC	26A39
Defines	Henstock–Kurzweil integral
Uses	Partition of an interval, Riemann sum
Changed	2 November 2026

Henstock–Kurzweil integral

Definition

Remarks

Reading path

Further reading

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