The *Henstock integral* (also attributed to *Kurzweil*, *Denjoy*, *Luzin*, and *Perron*, and sometimes called, neutrally but perhaps ambiguously, the *gauge integral*) is a way to define the integral of a (partial) function $f:\mathbb{R}\to \mathbb{R}$ which applies to more functions than either the Riemann integral or the Lebesgue integral and is in some ways better behaved as well.

However, the Lebesgue integral is more commonly used by working mathematicians because it fits more naturally into the general theory of measure, while the Riemann/Darboux integral is more commonly used in introductory calculus courses because its definition is simpler.

Let $[a,b]$ be a closed interval in $\mathbb{R}$ and let $f:[a,b]\to \mathbb{R}$. A *tagged partition* $P$ of $[a,b]$ is a finite sequence of points $a = u_0 \leq u_1 \leq \dots \leq u_n = b$ together with points $t_i \in [u_i, u_{i+1}]$ for all $i$. The *Riemann sum* of $f$ over such a tagged partition is

$\sum_P f = \sum_{i=0}^{n-1} f(t_i) \cdot (u_{i+1} - u_{i}).$

Define a *gauge* on $[a,b]$ to be any function $\delta: [a,b] \to (0,\infty]$. We say that a tagged partition is *$\delta$-fine* if $[u_i, u_{i+1}] \subset [t_i - \delta(t_i), t_i + \delta(t_i)]$.

Finally, we say that $I$ is the **integral** of $f$ on $[a,b]$, written $I = \int_a^b f = \int_{a}^b f(x) \,d x$, if for any $\epsilon\gt 0$ there exists a gauge $\delta$ such that

${\left| {\sum_P f - I} \right|} \lt \epsilon$

for any $\delta$-fine partition $P$. If such an $I$ exists, it must be unique, and we say that $f$ is (Henstock) **integrable** on $[a,b]$.

We can also write $\int_a f(x) \,d x$ for the semidefinite Henstock integral $\int_a^x f(t) \,d t$, and write $\int_a f$ for the function $x \mapsto \int_a f(x) \,d x$.

If we require a gauge to be a constant function, then we recover the definition of the Riemann integral.

Thus the Henstock integral may be seen as a non-uniform generalization of the Riemann integral. Whereas specifying a constant $\delta$ is tantamount to picking an entourage on $[a,b]$, specifying a gauge $\delta$ is tantamount to assigning a neighbourhood to each point in $[a,b]$. (Indeed, with either definition of integral, it would be equivalent to replace $\delta$ in the definition with an entourage or an assignment of neighbourhoods.) Similarly, the definition of uniformly continuous function becomes that of continuous function if you change $\delta$ from a constant to a gauge.

In constructive analysis, we must allow a gauge to take lower real values. (This is not necessary with the Riemann integral.) Otherwise, there may not be enough gauges, since these are rarely continuous. (The definition could also be made constructive by explicitly referring to an assignment of neighbourhoods to points or by replacing $\delta$ with an entire relation.)

The characteristic function $\chi_{\mathbb{Q}}$ of the rational numbers (as a subset of the real numbers) is a famous example of a function that’s Lebesgue integrable but not even locally Riemann integrable. It is Henstock integrable (with integral $0$) on any $[a,b]$ as follows: Enumerate the rationals in $[a,b]$ as $(q_i)_{i=0}^\infty$. Given $\epsilon \gt 0$, let $\delta(x)$ be $b-a$ if $x$ is irrational but $2^{-i-2}\epsilon(b-a)$ if $x$ is the rational $q_i$. Then $\sum_{x\in[a,b]} \chi_{\mathbb{Q}}(x) \,2\delta(x) = \epsilon$. Since a $\delta$-fine Riemann sum consists of just some of these terms, with $2\delta(x)$ replaced by a length that might be smaller, the value of the $\delta$-fine Riemann sum is thus always at most $\epsilon$.

The (even) function

$x\mapsto \frac{\sin(1/x^3)}{x},\quad x \in \mathbb{R}\setminus\{0\}$

is not Riemann or Lebesgue integrable on any interval containing 0, but it has the Henstock integral

$\int_{0} \frac{\sin (1/x^3)}{x}\,d x = \frac{1}{3}\left( \pi/2 - Si(1/x^3)\right)$

where $Si(t)$ is the sine integral? $\int_0 \frac{\sin(t)}{t}d t$ (which extends to an entire function on $\mathbb{C}$). This integral can also be found as an improper? Riemann integral.

For an example that is neither Lebesgue integrable nor improperly Riemann integrable (not even locally Riemann integrable), we can let $f(x)$ be $\sin(1/x^3)/x$ for irrational $x$ and $1$ for rational $x$. This one can still be done as an improper Lebesgue integral. (Are there any functions that are Henstock integrable but not locally Lebesgue integrable?)

The Henstock integral satisfies a very nice form of the fundamental theorem of calculus:

If $f$ is differentiable on $[a,b]$, then $f'$ is Henstock integrable on $[a,b]$, and $\int_a^b f'(x) d x = f(b) - f(a)$.

If $f$ is Henstock integrable on $[a,b]$, then $F(x) = \int_a f(x) d x$ is differentiable almost everywhere on $[a,b]$ and $F' = f|_{\dom F'}$.

For any $f$ we have

$\int_a^b f(x) d x = \lim_{c\to b^-} \int_a^c f(x) d x$

in the strong sense that if either side exists, then so does the other, and they are equal.

In particular, what is often taken as a *definition* of the improper? Riemann integral (of a potentially unbounded function on a finite interval) is actually a *theorem* for Henstock integrals. (However, we still need improper Henstock integrals to allow $a = -\infty$ or $b = \infty$.)

I need to check some of the claims below, but I'm out of time right now. They are definitely correct for the proper integrals. —Toby

Recall that $f\colon [a, b] \to \mathbb{R}$ is Riemann integrable iff $f$ is continuous almost everywhere and bounded; in this case, $f$ is also Henstock integrable, and the Riemann integral of $f$ equals its Henstock integral.

More generally, $f$ is improperly Riemann integrable iff $f$ is Henstock integrable and $f$ is locally Riemann integrable at all but finitely many points in $[a, b]$; then the improper Riemann integral of $f$ equals its Henstock integral.

Still more generally, $f\colon \mathbb{R} \to \mathbb{R}$ is improperly Riemann integrable iff $f$ is improperly Henstock integrable (meaning merely that

$\lim_{a \to -\infty, b \to \infty} \int_a^b f(x) \,\mathrm{d}x$

exists using Henstock integrals) and locally Riemann integrable except at a set of isolated points; then the improper Riemann integral of $f$ equals its improper Henstock integral.

Finally (and with incomparable generality), $f\colon \mathbb{R} \to \mathbb{R}$ is Lebesgue integrable iff ${|f|}$ is improperly Henstock integrable; then the Lebesgue integral of $f$ equals its improper Henstock integral (which is proper if the support of $f$ is bounded, of course).

Last revised on May 18, 2023 at 15:18:34. See the history of this page for a list of all contributions to it.