nLab Henstock integral

Idea

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.

Definition

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 $x \mapsto \int_a f(x) \,d x$.

Improper and infinite integrals

We don't usually try to define a Henstock integral on a non-closed interval. If $f\colon {[a, b[} \to \mathbb{R}$ is integrable on every $[a, c]$ with $a \leq c \lt b$, then we might define $\int_a^b f$ to be $\lim_{c\nearrow{b}} \int_a^c f$ as an improper integral. However, Hake's Theorem assures us that in this case, we can define $f(b)$ however we like and find that this limit exists if and only $\int_a^b f$ exists by the proper definition above (and then the two values are equal). So we don't bother with improper bounded integrals.

However, we do still want to define a Henstock integral on a non-bounded interval. If $f\colon {[a, \infty[} \to \mathbb{R}$ is integrable on every $[a, c]$ with $a \leq c$, then we can define $\int_a^\infty f$ to be $\lim_{c\nearrow\infty} \int_a^c f$. Another approach is to take a diffeomorphism $\phi\colon {[a, c[} \to {[a, \infty[}$ (such as $\phi(t) \coloneqq \frac{t - a t + a c - a}{c - t}$) and define $\int_a^\infty f$ to be $\int_a^c (f \circ \phi) \phi'$. We can again use Hake's Theorem (and the substitution theorem for Henstock integrals, which works as usual) to show that these definitions are equivalent. Finally, we can work directly on $[-\infty, +\infty]$, requiring $\delta(-\infty)$ and $\delta(+\infty)$ to be part of the data of a gauge, replace ${[t - \delta(t), t + \delta(t)]}$ (in the definition of a $\delta$-fine partition) with ${[-\infty, -{/\delta(-\infty)}]}$ when $t = -\infty$, and replace this with ${[{/\delta(+\infty)}, +\infty]}$ when $t = +\infty$; with some care, this can also be proved equivalent. (This is the approach taken in Leader 2001.)

Comparison to the Riemann definition

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.

Constructive version

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.)

Cousin's Lemma

Conceivably, there might exist an interval $[a,b]$ and a gauge $\delta$ such that no $\delta$-fine partition of $[a,b]$ exists at all; in that case, the integral of any function $f$ on $[a,b]$ would vacuously take every value $I$. This is ruled out by Cousin's Lemma, which states precisely that such a partition always exists. This lemma can be leveraged to prove that the integral is not only nontrivial but also (if it exists at all) unique. (Proof sketch: prove that if $I$ is an integral of $f$ on $[a, b]$ and $J$ is an integral of $g$ on $[a, b]$, then $I - J$ is an integral of $f - g$ on $[a, b]$, which can be done in the usual way without assuming uniqueness; use Cousin's Lemma to prove that any integral of $0$ on $[a, b]$ must be $0$; conclude that if $I$ and $J$ are both integrals of $f$ on $[a, b]$, then $I - J = 0$.)

Examples

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, however. (Are there any functions that are Henstock integrable but not improperly Lebesgue integrable? In other words, it's Henstock integrable, but the set of points on which it's not locally Lebesgue integrable, which is $\{0\}$ in this example, has an accumulation point.)

Properties

The fundamental theorem of calculus

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

Theorem

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

Theorem

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 in fact $F' = f$ almost everywhere (although possibly not everywhere $F$ is differentiable).

Hake's theorem

Theorem

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.

Since the theorem holds regardless of the value $f(c)$, this has as a corollary that the values of a function can be changed on any finite number of points without affecting the existence or value of any integral. (Can it be changed on any set of measure zero, like the Lebesgue integral?)

Recovery of Riemann and Lebesgue integrals

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

Finally (and with incomparable generality), $f\colon \mathbb{R} \to \mathbb{R}$ is Lebesgue integrable iff ${|f|}$ is Henstock integrable; then the Lebesgue integral of $f$ equals its Henstock integral.

References

Here is a comprehensive introduction the Henstock integral, including in multiple dimensions:

• Solomon Leader (2001). The Kurzweil–Henstock Integral and its Differentials: A Unified Theory of Integration on $\mathbb{R}$ and $\mathbb{R}^n$. Marcel Dekker, Inc., 2001.