Henstock–Kurzweil integral

Given a tagged partition P of [a, b], that is,

${\displaystyle a=u_{0}<u_{1}<\cdots <u_{n}=b}$

together with each subinterval's tag defined as a point

${\displaystyle t_{i}\in [u_{i-1},u_{i}],}$

we define the Riemann sum for a function ${\displaystyle f\colon [a,b]\to \mathbb {R} }$ to be

${\displaystyle \sum _{P}f=\sum _{i=1}^{n}f(t_{i})\Delta u_{i}.}$

where ${\displaystyle \Delta u_{i}:=u_{i}-u_{i-1}.}$ This is the summation of each subinterval's length (${\displaystyle \Delta u_{i}}$) multiplied by the function evaluated at that subinterval's tag (${\displaystyle f(t_{i})}$).

Given a positive function

${\displaystyle \delta \colon [a,b]\to (0,\infty ),}$

which we call a gauge, we say a tagged partition P is ${\displaystyle \delta }$-fine if

${\displaystyle \forall i\ \ [u_{i-1},u_{i}]\subset [t_{i}-\delta (t_{i}),t_{i}+\delta (t_{i})].}$

We now define a number I to be the Henstock–Kurzweil integral of f if for every ε > 0 there exists a gauge ${\displaystyle \delta }$ such that whenever P is ${\displaystyle \delta }$-fine, we have

${\displaystyle \left\vert I-\sum _{P}f\right\vert <\varepsilon .}$

If such an I exists, we say that f is Henstock–Kurzweil integrable on [a, b].

Cousin's theorem states that for every gauge ${\displaystyle \delta }$, such a ${\displaystyle \delta }$-fine partition P does exist, so this condition cannot be satisfied vacuously. The Riemann integral can be regarded as the special case where we only allow constant gauges.

Let f: [a, b] → R be any function.

Given a < c < b, f is Henstock–Kurzweil integrable on [a, b] if and only if it is Henstock–Kurzweil integrable on both [a, c] and [c, b]; in which case,

${\displaystyle \int _{a}^{b}f(x)\,dx=\int _{a}^{c}f(x)\,dx+\int _{c}^{b}f(x)\,dx.}$

Henstock–Kurzweil integrals are linear. Given integrable functions f, g and real numbers α, β, the expression αf + βg is integrable; for example,

${\displaystyle \int _{a}^{b}\left(\alpha f(x)+\beta g(x)\right)dx=\alpha \int _{a}^{b}f(x)\,dx+\beta \int _{a}^{b}g(x)\,dx.}$

If f is Riemann or Lebesgue integrable, then it is also Henstock–Kurzweil integrable, and calculating that integral gives the same result by all three formulations. The important Hake's theorem states that

${\displaystyle \int _{a}^{b}f(x)\,dx=\lim _{c\to b^{-}}\int _{a}^{c}f(x)\,dx}$

whenever either side of the equation exists, and likewise symmetrically for the lower integration bound. This means that if f is "improperly Henstock–Kurzweil integrable", then it is properly Henstock–Kurzweil integrable; in particular, improper Riemann or Lebesgue integrals of types such as

${\displaystyle \int _{0}^{1}{\frac {\sin(1/x)}{x}}\,dx}$

are also proper Henstock–Kurzweil integrals. To study an "improper Henstock–Kurzweil integral" with finite bounds would not be meaningful. However, it does make sense to consider improper Henstock–Kurzweil integrals with infinite bounds such as

${\displaystyle \int _{a}^{\infty }f(x)\,dx:=\lim _{b\to \infty }\int _{a}^{b}f(x)\,dx.}$

For many types of functions the Henstock–Kurzweil integral is no more general than Lebesgue integral. For example, if f is bounded with compact support, the following are equivalent:

• f is Henstock–Kurzweil integrable,
• f is Lebesgue integrable,
• f is Lebesgue measurable.

In general, every Henstock–Kurzweil integrable function is measurable, and f is Lebesgue integrable if and only if both f and |f| are Henstock–Kurzweil integrable. This means that the Henstock–Kurzweil integral can be thought of as a "non-absolutely convergent version of the Lebesgue integral". It also implies that the Henstock–Kurzweil integral satisfies appropriate versions of the monotone convergence theorem (without requiring the functions to be nonnegative) and dominated convergence theorem (where the condition of dominance is loosened to g(x) ≤ f_n(x) ≤ h(x) for some integrable g, h).

If F is differentiable everywhere (or with countably many exceptions), the derivative F′ is Henstock–Kurzweil integrable, and its indefinite Henstock–Kurzweil integral is F. (Note that F′ need not be Lebesgue integrable.) In other words, we obtain a simpler and more satisfactory version of the second fundamental theorem of calculus: each differentiable function is, up to a constant, the integral of its derivative:

${\displaystyle F(x)-F(a)=\int _{a}^{x}F'(t)\,dt.}$

Conversely, the Lebesgue differentiation theorem continues to hold for the Henstock–Kurzweil integral: if f is Henstock–Kurzweil integrable on [a, b], and

${\displaystyle F(x)=\int _{a}^{x}f(t)\,dt,}$

then F′(x) = f(x) almost everywhere in [a, b] (in particular, F is differentiable almost everywhere).

The space of all Henstock–Kurzweil-integrable functions is often endowed with the Alexiewicz norm, with respect to which it is barrelled but incomplete.

The gauge integral has increased utility when compared to the Riemann Integral in that the gauge integral of any function f: [a, b] → R which has a constant value c except possibly at a countable number of points ${\displaystyle C=\{c_{i}:i\in \mathbb {N} \}}$ can be calculated. Consider for example the piece-wise function

$${\displaystyle f(t)={\begin{cases}0,&{\text{if }}t\in [0,1]{\text{ and rational,}}\\1,&{\text{if }}t\in [0,1]{\text{ and irrational}}\end{cases}}}$$

This function is impossible to integrate using a Riemann Integral because it is impossible to make intervals ${\displaystyle [u_{i-1},u_{i}]}$ small enough to encapsulate the changing values of f(x). With the mapping nature of ${\displaystyle \delta }$-fine tagged partitions.

The value of the type of integral described above is equal to ${\displaystyle c(b-a)}$, where c is the constant value of the function, and a, b are the function's end points. To demonstrate this, let ${\displaystyle \varepsilon >0}$ be given and let ${\displaystyle D=\{(z_{j},J_{j}):1\leq j\leq n\}}$ be a ${\displaystyle \delta }$-fine tagged partition of ${\displaystyle [0,1]}$ with tags ${\displaystyle z_{j}}$ and intervals ${\displaystyle J_{j}}$, and let ${\displaystyle f(t)}$ be the piecewise function described above. Consider that

$${\displaystyle \left|\sum f(z_{j})l(J_{j})-1(1-0)\right|=\left|\sum [f(z_{j})-1]l(J_{j})\right|}$$

where ${\displaystyle l(J_{j})}$ represents the length of interval ${\displaystyle J_{j}}$. Note this equivalence is established because the summation of the differences in length of all intervals ${\displaystyle J_{j}}$ is equal to the length of the interval (or ${\displaystyle (1-0)}$).

By the definition of the gauge integral, we want to show that the above equation is less than any given ${\displaystyle \varepsilon }$. This produces two cases:

Case 1: ${\displaystyle z_{j}\notin C}$ (All tags of ${\displaystyle D}$ are irrational):

If none of the tags of the tagged partition ${\displaystyle D}$ are rational, then ${\displaystyle f(z_{j})}$ will always be 1 by the definition of ${\displaystyle f(t)}$, meaning ${\displaystyle [f(z_{j})-1]=0}$. If this term is zero, then for any interval length, the following inequality will be true:

$${\displaystyle \left|\sum [f(z_{j})-1]l(J_{j})\right|\leq \varepsilon ,}$$

So for this case, 1 is the integral of ${\displaystyle f(t)}$.

Case 2: ${\displaystyle z_{k}=c_{k}}$ (Some tag of ${\displaystyle D}$ is rational):

If a tag of ${\displaystyle D}$ is rational, then the function evaluated at that point will be 0, which is a problem. Since we know ${\displaystyle D}$ is ${\displaystyle \delta }$-fine, the inequality

$${\displaystyle \left|\sum [f(z_{j})-1]l(J_{j})\right|\leq \left|\sum [f(z_{j})-1]l(\delta (c_{k}))\right|}$$

holds because the length of any interval ${\displaystyle J_{j}}$ is shorter than its covering by the definition of being ${\displaystyle \delta }$-fine. If we can construct a gauge ${\displaystyle \delta }$ out of the right side of the inequality, then we can show the criteria are met for an integral to exist.

To do this, let ${\displaystyle \gamma _{k}=\varepsilon /[f(c_{k})-c]2^{k+2}}$ and set our covering gauges ${\displaystyle \delta (c_{k})=(c_{k}-\gamma _{k},c_{k}+\gamma _{k})}$, which makes

$${\displaystyle \left|\sum [f(z_{j})-c]l(J_{j})\right|<\varepsilon /2^{k+1}.}$$

From this, we have that

$${\displaystyle \left|\sum [f(z_{j})-1]l(J_{j})\right|\leq 2\sum \varepsilon /2^{k+1}=\varepsilon }$$

Because

$${\displaystyle 2\sum 1/2^{k+1}=1}$$

as a geometric series. This indicates that for this case, 1 is the integral of ${\displaystyle f(t)}$.

Since cases 1 and 2 are exhaustive, this shows that the integral of ${\displaystyle f(t)}$ is 1 and all properties from the above section hold.

Lebesgue integral on a line can also be presented in a similar fashion.

If we take the definition of the Henstock–Kurzweil integral from above, and we drop the condition

${\displaystyle t_{i}\in [u_{i-1},u_{i}],}$

then we get a definition of the McShane integral, which is equivalent to the Lebesgue integral. Note that the condition

${\displaystyle \forall i\ \ [u_{i-1},u_{i}]\subset [t_{i}-\delta (t_{i}),t_{i}+\delta (t_{i})]}$

does still apply, and we technically also require ${\textstyle t_{i}\in [a,b]}$ for ${\textstyle f(t_{i})}$ to be defined.

• Pfeffer integral
• Cauchy principal value
• Hadamard finite part integral

The following are additional resources on the web for learning more:

• "Kurzweil-Henstock integral", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• An Introduction to The Gauge Integral
• An Open Suggestion: To replace the Riemann integral with the gauge integral in calculus textbooks signed by Bartle, Henstock, Kurzweil, Schechter, Schwabik, and Výborný