Henstock–Kurzweil integral

Utility[edit]

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 piecewise function $${\displaystyle f(t)={\begin{cases}0,&{\text{if }}t\in [0,1]{\text{ and rational,}}\\1,&{\text{if }}t\in [0,1]{\text{ and irrational}}\end{cases}}}$$ which is equal to one minus the Dirichlet function on the interval.


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 endpoints. 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 consecutive 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.