McShane integral

Given a closed interval [a, b] of the real line, a free tagged partition ${\displaystyle P}$ of ${\displaystyle [a,b]}$ is a set

${\displaystyle \{(t_{i},[a_{i-1},a_{i}]):1\leq i\leq n\}}$

where

${\displaystyle a=a_{0}<a_{1}<\dots <a_{n}=b}$

and each tag ${\displaystyle t_{i}\in [a,b]}$.

The fact that the tags are allowed to be outside the subintervals is why the partition is called free. It's also the only difference between the definitions of the Henstock-Kurzweil integral and the McShane integral.

For a function ${\displaystyle f:[a,b]\to \mathbb {R} }$ and a free tagged partition ${\displaystyle P}$, define ${\displaystyle S(f,P)=\sum _{i=1}^{n}f(t_{i})(a_{i}-a_{i-1}).}$

A positive function ${\displaystyle \delta$ :[a,b]\to (0,+\infty )} is called a gauge in this context.

We say that a free tagged partition ${\displaystyle P}$ is ${\displaystyle \delta }$-fine if for all ${\displaystyle i=1,2,\dots ,n,}$

${\displaystyle [a_{i-1},a_{i}]\subseteq [t_{i}-\delta (t_{i}),t_{i}+\delta (t_{i})].}$

Intuitively, the gauge controls the widths of the subintervals. Like with the Henstock-Kurzweil integral, this provides flexibility (especially near problematic points) not given by the Riemann integral.

The value ${\displaystyle \int _{a}^{b}f}$ is the McShane integral of ${\displaystyle f:[a,b]\to \mathbb {R} }$ if for every ${\displaystyle \varepsilon >0}$ we can find a gauge ${\displaystyle \delta }$ such that for all ${\displaystyle \delta }$-fine free tagged partitions ${\displaystyle P}$ of ${\displaystyle [a,b]}$,

${\displaystyle \left|\int _{a}^{b}f-S(f,P)\right|<\varepsilon .}$

Let ${\displaystyle f:[a,b]\to \mathbb {R} }$ be such that ${\displaystyle f(a)=f(b)=0}$ and ${\textstyle f(x)=1}$ if ${\textstyle x\in ]a,b[.}$

As is well known, this function is Riemann integrable and the correspondent integral is equal to ${\displaystyle b-a.}$ We will show that this ${\displaystyle f}$ is also McShane integrable and that its integral assumes the same value.

For that purpose, for a given ${\displaystyle \varepsilon >0}$, let's choose the gauge ${\displaystyle \delta (t)}$ such that ${\displaystyle \delta (a)=\delta (b)=\varepsilon /4}$ and ${\displaystyle \delta (t)=b-a}$ if ${\textstyle t\in ]a,b[.}$

Any free tagged partition ${\displaystyle P=\{(t_{i},[a_{i-1},a_{i}]):i=1,...,n\}}$ of ${\displaystyle [a,b]}$ can be decomposed into sequences like

${\displaystyle (a,[x_{i_{j}-1},x_{i_{j}}])}$, for ${\displaystyle j=1,...,\lambda }$,

${\displaystyle (b,[x_{i_{k}-1},x_{i_{k}}])}$, for ${\displaystyle k=1,...,\mu }$, and

${\displaystyle (t_{i_{r}},[x_{i_{r}-1},x_{i_{r}}])}$, where ${\displaystyle r=1,...,\nu }$, such that ${\displaystyle t_{i_{r}}\in ]a,b[}$ ${\displaystyle (\lambda +\mu +\nu =n).}$

This way, we have the Riemann sum

${\displaystyle S(f,P)=\sum _{r=1}^{\nu }\displaystyle (x_{i_{r}}-x_{i_{r}-1})}$

and by consequence

${\displaystyle |S(P,f)-(b-a)|=\textstyle \sum _{j=1}^{\lambda }\displaystyle (x_{i_{j}}-x_{i_{j}-1})+\textstyle \sum _{k=1}^{\mu }\displaystyle (x_{i_{k}}-x_{i_{k}-1}).}$

Therefore if ${\displaystyle P}$ is a free tagged ${\displaystyle \delta }$-fine partition we have

${\displaystyle [x_{i_{j}-1},x_{i_{j}}]\subset [a-\delta (a),a+\delta (a)]}$, for every ${\displaystyle j=1,...,\lambda }$, and

${\displaystyle [x_{i_{k}-1},x_{i_{k}}]\subset [b-\delta (b),b+\delta (b)]}$, for every ${\displaystyle k=1,...,\mu }$.

Since each one of those intervals do not overlap the interior of all the remaining, we obtain

${\displaystyle |S(P,f)-(b-a)|<2\delta (a)+2\delta (b)={\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2}}=\varepsilon .}$

Thus ${\displaystyle f}$ is McShane integrable and

${\displaystyle \int _{a}^{b}f=b-a.}$

The next example proves the existence of a distinction between Riemann and McShane integrals.

Let ${\displaystyle d:[a,b]\rightarrow \mathbb {R} }$ the well known Dirichlet's function given by

${\displaystyle d(x)={\begin{cases}1,&{\text{if }}x{\text{ is rational,}}\\0,&{\text{if }}x{\text{ is irrational,}}\end{cases}}}$

which one knows to be not Riemann integrable. We will show that ${\displaystyle d}$ is integrable in the MacShane sense and that its integral is zero.

Denoting by ${\displaystyle \{r_{1},r_{2},...,r_{n},...\}}$ the set of all rational numbers of the interval ${\displaystyle [a,b]}$, for any ${\displaystyle \varepsilon >0}$ let's formulate the following gauge

${\displaystyle \delta (x)={\begin{cases}\varepsilon 2^{-n-1},&{\text{if }}x=r_{n}{\text{ and }}n=1,2,...,\\1,&{\text{if }}x{\text{ is irrational.}}\end{cases}}}$

For any ${\displaystyle \delta }$-fine free tagged partition ${\displaystyle P=\{(t_{i},[x_{i-1},x_{i}]):i=1,...,n\}}$ consider its Riemann sum

${\displaystyle S(P,f)=\textstyle \sum _{i=1}^{n}\displaystyle f(t_{i})(x_{i}-x_{i-1})}$.

Taking into account that ${\displaystyle f(t_{i})=0}$ whenever ${\displaystyle t_{i}}$ is irrational, we can exclude in the sequence of ordered pairs which constitute ${\displaystyle P}$, the pairs ${\displaystyle (t_{i},[x_{i-1},x_{i}])}$ where ${\displaystyle t_{i}}$ is irrational. The remainder are subsequences of the type ${\displaystyle (r_{k},[x_{i_{1}-1},x_{i_{1}}]),...,(r_{k},[x_{i_{k}-1},x_{i_{k}}])}$ such that ${\displaystyle [x_{i_{j}-1},x_{i_{j}}]\subset [r_{k}-\delta (r_{k}),r_{k}+\delta (r_{k})]}$, ${\displaystyle j=1,...,k.}$ Since each one of those intervals do not overlap the interior of the remaining, each one of these sequences gives rise in the Riemann sum to subsums of the type

${\textstyle \textstyle \sum _{j=1}^{k}\displaystyle f(r_{k})(x_{i_{j}}-x_{i_{j}-1})=\textstyle \sum _{j=1}^{k}\displaystyle (x_{i_{j}}-x_{i_{j}-1})\leq 2\delta (r_{k})={\frac {\varepsilon }{2^{n}}}}$.

Thus ${\textstyle 0\leq S(P,f)<\textstyle \sum _{n\geq 1}\displaystyle \varepsilon /2^{n}=\varepsilon }$, which proves that the Dirichlet's function is McShane integrable and that

${\displaystyle \int _{a}^{b}d=0.}$

If ${\displaystyle f:[a,b]\rightarrow \mathbb {R} }$ is McShane integrable on ${\displaystyle [a,b]}$ then ${\displaystyle |f|}$ is also McShane integrable on ${\displaystyle [a,b]}$ and${\textstyle |\int _{a}^{b}f|\leq \int _{a}^{b}|f|}$.

If ${\textstyle F:[a,b]\rightarrow \mathbb {R} }$ is differentiable on ${\displaystyle [a,b]}$, then ${\textstyle F'}$ is Henstock-Kurzweil integrable on ${\displaystyle [a,b]}$ and${\textstyle \int _{a}^{b}F'=F(b)-F(a)}$.

In order to illustrate these theorems we analyse the following example based upon Example 2.4.12.[4]

Let's consider the function:

${\displaystyle F(x)={\begin{cases}x^{2}\cos(\pi /x^{2}),&{\text{if }}x\neq 0,\\0,&{\text{if }}x=0.\end{cases}}}$

${\displaystyle F}$ is obviously differentiable at any ${\displaystyle x\neq 0}$ and differentiable, as well, at ${\displaystyle x=0}$, since${\textstyle \lim _{x\to 0}\left({\frac {F(x)}{x}}\right)=\lim _{x\to 0}\left(x\cos {\frac {\pi }{x^{2}}}\right)=0}$.

Moreover

${\textstyle F'(x)={\begin{cases}2x\cos(\pi /x^{2})+{\frac {2\pi }{x}}\sin(\pi /x^{2}),&{\text{if }}x\neq 0,\\0,&{\text{if }}x=0.\end{cases}}}$

As the function

${\displaystyle h(x)={\begin{cases}2x\cos(\pi /x^{2}),&{\text{if }}x\neq 0,\\0,&{\text{if }}x=0,\end{cases}}}$

is continuous and, by the Theorem 2, the function ${\displaystyle F'(x)}$ is Henstock-Kurzweil integrable on ${\displaystyle [0,1],}$ then by the properties 6 and 7, the same holds to the function

${\textstyle g_{0}(x)={\begin{cases}{\frac {1}{x}}\sin(\pi /x^{2}),&{\text{if }}x\neq 0,\\0,&{\text{if }}x=0.\end{cases}}}$

But the function

${\textstyle g(x)=|g_{0}(x)|={\begin{cases}{\frac {1}{x}}|\sin(\pi /x^{2})|,&{\text{if }}x\neq 0,\\0,&{\text{if }}x=0,\end{cases}}}$

is not integrable on ${\displaystyle [0,1]}$ for none of the mentioned integrals.

In fact, otherwise, denoting by ${\textstyle \int _{0}^{1}g(x)dx}$ anyone of such integrals, we should have necessarily ${\textstyle \int _{0}^{1}g(x)dx\geq \int _{1/{\sqrt {n}}}^{1}{\frac {1}{x}}|\sin(\pi /x^{2})|dx,}$ for any positive integer ${\displaystyle n}$. Then through the change of variable ${\textstyle x=1/{\sqrt {t}}}$, we should obtain taking into account the property 5:

${\displaystyle \int _{1/{\sqrt {n}}}^{1}{\frac {1}{x}}|\sin(\pi /x^{2})|dx={\frac {1}{2}}\int _{1}^{n}{\frac {1}{t}}|\sin(\pi t)|dt={\frac {1}{2}}\sum _{k=2}^{n}\int _{k-1}^{k}{\frac {1}{t}}|\sin(\pi t)|dt\geq }$

${\displaystyle \geq {\frac {1}{2}}\sum _{k=2}^{n}{\frac {1}{k}}\int _{k-1}^{k}|\sin(\pi t)|dt={\frac {1}{\pi }}\sum _{k=2}^{n}{\frac {1}{k}}}$.

As ${\displaystyle n}$ is an arbitrary positive integer and ${\textstyle \lim _{n\to \infty }\sum _{k=2}^{N}{\frac {1}{k}}=+\infty }$, we obtain a contradiction.

From this example we are able to conclude the following relevant consequences:

• I) Theorem 1 is no longer true for Henstock-Kurzweil integral since ${\displaystyle g_{0}}$ is Henstock-Kurzweil integrable and ${\displaystyle g}$ is not.

• II) Theorem 2 does not hold for McShane integral. Otherwise ${\displaystyle F'}$ should be McShane integrable as well as ${\displaystyle g_{0}}$ and by Theorem 1, as ${\displaystyle g}$, which is absurd.

• III) ${\displaystyle F'}$ is, this way, an example of a Henstock-Kurzweil integrable function which is not McShane integrable. That is, the class of McShane integrable functions is a strict subclass of the Henstock-Kurzweil integrable functions.

Let ${\displaystyle f:[a,b]\rightarrow \mathbb {R} }$. Then

${\displaystyle f}$ is McShane integrable ${\displaystyle \Leftrightarrow }$ ${\displaystyle f}$ is Lebesgue integrable.

The correspondent integrals coincide.

This fact enables to conclude that with the McShane integral one formulates a kind of unification of the integration theory around Riemann sums, which, after all, constitute the origin of that theory.

So far is not known an immediate proof of such theorem.

In Washek Pfeffer[4] [Ch. 4] it is stated through the development of the theory of McShane integral, including measure theory, in relationship with already known properties of Lebesgue integral.

Furtherly to the book by Russel Gordon,[3] on this subject we call the attention of the reader also to the works by Robert McLeod[5] [Ch. 8], [Ch. 10] and Douglas Kurtz and Charles Swartz.[2]

Another perspective of the McShane integral is that it can be looked as new formulation of the Lebesgue integral without using Measure Theory, as alternative to the courses of Frigyes Riesz and Bela Sz. Nagy[6] [Ch.II] or Serge Lang[7] [Ch.X, §4 Appendix] (see also[8]).