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Let $S$ be a closed geometric set in the plane. If $f$ is a real-valued function on $S,$ we would like to define what it means for $f$ to be “integrable” and then what the “integral” of $f$ is. To do this, we will simply mimic our development for integration of functions on a closed interval $[a,b].$
So, what should be a “step function” in this context? That is, what should is a “partition” of $S$ be in this context? Presumably a step function is going to be a function that is constant on the “elements” of a partition.Our idea is to replace the subintervals determined by a partition of the interval $[a,b]$ by geometric subsets of the geometric set $S.$
The overlap of two geometric sets ${S}_{1}$ and ${S}_{2}$ is defined to be the interior ${({S}_{1}\cap {S}_{2})}^{0}$ of their intersection. ${S}_{1}$ and ${S}_{2}$ are called nonoverlapping if this overlap ${({S}_{1}\cap {S}_{2})}^{0}$ is the empty set.
A partition of a closed geometric set $S$ in ${R}^{2}$ is a finite collection $\{{S}_{1},{S}_{2},...,{S}_{n}\}$ of nonoverlapping closed geometric sets for which ${\cup}_{i=1}^{n}{S}_{i}=S;$ i.e., the union of the ${S}_{i}$ 's is all of the geometric set $S.$
The open subsets $\left\{{S}_{i}^{0}\right\}$ are called the elements of the partition.
A step function on the closed geometric set $S$ is a real-valued function $h$ on $S$ for which there exists a partition $P=\left\{{S}_{i}\right\}$ of $S$ such that $h\left(z\right)={a}_{i}$ for all $z\in {S}_{i}^{0};$ i.e., $h$ is constant on each element of the partition $P.$
REMARK One example of a partition of a geometric set, though not at all the most general kind, is the following.Suppose the geometric set $S$ is determined by the interval $[a,b]$ and the two bounding functions $u$ and $l.$ Let $\{{x}_{0}<{x}_{1}<...<{x}_{n}\}$ be a partition of the interval $[a,b].$ We make a partition $\left\{{S}_{i}\right\}$ of $S$ by constructing vertical lines at the points ${x}_{i}$ from $l\left({x}_{i}\right)$ to $u\left({x}_{i}\right).$ Then ${S}_{i}$ is the geometric set determined by the interval $[{x}_{i-1},{x}_{i}]$ and the two bounding functions ${u}_{i}$ and ${l}_{i}$ that are the restrictions of $u$ and $l$ to the interval $[{x}_{i-1},{x}_{i}].$
A step function is constant on the open geometric sets that form the elements of some partition. We say nothing about the values of $h$ on the “boundaries” of these geometric sets. For a step function $h$ on an interval $[a,b],$ we do not worry about the finitely many values of $h$ at the endpoints of the subintervals. However, in the plane, we are ignoring the values on the boundaries, which are infinite sets. As a consequence, a step function on a geometric set may very well have an infinite range,and may not even be a bounded function, unlike the case for a step function on an interval.The idea is that the boundaries of geometric sets are “negligible” sets as far as area is concerned, so that the values of a function on these boundaries shouldn't affect the integral (average value) of the function.
Before continuing our development of the integral of functions in the plane, we digress to present an analog of [link] to functions that are continuous on a closed geometric set.
Let $f$ be a continuous real-valued function whose domain is a closed geometric set $S.$ Then there exists a sequence $\left\{{h}_{n}\right\}$ of step functions on $S$ that converges uniformly to $f.$
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