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When we defined the basic integral as the limit of a Riemannian sum we made an implicit assumption about the geometry of the realm on which we work out that sum. We say that the integral calculates the area under a curve on a graph and we assume that we know what we mean by area. In fact, we need to take a closer look at the concept of area.
Technically speaking, area does not properly exist for any figure except a set of squares. When we claim that we have calculated the area of a non-square figure we actually offer a count of identical non-overlapping squares that fill the figure exactly. If the figure has straight sides that do not meet at a right angle or has any curved sides, then a grid of squares will not fill the figure exactly. We thus have two ways to count up the area of the figure when we lay a regular grid of squares over the figure B we can exclude from the count the squares that cross the figure=s boundary or we can include in our count the squares that cross the figure=s boundary. Those two counts will differ from each other, but if we make the squares on our grid smaller, calibrating our counts to a fixed unit (such as, let=s say, a square centimeter), then we make the difference between the counts smaller. If we make the squares on our grid minuscule, of linear extent dx on each side, then as we make dx approach zero the difference approaches zero as well, which necessitates that the two counts approach a single value. That limit gives us what mathematicians call the Peano-Jordan content of the figure, which is named for Giuseppe Peano (1858 Aug 27 - 1932 Apr 20) and Marie Ennemond Camille Jordan (1838 Jan 05 - 1922 Jan 22).
If we have a figure with at least three straight sides such that one side coincides with the x-axis of our coordinate grid and two of the sides meet the endpoints of that side at a right angle, then we have a figure whose area we calculate with the Riemann integral. That figure thus gives us the Riemann content as a special case of the Peano-Jordan content of the more general figure.
Just as the Riemann content gives us a special case of Peano-Jordan content, so the Peano-Jordan content gives us a special case of Lebesgue content (or measure if we refer it to some standard unit). Each square on our grid constitutes a set of points with a specific Peano-Jordan content, the same for each square. In general we can use non-overlapping sets of different magnitudes. If we have a set S with non-overlapping (disjoint) subsets A and B, then we have the measure m(A)$0, m(B)$0, and m(AcB)=m(A)+m(B). For a simple example, if A represents some random event as a subset of the point set S of all elementary events, then 1$m(A)$0 represents the measure of the probability of A.
Imagine that we have a Lebesgue-measurable set S of finite measure (i.e. m(S)<4) that we have subdivided into non-overlapping subsets σi by the criterion to follow. Imagine also that we have a bounded, measurable function y=f(x) defined on S, which function has a minimum value (lower bound) L and maximum value (upper bound) U that define the interval [L,U] which we subdivide into N parts in accordance with
L=y0, y1, y2, ... yN-1, yN=U.
We define each subsetσi as the set of all points x for which
for i=1, 2, ... N-1 andσN as the set of all points x for which
(Eq= n 3)
With all of that in mind, we now define the Lebesgue integral as the limit as the greatest ofδi=yi-yi-1 goes toward zero of the Lebesgue sum over the measure of S;
In the Lebesgue integral thus defined, the measures of the subsetσi take the place of the differential increment in the coordinates that we see in the Riemann integral. The value of yi-1 or yi provides a kind of average for the value of f(x) and that average becomes a more accurate representation of the function as we invoke the limit. As in the Riemann integral, the value of N goes toward infinity as we shrink the subsets σi. In this case, though, the analogues of the differential have different magnitudes, so we must specify that the largest of the subsets go toward zero, the smaller ones shrinking as well.
That formulation enables us to define the Lebesgue-Stieltjes integral, the extension of the Lebesgue integral created by Thomas Joannes Stieltjes (1856 Dec 29 B 1894 Dec 31). We have an infinite set of points x comprising an interval [a,b] and on that interval we have a measurable function f(x) and a monotonically increasing functionφ(x). We define a new function F(ψ) with ψ taking values defined by
With that function we have F(ψ)=f(x) for every point in the interval [a,b] for which ψ=φ(x). If we have ψz… φ(x) for any xz, then we have a unique point of discontinuity such that
In that case we have F(ψz)=f(xz). If the first of the following integrals exists, then we have as true to mathematics the statement that
in which the integral on the left is a Lebesgue integral and the integral on the right is the Lebesgue-Stieltjes integral. If F(ψ) is measurable on the interval [φ(a),φ(b)], if f(x) is measurable on the interval [a,b], and we have
for some functionθ measurable on the interval [a,b], then we have the Lebesgue-Stieltjes integral as
in which the integral on the right is a Lebesgue integral.
Thus we extend the concept of the basic Riemann integral into a more general calculation.
We also have a Riemann-Stieltjes integral, a generalization of the basic integral that Bernard Riemann (1826 Sep 17 B 1866 Jul 20) defined as the limit approached by a Riemann sum as the subdivisions of the domain of summation/integration shrink to arbitrary smallness. Thomas Joannes Stieltjes extended that concept to an integral of one function with respect to some other function.
Let f(x) andφ(x) denote two real-valued, bounded functions (that is, functions whose values do not exceed certain fixed finite limits; essentially, functions that do not go toward infinity). Let those functions exist on a closed interval [a, b] and partition that interval according to
The Stieltjes integral then corresponds to
in whichξi denotes any point that lies in the subinterval [xi, xi+1]. We note the caveat that if f(x) and φ(x) each have a discontinuity at the same point x, the integral does not exist. If the integral exists and we have difficulty in solving it, we can resort to integration by parts,
At the end of his proof of the ergodic theorem George Birkhoff equated a Stieltjes integral ofλ with respect to the measure of the set Sλ to the integral of recurrence times over the set Sλ directly. From that integral we extract the conventional physicists= statement of the ergodic theorem: the average value of a measurable quantity over a system at any given instant equals the average value of that quantity measured of some small part of the system over a suitably long time.
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