Difference between revisions of "Riemann Integrable Functions"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
Line 171: Line 171:
 
Thus the partition divides {{math|[''a'', ''b'']}} to two kinds of intervals:
 
Thus the partition divides {{math|[''a'', ''b'']}} to two kinds of intervals:
  
*Intervals of the latter kind (themselves subintervals of some {{math|''J''(''ε''<sub>1</sub>)<sub>''i''</sub>}}). In each of these, {{mvar|f}} oscillates by less than {{math|''ε''<sub>1</sub>}}. Since the total length of these is not larger than {{math|''b'' − ''a''}}, they together contribute at most {{math|1=''ε''{{su|b=1|p=∗}}(''b'' − ''a'') = ''ε''/2}} to the difference between the upper and lower sums of the partition.
+
*Intervals of the latter kind (themselves subintervals of some {{math|''J''(''ε''<sub>1</sub>)<sub>''i''</sub>}}). In each of these, {{mvar|f}} oscillates by less than {{math|''ε''<sub>1</sub>}}. Since the total length of these is not larger than {{math|''b'' − ''a''}}, they together contribute at most {{math|1=''ε''<sub>1</sub><sup>*</sup>(''b'' − ''a'') = ''ε''/2}} to the difference between the upper and lower sums of the partition.
*The intervals {{math|{''I''(''ε'')<sub>''i''</sub>}|}}. These have total length smaller than {{math|''ε''<sub>2</sub>}}, and {{mvar|f}} oscillates on them by no more than {{math|''M'' − ''m''}}. Thus together they contribute less than {{math|1=''ε''{{su|b=2|p=∗}}(''M'' − ''m'') = ''ε''/2}} to the difference between the upper and lower sums of the partition.
+
*The intervals {{math|{''I''(''ε'')<sub>''i''</sub>}|}}. These have total length smaller than {{math|''ε''<sub>2</sub>}}, and {{mvar|f}} oscillates on them by no more than {{math|''M'' − ''m''}}. Thus together they contribute less than {{math|1=''ε''<sub>2</sub><sup>*</sup>(''M'' − ''m'') = ''ε''/2}} to the difference between the upper and lower sums of the partition.
  
 
In total, the difference between the upper and lower sums of the partition is smaller than {{mvar|ε}}, as required.
 
In total, the difference between the upper and lower sums of the partition is smaller than {{mvar|ε}}, as required.

Revision as of 15:12, 9 November 2021

Overview

Let f be a non-negative real-valued function on the interval [a, b], and let

be the region of the plane under the graph of the function f and above the interval [a, b] (see the figure on the top right). We are interested in measuring the area of S. Once we have measured it, we will denote the area by:

The basic idea of the Riemann integral is to use very simple approximations for the area of S. By taking better and better approximations, we can say that "in the limit" we get exactly the area of S under the curve.

Where f can be both positive and negative, the definition of S is modified so that the integral corresponds to the signed area under the graph of f: that is, the area above the x-axis minus the area below the x-axis.

Definition

Partitions of an interval

A partition of an interval [a, b] is a finite sequence of numbers of the form

Each [xi, xi + 1] is called a sub-interval of the partition. The mesh or norm of a partition is defined to be the length of the longest sub-interval, that is,

A tagged partition P(x, t) of an interval [a, b] is a partition together with a finite sequence of numbers t0, ..., tn − 1 subject to the conditions that for each i, ti ∈ [xi, xi + 1]. In other words, it is a partition together with a distinguished point of every sub-interval. The mesh of a tagged partition is the same as that of an ordinary partition.

Suppose that two partitions P(x, t) and Q(y, s) are both partitions of the interval [a, b]. We say that Q(y, s) is a refinement of P(x, t) if for each integer i, with i ∈ [0, n], there exists an integer r(i) such that {{{1}}} and such that {{{1}}} for some j with j ∈ [r(i), r(i + 1)). Said more simply, a refinement of a tagged partition breaks up some of the sub-intervals and adds tags to the partition where necessary, thus it "refines" the accuracy of the partition.

We can turn the set of all tagged partitions into a directed set by saying that one tagged partition is greater than or equal to another if the former is a refinement of the latter.

Riemann sum

Let f be a real-valued function defined on the interval [a, b]. The Riemann sum of f with respect to the tagged partition x0, ..., xn together with t0, ..., tn − 1 is

Each term in the sum is the product of the value of the function at a given point and the length of an interval. Consequently, each term represents the (signed) area of a rectangle with height f(ti) and width xi + 1xi. The Riemann sum is the (signed) area of all the rectangles.

Closely related concepts are the lower and upper Darboux sums. These are similar to Riemann sums, but the tags are replaced by the infimum and supremum (respectively) of f on each sub-interval:

If f is continuous, then the lower and upper Darboux sums for an untagged partition are equal to the Riemann sum for that partition, where the tags are chosen to be the minimum or maximum (respectively) of f on each subinterval. (When f is discontinuous on a subinterval, there may not be a tag that achieves the infimum or supremum on that subinterval.) The Darboux integral, which is similar to the Riemann integral but based on Darboux sums, is equivalent to the Riemann integral.

Riemann integral

Loosely speaking, the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer. If the limit exists then the function is said to be integrable (or more specifically Riemann-integrable). The Riemann sum can be made as close as desired to the Riemann integral by making the partition fine enough.[1]

One important requirement is that the mesh of the partitions must become smaller and smaller, so that in the limit, it is zero. If this were not so, then we would not be getting a good approximation to the function on certain subintervals. In fact, this is enough to define an integral. To be specific, we say that the Riemann integral of f equals s if the following condition holds:

For all ε > 0, there exists δ > 0 such that for any tagged partition x0, ..., xn and t0, ..., tn − 1 whose mesh is less than δ, we have

Unfortunately, this definition is very difficult to use. It would help to develop an equivalent definition of the Riemann integral which is easier to work with. We develop this definition now, with a proof of equivalence following. Our new definition says that the Riemann integral of f equals s if the following condition holds:

For all ε > 0, there exists a tagged partition y0, ..., ym and r0, ..., rm − 1 such that for any tagged partition x0, ..., xn and t0, ..., tn − 1 which is a refinement of y0, ..., ym and r0, ..., rm − 1, we have

Both of these mean that eventually, the Riemann sum of f with respect to any partition gets trapped close to s. Since this is true no matter how close we demand the sums be trapped, we say that the Riemann sums converge to s. These definitions are actually a special case of a more general concept, a net.

As we stated earlier, these two definitions are equivalent. In other words, s works in the first definition if and only if s works in the second definition. To show that the first definition implies the second, start with an ε, and choose a δ that satisfies the condition. Choose any tagged partition whose mesh is less than δ. Its Riemann sum is within ε of s, and any refinement of this partition will also have mesh less than δ, so the Riemann sum of the refinement will also be within ε of s.

To show that the second definition implies the first, it is easiest to use the Darboux integral. First, one shows that the second definition is equivalent to the definition of the Darboux integral. Now we will show that a Darboux integrable function satisfies the first definition. Fix ε, and choose a partition y0, ..., ym such that the lower and upper Darboux sums with respect to this partition are within ε/2 of the value s of the Darboux integral. Let

If {{{1}}}, then f is the zero function, which is clearly both Darboux and Riemann integrable with integral zero. Therefore, we will assume that r > 0. If m > 1, then we choose δ such that

If {{{1}}}, then we choose δ to be less than one. Choose a tagged partition x0, ..., xn and t0, ..., tn − 1 with mesh smaller than δ. We must show that the Riemann sum is within ε of s.

To see this, choose an interval [xi, xi + 1]. If this interval is contained within some [yj, yj + 1], then

where mj and Mj are respectively, the infimum and the supremum of f on [yj, yj + 1]. If all intervals had this property, then this would conclude the proof, because each term in the Riemann sum would be bounded by a corresponding term in the Darboux sums, and we chose the Darboux sums to be near s. This is the case when {{{1}}}, so the proof is finished in that case.

Therefore, we may assume that m > 1. In this case, it is possible that one of the [xi, xi + 1] is not contained in any [yj, yj + 1]. Instead, it may stretch across two of the intervals determined by y0, ..., ym. (It cannot meet three intervals because δ is assumed to be smaller than the length of any one interval.) In symbols, it may happen that

(We may assume that all the inequalities are strict because otherwise we are in the previous case by our assumption on the length of δ.) This can happen at most m − 1 times.

To handle this case, we will estimate the difference between the Riemann sum and the Darboux sum by subdividing the partition x0, ..., xn at yj + 1. The term f(ti)(xi + 1xi) in the Riemann sum splits into two terms:

Suppose, without loss of generality, that ti ∈ [yj, yj + 1]. Then

so this term is bounded by the corresponding term in the Darboux sum for yj. To bound the other term, notice that

It follows that, for some (indeed any) ti* ∈ [yj + 1, xi + 1],

Since this happens at most m − 1 times, the distance between the Riemann sum and a Darboux sum is at most ε/2. Therefore, the distance between the Riemann sum and s is at most ε.

Examples

Let be the function which takes the value 1 at every point. Any Riemann sum of f on [0, 1] will have the value 1, therefore the Riemann integral of f on [0, 1] is 1.

Let be the indicator function of the rational numbers in [0, 1]; that is, takes the value 1 on rational numbers and 0 on irrational numbers. This function does not have a Riemann integral. To prove this, we will show how to construct tagged partitions whose Riemann sums get arbitrarily close to both zero and one.

To start, let x0, ..., xn and t0, ..., tn − 1 be a tagged partition (each ti is between xi and xi + 1). Choose ε > 0. The ti have already been chosen, and we can't change the value of f at those points. But if we cut the partition into tiny pieces around each ti, we can minimize the effect of the ti. Then, by carefully choosing the new tags, we can make the value of the Riemann sum turn out to be within ε of either zero or one.

Our first step is to cut up the partition. There are n of the ti, and we want their total effect to be less than ε. If we confine each of them to an interval of length less than ε/n, then the contribution of each ti to the Riemann sum will be at least 0 · ε/n and at most 1 · ε/n. This makes the total sum at least zero and at most ε. So let δ be a positive number less than ε/n. If it happens that two of the ti are within δ of each other, choose δ smaller. If it happens that some ti is within δ of some xj, and ti is not equal to xj, choose δ smaller. Since there are only finitely many ti and xj, we can always choose δ sufficiently small.

Now we add two cuts to the partition for each ti. One of the cuts will be at tiδ/2, and the other will be at ti + δ/2. If one of these leaves the interval [0, 1], then we leave it out. ti will be the tag corresponding to the subinterval

If ti is directly on top of one of the xj, then we let ti be the tag for both intervals:

We still have to choose tags for the other subintervals. We will choose them in two different ways. The first way is to always choose a rational point, so that the Riemann sum is as large as possible. This will make the value of the Riemann sum at least 1 − ε. The second way is to always choose an irrational point, so that the Riemann sum is as small as possible. This will make the value of the Riemann sum at most ε.

Since we started from an arbitrary partition and ended up as close as we wanted to either zero or one, it is false to say that we are eventually trapped near some number s, so this function is not Riemann integrable. However, it is Lebesgue integrable. In the Lebesgue sense its integral is zero, since the function is zero almost everywhere. But this is a fact that is beyond the reach of the Riemann integral.

There are even worse examples. is equivalent (that is, equal almost everywhere) to a Riemann integrable function, but there are non-Riemann integrable bounded functions which are not equivalent to any Riemann integrable function. For example, let C be the Smith–Volterra–Cantor set, and let IC be its indicator function. Because C is not Jordan measurable, IC is not Riemann integrable. Moreover, no function g equivalent to IC is Riemann integrable: g, like IC, must be zero on a dense set, so as in the previous example, any Riemann sum of g has a refinement which is within ε of 0 for any positive number ε. But if the Riemann integral of g exists, then it must equal the Lebesgue integral of IC, which is 1/2. Therefore, g is not Riemann integrable.

Properties

Linearity

The Riemann integral is a linear transformation; that is, if f and g are Riemann-integrable on [a, b] and α and β are constants, then

Because the Riemann integral of a function is a number, this makes the Riemann integral a linear functional on the vector space of Riemann-integrable functions.

Integrability

A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). This is the Lebesgue integrability condition (of characterization of the Riemann integrable functions). It has been proven independently by Giuseppe Vitali and by Henri Lebesgue in 1907, and uses the notion of measure zero, but makes use of neither Lebesgue's general measure or integral.

The integrability condition can be proven in various ways, one of which is sketched below.

In particular, any set that is at most countable has Lebesgue measure zero, and thus a bounded function (on a compact interval) with only finitely or countably many discontinuities is Riemann integrable.

An indicator function of a bounded set is Riemann-integrable if and only if the set is Jordan measurable. The Riemann integral can be interpreted measure-theoretically as the integral with respect to the Jordan measure.

If a real-valued function is monotone on the interval [a, b] it is Riemann-integrable, since its set of discontinuities is at most countable, and therefore of Lebesgue measure zero.

If a real-valued function on [a, b] is Riemann-integrable, it is Lebesgue-integrable. That is, Riemann-integrability is a stronger (meaning more difficult to satisfy) condition than Lebesgue-integrability.

For Lebesgue-Vitali seems that all type of discontinuities have the same weight on the obstruction that a real-valued bounded function be Riemann integrable on [a, b]. However this not the case. In fact, certain discontinuities have absolutely no role on the Riemann integrability of the function. That is a consequence of the classification of the discontinuities of a function.

If fn is a uniformly convergent sequence on [a, b] with limit f, then Riemann integrability of all fn implies Riemann integrability of f, and

However, the Lebesgue monotone convergence theorem (on a monotone pointwise limit) does not hold. In Riemann integration, taking limits under the integral sign is far more difficult to logically justify than in Lebesgue integration.

Licensing

Content obtained and/or adapted from:

  • Template:Cite book