Riemann Integrable Functions

Overview

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

$S=\left\{(x,y)\,:\ a\leq x\leq b,0<y<f(x)\right\}$

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:

$\int _{a}^{b}f(x)\,dx.$

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

$a=x_{0}<x_{1}<x_{2}<\dots <x_{n}=b$

Each [x_i, x_{i + 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,

$\max \left(x_{i+1}-x_{i}\right),\quad i\in [0,n-1].$

A tagged partition P(x, t) of an interval [a, b] is a partition together with a finite sequence of numbers t₀, ..., t_{n − 1} subject to the conditions that for each i, t_i ∈ [x_i, x_{i + 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, ..., x_n together with t₀, ..., t_{n − 1} is

$\sum _{i=0}^{n-1}f(t_{i})\left(x_{i+1}-x_{i}\right).$

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(t_i) and width x_{i + 1} − x_i. 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:

${\begin{aligned}L(f,P)&=\sum _{i=0}^{n-1}\inf _{t\in [x_{i},x_{i+1}]}f(t)(x_{i+1}-x_{i}),\\U(f,P)&=\sum _{i=0}^{n-1}\sup _{t\in [x_{i},x_{i+1}]}f(t)(x_{i+1}-x_{i}).\end{aligned}}$

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.

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, ..., x_n and t₀, ..., t_{n − 1} whose mesh is less than δ, we have

$\left|\left(\sum _{i=0}^{n-1}f(t_{i})(x_{i+1}-x_{i})\right)-s\right|<\varepsilon .$

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, ..., y_m and r₀, ..., r_{m − 1} such that for any tagged partition x₀, ..., x_n and t₀, ..., t_{n − 1} which is a refinement of y₀, ..., y_m and r₀, ..., r_{m − 1}, we have

$\left|\left(\sum _{i=0}^{n-1}f(t_{i})(x_{i+1}-x_{i})\right)-s\right|<\varepsilon .$

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, ..., y_m 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

$r=2\sup _{x\in [a,b]}|f(x)|.$

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

$\delta <\min \left\{{\frac {\varepsilon }{2r(m-1)}},\left(y_{1}-y_{0}\right),\left(y_{2}-y_{1}\right),\cdots ,\left(y_{m}-y_{m-1}\right)\right\}$

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

To see this, choose an interval [x_i, x_{i + 1}]. If this interval is contained within some [y_j, y_{j + 1}], then

$m_{j}<f(t_{i})<M_{j}$

where m_j and M_j are respectively, the infimum and the supremum of f on [y_j, y_{j + 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 [x_i, x_{i + 1}] is not contained in any [y_j, y_{j + 1}]. Instead, it may stretch across two of the intervals determined by y₀, ..., y_m. (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

$y_{j}<x_{i}<y_{j+1}<x_{i+1}<y_{j+2}.$

(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, ..., x_n at y_{j + 1}. The term f(t_i)(x_{i + 1} − x_i) in the Riemann sum splits into two terms:

$f\left(t_{i}\right)\left(x_{i+1}-x_{i}\right)=f\left(t_{i}\right)\left(x_{i+1}-y_{j+1}\right)+f\left(t_{i}\right)\left(y_{j+1}-x_{i}\right).$

Suppose, without loss of generality, that t_i ∈ [y_j, y_{j + 1}]. Then

$m_{j}<f(t_{i})<M_{j},$

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

$x_{i+1}-y_{j+1}<\delta <{\frac {\varepsilon }{2r(m-1)}},$

It follows that, for some (indeed any) t_i^* ∈ [y_{j + 1}, x_{i + 1}],

$\left|f\left(t_{i}\right)-f\left(t_{i}^{*}\right)\right|\left(x_{i+1}-y_{j+1}\right)<{\frac {\varepsilon }{2(m-1)}}.$

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 $f:[0,1]\to \mathbb {R}$ 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 $I_{\mathbb {Q} }:[0,1]\to \mathbb {R}$ be the indicator function of the rational numbers in [0, 1]; that is, $I_{\mathbb {Q} }$ 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, ..., x_n and t₀, ..., t_{n − 1} be a tagged partition (each t_i is between x_i and x_{i + 1}). Choose ε > 0. The t_i 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 t_i, we can minimize the effect of the t_i. 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 t_i, 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 t_i 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 t_i are within δ of each other, choose δ smaller. If it happens that some t_i is within δ of some x_j, and t_i is not equal to x_j, choose δ smaller. Since there are only finitely many t_i and x_j, we can always choose δ sufficiently small.

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

$\left[t_{i}-{\frac {\delta }{2}},t_{i}+{\frac {\delta }{2}}\right].$

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

$\left[t_{i}-{\frac {\delta }{2}},x_{j}\right],\quad {\text{and}}\quad \left[x_{j},t_{i}+{\frac {\delta }{2}}\right].$

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. $I_{\mathbb {Q} }$ 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 I_C be its indicator function. Because C is not Jordan measurable, I_C is not Riemann integrable. Moreover, no function g equivalent to I_C is Riemann integrable: g, like I_C, 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 I_C, 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

$\int _{a}^{b}(\alpha f(x)+\beta g(x))\,dx=\alpha \int _{a}^{b}f(x)\,dx+\beta \int _{a}^{b}g(x)\,dx.$

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.

Proof

The proof is easiest using the Darboux integral definition of integrability (formally, the Riemann condition for integrability) – a function is Riemann integrable if and only if the upper and lower sums can be made arbitrarily close by choosing an appropriate partition. One direction can be proven using the oscillation definition of continuity: For every positive ε, Let Xε be the set of points in [a, b] with oscillation of at least ε. Since every point where f is discontinuous has a positive oscillation and vice versa, the set of points in [a, b], where f is discontinuous is equal to the union over {X1/n} for all natural numbers n. If this set does not have a zero Lebesgue measure, then by countable additivity of the measure there is at least one such n so that X1/n does not have a zero measure. Thus there is some positive number c such that every countable collection of open intervals covering X1/n has a total length of at least c. In particular this is also true for every such finite collection of intervals. This remains true also for X1/n less a finite number of points (as a finite number of points can always be covered by a finite collection of intervals with arbitrarily small total length). For every partition of [a, b], consider the set of intervals whose interiors include points from X1/n. These interiors consist of a finite open cover of X1/n, possibly up to a finite number of points (which may fall on interval edges). Thus these intervals have a total length of at least c. Since in these points f has oscillation of at least 1/n, the infimum and supremum of f in each of these intervals differ by at least 1/n. Thus the upper and lower sums of f differ by at least c/n. Since this is true for every partition, f is not Riemann integrable. We now prove the converse direction using the sets Xε defined above. For every ε, Xε is compact, as it is bounded (by a and b) and closed: For every series of points in Xε that is converging in [a, b], its limit is in Xε as well. This is because every neighborhood of the limit point is also a neighborhood of some point in Xε, and thus f has an oscillation of at least ε on it. Hence the limit point is in Xε. Now, suppose that f is continuous almost everywhere. Then for every ε, Xε has zero Lebesgue measure. Therefore, there is a countable collections of open intervals in [a, b] which is an open cover of Xε, such that the sum over all their lengths is arbitrarily small. Since Xε is compact, there is a finite subcover – a finite collections of open intervals in [a, b] with arbitrarily small total length that together contain all points in Xε. We denote these intervals {I(ε)i}, for 1 ≤ i ≤ k, for some natural k. The complement of the union of these intervals is itself a union of a finite number of intervals, which we denote {J(ε)i} (for 1 ≤ i ≤ k − 1 and possibly for i = k, k + 1 as well). We now show that for every ε > 0, there are upper and lower sums whose difference is less than ε, from which Riemann integrability follows. To this end, we construct a partition of [a, b] as follows: Denote ε1 = ε / 2(b − a) and ε2 = ε / 2(M − m), where m and M are the infimum and supremum of f on [a, b]. Since we may choose intervals {I(ε1)i} with arbitrarily small total length, we choose them to have total length smaller than ε2. Each of the intervals {J(ε1)i} has an empty intersection with Xε1, so each point in it has a neighborhood with oscillation smaller than ε1. These neighborhoods consist of an open cover of the interval, and since the interval is compact there is a finite subcover of them. This subcover is a finite collection of open intervals, which are subintervals of J(ε1)i (except for those that include an edge point, for which we only take their intersection with J(ε1)i). We take the edge points of the subintervals for all J(ε1)i − s, including the edge points of the intervals themselves, as our partition. Thus the partition divides [a, b] to two kinds of intervals: Intervals of the latter kind (themselves subintervals of some J(ε1)i). In each of these, f oscillates by less than ε1. Since the total length of these is not larger than b − a, they together contribute at most ε1(b − a) = ε/2 to the difference between the upper and lower sums of the partition. The intervals {I(ε)i}. These have total length smaller than ε2, and f oscillates on them by no more than M − m. Thus together they contribute less than ε2(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 ε, as required.

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

$\int _{a}^{b}f\,dx=\int _{a}^{b}{\lim _{n\to \infty }{f_{n}}\,dx}=\lim _{n\to \infty }\int _{a}^{b}f_{n}\,dx.$

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:

• Riemann integral, Wikipedia under a CC BY-SA license