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In the branch ofmathematics known asreal analysis, theRiemann integral, created byBernhard Riemann, was the first rigorous definition of theintegral of afunction on aninterval. It was presented to the faculty at theUniversity of Göttingen in 1854, but not published in a journal until 1868.[1] For many functions and practical applications, the Riemann integral can be evaluated by thefundamental theorem of calculus or approximated bynumerical integration, or simulated usingMonte Carlo integration.
Imagine you have a curve on a graph, and the curve stays above the x-axis between two points, a and b. The area under that curve, from a to b, is what we want to figure out. This area can be described as the set of all points (x, y) on the graph that follow these rules: a ≤ x ≤ b (the x-coordinate is between a and b) and 0 < y < f(x) (the y-coordinate is between 0 and the height of the curve f(x)). Mathematically, this region can be expressed inset-builder notation as
To measure this area, we use aRiemann integral, which is written as:
This notation means “the integral of f(x) from a to b,” and it represents the exact area under the curve f(x) and above the x-axis, between x = a and x = b.
The idea behind the Riemann integral is to break the area into small, simple shapes (like rectangles), add up their areas, and then make the rectangles smaller and smaller to get a better estimate. In the end, when the rectangles are infinitely small, the sum gives the exact area, which is what the integral represents.
If the curve dips below the x-axis, the integral gives asigned area. This means the integral adds the part above the x-axis as positive and subtracts the part below the x-axis as negative. So, the result of can be positive, negative, or zero, depending on how much of the curve is above or below the x-axis.
Apartition of an interval[a,b] is a finite sequence of numbers of the form
Each[xi,xi + 1] is called asub-interval of the partition. Themesh ornorm of a partition is defined to be the length of the longest sub-interval, that is,
Atagged partitionP(x,t) of an interval[a,b] is a partition together with a choice of a sample point within each sub-interval: that is, numberst0, ...,tn − 1 withti ∈ [xi,xi + 1] for eachi. The mesh of a tagged partition is the same as that of an ordinary partition.
Suppose that two partitionsP(x,t) andQ(y,s) are both partitions of the interval[a,b]. We say thatQ(y,s) is arefinement ofP(x,t) if for each integeri, withi ∈ [0,n], there exists an integerr(i) such thatxi =yr(i) and such thatti =sj for somej withj ∈ [r(i),r(i + 1)]. That is, a tagged partition breaks up some of the sub-intervals and adds sample points where necessary, "refining" the accuracy of the partition.
We can turn the set of all tagged partitions into adirected set by saying that one tagged partition is greater than or equal to another if the former is a refinement of the latter.
Letf be a real-valued function defined on the interval[a,b]. TheRiemann sum off with respect to a tagged partitionP(x,t) of[a,b] is[2]
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 heightf(ti) and widthxi + 1 −xi. The Riemann sum is the (signed) area of all the rectangles.
Closely related concepts are thelower and upper Darboux sums. These are similar to Riemann sums, but the tags are replaced by theinfimum and supremum (respectively) off on each sub-interval:
Iff 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) off on each subinterval. (Whenf is discontinuous on a subinterval, there may not be a tag that achieves the infimum or supremum on that subinterval.) TheDarboux integral, which is similar to the Riemann integral but based on Darboux sums, is equivalent to the Riemann integral.
Loosely speaking, the Riemann integral is the limit of theRiemann sums of a function as the partitions get finer. If the limit exists then the function is said to beintegrable (or more specificallyRiemann-integrable). The Riemann sum can be made as close as desired to the Riemann integral by making the partition fine enough.[3]
One important requirement is that the mesh of the partitions must become smaller and smaller, so that it has the limit 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 off exists and equalss if the following condition holds:
For allε > 0, there existsδ > 0 such that for anytagged partitionx0, ...,xn andt0, ...,tn − 1 whose mesh is less thanδ, 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 off exists and equalss if the following condition holds:
For allε > 0, there exists a tagged partitiony0, ...,ym andr0, ...,rm − 1 such that for any tagged partitionx0, ...,xn andt0, ...,tn − 1 which is a refinement ofy0, ...,ym andr0, ...,rm − 1, we have
Both of these mean that eventually, the Riemann sum off with respect to any partition gets trapped close tos. Since this is true no matter how close we demand the sums be trapped, we say that the Riemann sums converge tos. These definitions are actually a special case of a more general concept, anet.
As we stated earlier, these two definitions are equivalent. In other words,s works in the first definition if and only ifs works in the second definition. To show that the first definition implies the second, start with anε, and choose aδ that satisfies the condition. Choose any tagged partition whose mesh is less thanδ. Its Riemann sum is withinε ofs, and any refinement of this partition will also have mesh less thanδ, so the Riemann sum of the refinement will also be withinε ofs.
To show that the second definition implies the first, it is easiest to use theDarboux integral. First, one shows that the second definition is equivalent to the definition of the Darboux integral; for this see theDarboux integral article. Now we will show that a Darboux integrable function satisfies the first definition. Fixε, and choose a partitiony0, ...,ym such that the lower and upper Darboux sums with respect to this partition are withinε/2 of the values of the Darboux integral. Let
Ifr = 0, thenf is the zero function, which is clearly both Darboux and Riemann integrable with integral zero. Therefore, we will assume thatr > 0. Ifm > 1, then we chooseδ such that
Ifm = 1, then we chooseδ to be less than one. Choose a tagged partitionx0, ...,xn andt0, ...,tn − 1 with mesh smaller thanδ. We must show that the Riemann sum is withinε ofs.
To see this, choose an interval[xi,xi + 1]. If this interval is contained within some[yj,yj + 1], thenwheremj andMj are respectively, the infimum and the supremum off 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 nears. This is the case whenm = 1, so the proof is finished in that case.
Therefore, we may assume thatm > 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 byy0, ...,ym. (It cannot meet three intervals becauseδ 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 ofδ.) This can happen at mostm − 1 times.
To handle this case, we will estimate the difference between the Riemann sum and the Darboux sum by subdividing the partitionx0, ...,xn atyj + 1. The termf(ti)(xi + 1 −xi) in the Riemann sum splits into two terms:
Suppose,without loss of generality, thatti ∈ [yj,yj + 1]. Thenso this term is bounded by the corresponding term in the Darboux sum foryj. To bound the other term, notice that
It follows that, for some (indeed any)t*
i ∈ [yj + 1,xi + 1],
Since this happens at mostm − 1 times, the distance between the Riemann sum and a Darboux sum is at mostε/2. Therefore, the distance between the Riemann sum ands is at most ε.
Let be the function which takes the value 1 at every point. Any Riemann sum off on[0, 1] will have the value 1, therefore the Riemann integral off on[0, 1] is 1.
Let be theindicator 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, letx0, ...,xn andt0, ...,tn − 1 be a tagged partition (eachti is betweenxi andxi + 1). Chooseε > 0. Theti have already been chosen, and we can't change the value off at those points. But if we cut the partition into tiny pieces around eachti, we can minimize the effect of theti. Then, by carefully choosing the new tags, we can make the value of the Riemann sum turn out to be withinε of either zero or one.
Our first step is to cut up the partition. There aren of theti, and we want their total effect to be less thanε. If we confine each of them to an interval of length less thanε/n, then the contribution of eachti to the Riemann sum will be at least0 ·ε/n and at most1 ·ε/n. This makes the total sum at least zero and at mostε. So letδ be a positive number less thanε/n. If it happens that two of theti are withinδ of each other, chooseδ smaller. If it happens that someti is withinδ of somexj, andti is not equal toxj, chooseδ smaller. Since there are only finitely manyti andxj, we can always chooseδ sufficiently small.
Now we add two cuts to the partition for eachti. One of the cuts will be atti −δ/2, and the other will be atti +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
Ifti is directly on top of one of thexj, then we letti 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 arational point, so that the Riemann sum is as large as possible. This will make the value of the Riemann sum at least1 −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 mostε.
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 numbers, so this function is not Riemann integrable. However, it isLebesgue integrable. In the Lebesgue sense its integral is zero, since the function is zeroalmost 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, letC be theSmith–Volterra–Cantor set, and letIC be its indicator function. BecauseC is notJordan measurable,IC is not Riemann integrable. Moreover, no functiong equivalent toIC is Riemann integrable:g, likeIC, must be zero on a dense set, so as in the previous example, any Riemann sum ofg has a refinement which is withinε of 0 for any positive number ε. But if the Riemann integral ofg exists, then it must equal the Lebesgue integral ofIC, which is1/2. Therefore,g is not Riemann integrable.
It is popular to define the Riemann integral as theDarboux integral. This is because the Darboux integral is technically simpler and because a function is Riemann-integrable if and only if it is Darboux-integrable.
Some calculus books do not use general tagged partitions, but limit themselves to specific types of tagged partitions. If the type of partition is limited too much, some non-integrable functions may appear to be integrable.
One popular restriction is the use of "left-hand" and "right-hand" Riemann sums. In a left-hand Riemann sum,ti =xi for alli, and in a right-hand Riemann sum,ti =xi + 1 for alli. Alone this restriction does not impose a problem: we can refine any partition in a way that makes it a left-hand or right-hand sum by subdividing it at eachti. In more formal language, the set of all left-hand Riemann sums and the set of all right-hand Riemann sums iscofinal in the set of all tagged partitions.
Another popular restriction is the use of regular subdivisions of an interval. For example, thenth regular subdivision of[0, 1] consists of the intervals
Again, alone this restriction does not impose a problem, but the reasoning required to see this fact is more difficult than in the case of left-hand and right-hand Riemann sums.
However, combining these restrictions, so that one uses only left-hand or right-hand Riemann sums on regularly divided intervals, is dangerous. If a function is known in advance to be Riemann integrable, then this technique will give the correct value of the integral. But under these conditions theindicator function will appear to be integrable on[0, 1] with integral equal to one: Every endpoint of every subinterval will be a rational number, so the function will always be evaluated at rational numbers, and hence it will appear to always equal one. The problem with this definition becomes apparent when we try to split the integral into two pieces. The following equation ought to hold:
If we use regular subdivisions and left-hand or right-hand Riemann sums, then the two terms on the left are equal to zero, since every endpoint except 0 and 1 will be irrational, but as we have seen the term on the right will equal 1.
As defined above, the Riemann integral avoids this problem by refusing to integrate The Lebesgue integral is defined in such a way that all these integrals are 0.
The Riemann integral is a linear transformation; that is, iff andg are Riemann-integrable on[a,b] andα andβ are constants, then
Because the Riemann integral of a function is a number, this makes the Riemann integral alinear functional on thevector space of Riemann-integrable functions.
Abounded function on acompact interval[a,b] is Riemann integrable if and only if it iscontinuousalmost everywhere (the set of itspoints of discontinuity hasmeasure zero, in the sense ofLebesgue measure). This is theLebesgue-Vitali theorem (of characterization of the Riemann integrable functions). It has been proven independently byGiuseppe Vitali and byHenri Lebesgue in 1907, and uses the notion ofmeasure zero, but makes use of neither Lebesgue's general measure or integral.
The integrability condition can be proven in various ways,[4][5][6][7] one of which is sketched below.
| Proof |
|---|
| The proof is easiest using theDarboux 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 theoscillation definition of continuity:[8] For every positiveε, LetXε be the set of points in[a,b] with oscillation of at leastε. Since every point wheref is discontinuous has a positive oscillation and vice versa, the set of points in[a,b], wheref is discontinuous is equal to the union over{X1/n} for all natural numbersn. If this set does not have zeroLebesgue measure, then bycountable additivity of the measure there is at least one suchn so thatX1/n does not have a zero measure. Thus there is some positive numberc such that everycountable collection of open intervalscoveringX1/n has a total length of at leastc. In particular this is also true for every such finite collection of intervals. This remains true also forX1/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 everypartition of[a,b], consider the set of intervals whose interiors include points fromX1/n. These interiors consist of a finite open cover ofX1/n, possibly up to a finite number of points (which may fall on interval edges). Thus these intervals have a total length of at leastc. Since in these pointsf has oscillation of at least1/n, theinfimum and supremum off in each of these intervals differ by at least1/n. Thus the upper and lower sums off differ by at leastc/n. Since this is true for every partition,f is not Riemann integrable. We now prove the converse direction using the setsXε defined above.[9] For everyε,Xε iscompact, as it is bounded (bya andb) and closed:
Now, suppose thatf is continuousalmost everywhere. Then for everyε,Xε has zeroLebesgue measure. Therefore, there is a countable collections of open intervals in[a,b] which is anopen cover ofXε, such that the sum over all their lengths is arbitrarily small.SinceXε is compact, there is a finitesubcover – a finite collections of open intervals in[a,b] with arbitrarily small total length that together contain all points inXε. We denote these intervals{I(ε)i}, for1 ≤i ≤k, for some naturalk. Thecomplement of the union of these intervals is itself a union of a finite number of intervals, which we denote{J(ε)i} (for1 ≤i ≤k − 1 and possibly fori =k,k + 1 as well). We now show that for everyε > 0, there areupper and lower sums whose difference is less thanε, from which Riemann integrability follows. To this end, we construct apartition of[a,b] as follows: Denoteε1 =ε / 2(b −a) andε2 =ε / 2(M −m), wherem andM are theinfimum and supremum off on[a,b]. Since we may choose intervals{I(ε1)i} with arbitrarily small total length, we choose them to have total length smaller thanε2. Each of the intervals{J(ε1)i} has an empty intersection withXε1, so each point in it has a neighborhood with oscillation smaller thanε1. These neighborhoods consist of anopen 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 ofJ(ε1)i (except for those that include an edge point, for which we only take their intersection withJ(ε1)i). We take the edge points of the subintervals for allJ(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:
In total, the difference between the upper and lower sums of the partition is smaller thanε, as required. |
In particular, any set that is at mostcountable hasLebesgue measure zero, and thus a bounded function (on a compact interval) with only finitely or countably many discontinuities is Riemann integrable. Another sufficient criterion to Riemann integrability over[a,b], but which does not involve the concept of measure, is the existence of a right-hand (or left-hand) limit atevery point in[a,b) (or(a,b]).[10]
Anindicator function of a bounded set is Riemann-integrable if and only if the set isJordan measurable. The Riemann integral can be interpretedmeasure-theoretically as the integral with respect to the Jordan measure.
If a real-valued function ismonotone 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 isLebesgue integrable. That is, Riemann-integrability is astronger (meaning more difficult to satisfy) condition than Lebesgue-integrability. The converse does not hold; not all Lebesgue-integrable functions are Riemann integrable.
The Lebesgue–Vitali theorem does not imply that all type of discontinuities have the same weight on the obstruction that a real-valued bounded function be Riemann integrable on[a,b]. In fact, certain discontinuities have absolutely no role on the Riemann integrability of the function—a consequence of theclassification of the discontinuities of a function.[citation needed]
Iffn is auniformly convergent sequence on[a,b] with limitf, then Riemann integrability of allfn implies Riemann integrability off, and
However, theLebesgue monotone convergence theorem (on a monotone pointwise limit) does not hold for Riemann integrals. Thus, in Riemann integration, taking limits under the integral sign is far more difficult to logically justify than in Lebesgue integration.[11]
It is easy to extend the Riemann integral to functions with values in the Euclidean vector space for anyn. The integral is defined component-wise; in other words, iff = (f1, ...,fn) then
In particular, since the complex numbers are a realvector space, this allows the integration of complex valued functions.
The Riemann integral is only defined on bounded intervals, and it does not extend well to unbounded intervals. The simplest possible extension is to define such an integral as alimit, in other words, as animproper integral:
This definition carries with it some subtleties, such as the fact that it is not always equivalent to compute theCauchy principal value
For example, consider thesign functionf(x) = sgn(x) which is 0 atx = 0, 1 forx > 0, and −1 forx < 0. By symmetry,always, regardless ofa. But there are many ways for the interval of integration to expand to fill the real line, and other ways can produce different results; in other words, the multivariate limit does not always exist. We can compute
In general, this improper Riemann integral is undefined. Even standardizing a way for the interval to approach the real line does not work because it leads to disturbingly counterintuitive results. If we agree (for instance) that the improper integral should always bethen the integral of the translationf(x − 1) is −2, so this definition is not invariant under shifts, a highly undesirable property. In fact, not only does this function not have an improper Riemann integral, its Lebesgue integral is also undefined (it equals∞ − ∞).
Unfortunately, the improper Riemann integral is not powerful enough. The most severe problem is that there are no widely applicable theorems for commuting improper Riemann integrals with limits of functions. In applications such asFourier series it is important to be able to approximate the integral of a function using integrals of approximations to the function. For proper Riemann integrals, a standard theorem states that iffn is a sequence of functions thatconverge uniformly tof on a compact set[a,b], then
On non-compact intervals such as the real line, this is false. For example, takefn(x) to ben−1 on[0,n] and zero elsewhere. For alln we have:
The sequence(fn) converges uniformly to the zero function, and clearly the integral of the zero function is zero. Consequently,
This demonstrates that for integrals on unbounded intervals, uniform convergence of a function is not strong enough to allow passing a limit through an integral sign. This makes the Riemann integral unworkable in applications (even though the Riemann integral assigns both sides the correct value), because there is no other general criterion for exchanging a limit and a Riemann integral, and without such a criterion it is difficult to approximate integrals by approximating their integrands.
A better route is to abandon the Riemann integral for theLebesgue integral. The definition of the Lebesgue integral is not obviously a generalization of the Riemann integral, but it is not hard to prove that every Riemann-integrable function is Lebesgue-integrable and that the values of the two integrals agree whenever they are both defined. Moreover, a functionf defined on a bounded interval is Riemann-integrable if and only if it is bounded and the set of points wheref is discontinuous has Lebesgue measure zero.
An integral which is in fact a direct generalization of the Riemann integral is theHenstock–Kurzweil integral.
Another way of generalizing the Riemann integral is to replace the factorsxk + 1 −xk in the definition of a Riemann sum by something else; roughly speaking, this gives the interval of integration a different notion of length. This is the approach taken by theRiemann–Stieltjes integral.
Inmultivariable calculus, the Riemann integrals for functions from aremultiple integrals.
The Riemann integral is unsuitable for many theoretical purposes. Some of the technical deficiencies in Riemann integration can be remedied with theRiemann–Stieltjes integral, and most disappear with theLebesgue integral, though the latter does not have a satisfactory treatment ofimproper integrals. Thegauge integral is a generalisation of the Lebesgue integral that is at the same time closer to the Riemann integral. These more general theories allow for the integration of more "jagged" or "highly oscillating" functions whose Riemann integral does not exist; but the theories give the same value as the Riemann integral when it does exist.
In educational settings, theDarboux integral offers a simpler definition that is easier to work with; it can be used to introduce the Riemann integral. The Darboux integral is defined whenever the Riemann integral is, and always gives the same result. Conversely, thegauge integral is a simple but more powerful generalization of the Riemann integral and has led some educators to advocate that it should replace the Riemann integral in introductory calculus courses.[12]
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