En la rama de las matemáticas conocida como análisis real , la integral de Riemann , creada por Bernhard Riemann , fue la primera definición rigurosa de la integral de una función en un intervalo . Se presentó a la facultad de la Universidad de Göttingen en 1854, pero no se publicó en una revista hasta 1868. [1] Para muchas funciones y aplicaciones prácticas, la integral de Riemann puede evaluarse mediante el teorema fundamental del cálculo o aproximarse mediante integración numérica. .
La integral de Riemann no es adecuada para muchos propósitos teóricos. Algunas de las deficiencias técnicas en la integración de Riemann se pueden remediar con la integral de Riemann-Stieltjes y la mayoría desaparecen con la integral de Lebesgue , aunque esta última no tiene un tratamiento satisfactorio de integrales impropias . La integral de gauge es una generalización de la integral de Lebesgue que está a la vez más cerca de la integral de Riemann. Estas teorías más generales permiten la integración de funciones más "irregulares" o "altamente oscilantes" cuya integral de Riemann no existe; pero las teorías dan el mismo valor que la integral de Riemann cuando existe.
En entornos educativos, la integral Darboux ofrece una definición más simple con la que es más fácil trabajar; se puede utilizar para introducir la integral de Riemann. La integral de Darboux se define siempre que la integral de Riemann es, y siempre da el mismo resultado. Por el contrario, la integral de gauge es una generalización simple pero más poderosa de la integral de Riemann y ha llevado a algunos educadores a defender que debería reemplazar la integral de Riemann en los cursos introductorios de cálculo. [2]
Descripción general
Sea f una función de valor real no negativo en el intervalo [ a , b ] , y sea
sea la región del plano debajo de la gráfica de la función f y por encima del intervalo [ a , b ] (vea la figura en la parte superior derecha). Estamos interesados en la medición del área de S . Una vez la hayamos medido, denotaremos el área por:
La idea básica de la integral de Riemann es utilizar aproximaciones muy simples para el área de S . Al tomar cada vez mejores aproximaciones, podemos decir que "en el límite" obtenemos exactamente el área de S debajo de la curva.
Donde f puede ser tanto positiva como negativa, la definición de S se modifica para que la integral corresponda al área con signo debajo de la gráfica de f : es decir, el área sobre el eje x menos el área debajo del eje x .
Definición
Particiones de un intervalo
Una partición de un intervalo [ a , b ] es una secuencia finita de números de la forma
Cada [ x i , x i + 1 ] se denomina subintervalo de la partición. La malla o norma de una partición se define como la longitud del subintervalo más largo, es decir,
Una partición etiquetada P ( x , t ) de un intervalo [ a , b ] es una partición junto con una secuencia finita de números t 0 , ..., t n - 1 sujeto a las condiciones que para cada i , t i ∈ [ x i , x i + 1 ] . En otras palabras, es una partición junto con un punto distinguido de cada subintervalo. La malla de una partición etiquetada es la misma que la de una partición ordinaria.
Suponga que dos particiones P ( x , t ) y Q ( y , s ) son ambas particiones del intervalo [ a , b ] . Decimos que Q ( y , s ) es un refinamiento de P ( x , t ) si para cada entero i , con i ∈ [0, n ] , existe un entero r ( i ) tal que x i = y r ( i ) y tal que t i = s j para algún j con j ∈ [ r ( i ), r ( i + 1)) . Dicho de manera más simple, un refinamiento de una partición etiquetada rompe algunos de los subintervalos y agrega etiquetas a la partición donde sea necesario, por lo que "refina" la precisión de la partición.
Podemos convertir el conjunto de todas las particiones etiquetadas en un conjunto dirigido diciendo que una partición etiquetada es mayor o igual que otra si la primera es un refinamiento de la última.
Sumas de Riemann
Sea f una función de valor real definida en el intervalo [ a , b ] . La suma de Riemann de f con respecto a la partición etiquetada x 0 , ..., x n junto con t 0 , ..., t n - 1 es [3]
Cada término de la suma es el producto del valor de la función en un punto dado y la longitud de un intervalo. En consecuencia, cada término representa el área (con signo) de un rectángulo con altura f ( t i ) y ancho x i + 1 - x i . La suma de Riemann es el área (con signo) de todos los rectángulos.
Conceptos estrechamente relacionados son las sumas Darboux inferior y superior . Son similares a las sumas de Riemann, pero las etiquetas se reemplazan por el infimum y supremum (respectivamente) de f en cada subintervalo:
Si f es continua, entonces las sumas Darboux inferior y superior para una partición sin etiquetar son iguales a la suma de Riemann para esa partición, donde las etiquetas se eligen para que sean el mínimo o máximo (respectivamente) de f en cada subintervalo. (Cuando f es discontinua en un subintervalo, puede que no haya una etiqueta que alcance el mínimo o el superior en ese subintervalo). La integral de Darboux , que es similar a la integral de Riemann pero basada en sumas de Darboux, es equivalente a la integral de Riemann.
Integral de Riemann
Hablando libremente, la integral de Riemann es el límite de las sumas de Riemann de una función a medida que las particiones se vuelven más finas. Si existe el límite, se dice que la función es integrable (o más específicamente integrable de Riemann ). La suma de Riemann se puede hacer tan cerca como se desee de la integral de Riemann haciendo que la partición sea lo suficientemente fina. [4]
Un requisito importante es que la malla de las particiones debe hacerse cada vez más pequeña, de modo que en el límite sea cero. Si esto no fuera así, entonces no obtendríamos una buena aproximación a la función en ciertos subintervalos. De hecho, esto es suficiente para definir una integral. Para ser específicos, decimos que la integral de Riemann de f es igual a s si se cumple la siguiente condición:
Para todo ε > 0 , existe δ > 0 tal que para cualquier partición etiquetada x 0 , ..., x n y t 0 , ..., t n - 1 cuya malla es menor que δ , tenemos
Desafortunadamente, esta definición es muy difícil de usar. Ayudaría a desarrollar una definición equivalente de la integral de Riemann con la que sea más fácil trabajar. Desarrollamos esta definición ahora, con una prueba de equivalencia a continuación. Nuestra nueva definición dice que la integral de Riemann de f es igual a s si se cumple la siguiente condición:
Para todos ε > 0 , existe una partición etiquetada y 0 , ..., y m y r 0 , ..., r m - 1 tal que para cualquier partición etiquetada x 0 , ..., x n y t 0 , ..., t n - 1 , que es un refinamiento de y 0 , ..., y m y r 0 , ..., r m - 1 , tenemos
Ambos significan que eventualmente, la suma de Riemann de f con respecto a cualquier partición queda atrapada cerca de s . Dado que esto es cierto sin importar cuán cerca exijamos que las sumas queden atrapadas, decimos que las sumas de Riemann convergen en s . Estas definiciones son en realidad un caso especial de un concepto más general, una red .
Como dijimos anteriormente, estas dos definiciones son equivalentes. En otras palabras, s funciona en la primera definición si y solo si s funciona en la segunda definición. Para mostrar que la primera definición implica la segunda, comience con un ε y elija un δ que satisfaga la condición. Elija cualquier partición etiquetada cuya malla sea menor que δ . Su suma de Riemann está dentro de ε de s , y cualquier refinamiento de esta partición también tendrá una malla menor que δ , por lo que la suma de Riemann del refinamiento también estará dentro de ε de s .
Para mostrar que la segunda definición implica la primera, es más fácil usar la integral de Darboux . Primero, se muestra que la segunda definición es equivalente a la definición de la integral de Darboux; para esto, consulte el artículo de Darboux Integral . Ahora mostraremos que una función integrable de Darboux satisface la primera definición. Fije ε y elija una partición y 0 , ..., y m tal que las sumas de Darboux superior e inferior con respecto a esta partición estén dentro de ε / 2 del valor s de la integral de Darboux. Dejar
Si r = 0 , entonces f es la función cero, que es claramente integrable tanto de Darboux como de Riemann con la integral cero. Por tanto, asumiremos que r > 0 . Si m > 1 , entonces elegimos δ tal que
Si m = 1 , entonces elegimos que δ sea menor que uno. Elija una partición etiquetada x 0 , ..., x n y t 0 , ..., t n - 1 con una malla menor que δ . Debemos demostrar que la suma de Riemann está dentro de ε de s .
Para ver esto, elija un intervalo [ x i , x i + 1 ] . Si este intervalo está contenido dentro de algunos [ y j , y j + 1 ] , entonces
donde m j y M j son, respectivamente, el mínimo y el superior de f en [ y j , y j + 1 ] . Si todos los intervalos tuvieran esta propiedad, entonces esto concluiría la demostración, porque cada término en la suma de Riemann estaría acotado por un término correspondiente en las sumas de Darboux, y elegimos las sumas de Darboux para que estuvieran cerca de s . Este es el caso cuando m = 1 , por lo que la demostración está terminada en ese caso.
Por tanto, podemos suponer que m > 1 . En este caso, es posible que uno de los [ x i , x i + 1 ] no esté contenido en ningún [ y j , y j + 1 ] . En cambio, puede extenderse a lo largo de dos de los intervalos determinados por y 0 , ..., y m . (No puede cumplir con tres intervalos porque se supone que δ es menor que la longitud de cualquier intervalo). En símbolos, puede suceder que
(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 + 1 − xi) 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) t*
i ∈ [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.
Ejemplos de
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 ti − δ/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.
Conceptos similares
It is popular to define the Riemann integral as the Darboux 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 all i, and in a right-hand Riemann sum, ti = xi + 1 for all i. 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 each ti. In more formal language, the set of all left-hand Riemann sums and the set of all right-hand Riemann sums is cofinal in the set of all tagged partitions.
Another popular restriction is the use of regular subdivisions of an interval. For example, the nth 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 the indicator 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.
Propiedades
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.
Integrabilidad
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-Vitali theorem (of characterization of the Riemann integrable functions). It has been proved 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,[5][6][7][8] 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:[9] 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.[10] 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 ε∗
1(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 ε∗
2(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.[11] 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.
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.[12]
Generalizaciones
It is easy to extend the Riemann integral to functions with values in the Euclidean vector space for any n. The integral is defined component-wise; in other words, if f = (f1, ..., fn) then
In particular, since the complex numbers are a real vector 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 a limit, in other words, as an improper integral:
This definition carries with it some subtleties, such as the fact that it is not always equivalent to compute the Cauchy principal value
For example, consider the sign function f(x) = sgn(x) which is 0 at x = 0, 1 for x > 0, and −1 for x < 0. By symmetry,
always, regardless of a. 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 be
then the integral of the translation f(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 as Fourier 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 if fn is a sequence of functions that converge uniformly to f on a compact set [a, b], then
On non-compact intervals such as the real line, this is false. For example, take fn(x) to be n−1 on [0, n] and zero elsewhere. For all n 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 the Lebesgue 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 function f defined on a bounded interval is Riemann-integrable if and only if it is bounded and the set of points where f is discontinuous has Lebesgue measure zero.
An integral which is in fact a direct generalization of the Riemann integral is the Henstock–Kurzweil integral.
Another way of generalizing the Riemann integral is to replace the factors xk + 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 the Riemann–Stieltjes integral.
In multivariable calculus, the Riemann integrals for functions from are multiple integrals.
Ver también
- Area
- Antiderivative
- Lebesgue integration
Notas
- ^ The Riemann integral was introduced in Bernhard Riemann's paper "Über die Darstellbarkeit einer Function durch eine trigonometrische Reihe" (On the representability of a function by a trigonometric series; i.e., when can a function be represented by a trigonometric series). This paper was submitted to the University of Göttingen in 1854 as Riemann's Habilitationsschrift (qualification to become an instructor). It was published in 1868 in Abhandlungen der Königlichen Gesellschaft der Wissenschaften zu Göttingen (Proceedings of the Royal Philosophical Society at Göttingen), vol. 13, pages 87-132. (Available online here.) For Riemann's definition of his integral, see section 4, "Über den Begriff eines bestimmten Integrals und den Umfang seiner Gültigkeit" (On the concept of a definite integral and the extent of its validity), pages 101–103.
- ^ "An Open Letter to Authors of Calculus Books". Retrieved 27 February 2014.
- ^ Krantz, Steven G. (1991). Real Analysis and Foundations. CRC Press. p. 173.; 2005 edition. ISBN 9781584884835.
- ^ Taylor, Michael E. (2006). Measure Theory and Integration. American Mathematical Society. p. 1. ISBN 9780821872468.
- ^ Apostol 1974, pp. 169–172
- ^ Brown, A. B. (September 1936). "A Proof of the Lebesgue Condition for Riemann Integrability". The American Mathematical Monthly. 43 (7): 396–398. doi:10.2307/2301737. ISSN 0002-9890. JSTOR 2301737.
- ^ Basic real analysis, by Houshang H. Sohrab, section 7.3, Sets of Measure Zero and Lebesgue’s Integrability Condition, pp. 264–271
- ^ Introduction to Real Analysis, updated April 2010, William F. Trench, 3.5 "A More Advanced Look at the Existence of the Proper Riemann Integral", pp. 171–177
- ^ Lebesgue’s Condition, John Armstrong, December 15, 2009, The Unapologetic Mathematician
- ^ Jordan Content Integrability Condition, John Armstrong, December 9, 2009, The Unapologetic Mathematician
- ^ PlanetMath Volume
- ^ Cunningham, Frederick Jr. (1967). "Taking limits under the integral sign". Mathematics Magazine. 40: 179–186. doi:10.2307/2688673.
Referencias
- Shilov, G. E., and Gurevich, B. L., 1978. Integral, Measure, and Derivative: A Unified Approach, Richard A. Silverman, trans. Dover Publications. ISBN 0-486-63519-8.
- Apostol, Tom (1974), Mathematical Analysis, Addison-Wesley
enlaces externos
- Media related to Riemann integral at Wikimedia Commons
- "Riemann integral", Encyclopedia of Mathematics, EMS Press, 2001 [1994]