黎曼积分

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在实分析中, 由黎曼创立的黎曼积分首次对函数在给定区间上的积分给出了一个精确定义。黎曼积分在技术上的某些不足之处可由后来的黎曼－斯蒂尔杰斯积分和勒贝格积分得到修补。

[编辑] 概念

作为曲线与坐标轴所夹面积的黎曼积分

对于一在区间 $[a, b]$ 上之给定非负函数 $f (x)$ ，我们想要确定 $f (x)$ 所代表的曲线与 $X$ 坐标轴所夹图形的面积，我们可以将此记为

$S=\int_{a}^{b} f(x)dx.$

黎曼积分的核心思想就是试图通过无限逼近来确定这个积分值。同时清注意，如 $f (x)$ 取负值，则相应的面积值 $S$ 亦取负值。

A sequence of Riemann sums. The numbers in the upper right are the areas of the grey rectangles. They converge to the integral of the function.

[编辑] 定义

[编辑] 区间分割

A partition of an interval $[a, b]$ is a finite sequence $a = x_0 < x_1 < x_2 < \ldots < x_n = b$ . Each $[x i, x i + 1]$ is called a subinterval of the partition. The mesh of a partition is defined to be the length of the longest subinterval $[x i, x i + 1]$ , that is, it is $max(x i + 1 - x i)$ where $0 \le i \le n - 1$ . It is also called the norm of the partition.

A tagged partition of an interval is a partition of an interval together with a finite sequence of numbers $t_0, \ldots, t_{n-1}$ subject to the conditions that for each $i$ , $x_i \le t_i \le x_{i+1}$ . In other words, it is a partition together with a distinguished point of every subinterval. The mesh of a tagged partition is defined the same as for an ordinary partition.

Suppose that $x_0,\ldots,x_n$ together with $t_0,\ldots,t_{n-1}$ are a tagged partition of $[a, b]$ , and that $y_0,\ldots,y_m$ together with $s_0,\ldots,s_{m-1}$ are another tagged partition of $[a, b]$ . We say that $y_0,\ldots,y_m$ and $s_0,\ldots,s_{m-1}$ together are a refinement of $x_0,\ldots,x_n$ together with $t_0,\ldots,t_{n-1}$ if for each integer $i$ with $0 \le i \le n$ , there is an integer $r (i)$ such that $x i = y r (i)$ and such that $t i = s j$ for some $j$ with $r(i) \le j \le r(i+1)$ . Said more simply, a refinement of a tagged partition takes the starting partition and adds more tags, but does not take any away.

We can define a partial order on the set of all tagged partitions by saying that one tagged partition is bigger than another if the bigger one is a refinement of the smaller one.

[编辑] 黎曼和

Choose a real-valued function $f$ which is defined on the interval $[a, b]$ . The Riemann sum of $f$ with respect to the tagged partition $x_0,\ldots,x_n$ together with $t_0,\ldots,t_{n-1}$ is:

$\sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i)$

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 area of a rectangle with height $f (t i)$ and length $x i + 1 - x i$ . The Riemann sum is the signed area under all the rectangles.

[编辑] 积分

Loosely speaking, the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer and finer. However, being precise about what is meant by "finer and finer" is somewhat tricky.

One important fact 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 $x_0,\ldots,x_n$ and $t_0,\ldots,t_{n-1}$ whose mesh is less than

δ

, we have

$\left|\sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i) - s\right| < \epsilon.\,$

However, there is an unfortunate problem with this definition: it is very difficult to work with. So we will make an alternate definition of the Riemann integral which is easier to work with, then prove that it is the same as the definition we have just made. 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 $x_0,\ldots,x_n$ and $t_0,\ldots,t_{n-1}$ such that for any refinement $y_0,\ldots,y_m$ and $s_0,\ldots,s_{m-1}$ of $x_0,\ldots,x_n$ and $t_0,\ldots,t_{n-1}$ , we have

$\left|\sum_{i=0}^{m-1} f(s_i) (y_{i+1}-y_i) - s\right| < \epsilon.\,$

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; for this see the page on Darboux integration. Now we will show that a Darboux integrable function satisfies the first definition. Choose a partition $x_0, \ldots, x_n$ such that the lower and upper Darboux sums with respect to this partition are within $\frac{\epsilon}{2}$ of the value $s$ of the Darboux integral. Let $r$ equal $\max_{0 \le i \le n-1} M_i-m_i$ , where $M i$ and $m i$ are the supremum and infimum, respectively, of $f$ on $[x i, x i + 1]$ , and let $δ$ be less than both $\frac{\epsilon}{2rn}$ and $\min_{0 \le i \le n-1} x_{i+1}-x_i$ . Then it is not hard to see that the Riemann sum of $f$ with respect to any tagged partition of mesh less than $δ$ will be within $\frac{\epsilon}{2}$ of the upper or lower Darboux sum, so it will be within $ε$ of $s$ .

[编辑] Things that masquerade as the Riemann integral

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, $t i = x i$ for all $i$ , and in a right-hand Riemann sum, $t i = x i + 1$ 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 $t i$ . 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 $n'$ th regular subdivision of $[0,1]$ consists of the intervals $[0, 1/n], [1/n, 2/n], \ldots, [(n-1)/n, 1]$ . 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 $I_{\mathbb{Q}}$ 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:

$\int_0^{\sqrt{2}-1}\! I_\mathbf{Q}(x) \,\mathrm{d}x + \int_{\sqrt{2}-1}^1\! I_\mathbf{Q}(x) \,\mathrm{d}x = \int_0^1\! I_\mathbf{Q}(x) \,\mathrm{d}x .$

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 $I_{\mathbb{Q}}$ . The Lebesgue integral is defined in such a way that all these integrals are 0.

[编辑] Facts about the Riemann integral

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

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

A real-valued function $f$ on $[a, b]$ is Riemann-integrable if and only if it is bounded and continuous almost everywhere.

If a real-valued function on $[a, b]$ is Riemann-integrable, it is Lebesgue-integrable.

If $f n$ is a uniformly convergent sequence on $[a, b]$ with limit $f$ , then

$\int_{a}^{b} f\, dx = \int_a^b{\lim_{n \to \infty}{f_n}\, dx} = \lim_{n \to \infty} \int_{a}^{b} f_n\, dx.$

If a real-valued function is monotone on the interval $[a, b],$ it is Riemann-integrable.

[编辑] Generalizations of the Riemann integral

It is easy to extend the Riemann integral to functions with values in the Euclidean vector space $\mathbb{R}^n$ for any $n$ . The integral is defined by linearity; in other words, if $\mathbf{f} = (f_1, \dots, f_n)$ , then $\int\mathbf{f} = (\int f_1,\,\dots, \int f_n)$ . 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. We could set:

$\int_{-\infty}^\infty f(t)\,dt = \lim_{x\to\infty}\int_{-x}^x f(t)\,dt.$

Unfortunately, this does not work well. Translation invariance, the fact that the Riemann integral of the function should not change if we move the function left or right, is lost. For example, let $f (x) = 1$ for all $x > 0$ , $f (0) = 0$ , and $f (x) = - 1$ for all $x < 0$ . Then,

$\int_{-x}^x f(t)\,dt = \int_{-x}^0 f(t)\,dt + \int_0^x f(t)\,dt = -x + x = 0$

for all $x$ . But if we shift $f (x)$ to the right by one unit to get $f (x - 1)$ , we get

$\int_{-x}^x f(t-1)\,dt = \int_{-x}^1 f(t-1)\,dt + \int_1^x f(t-1)\,dt = -(x+1) + (x-1) = -2$

for all $x > 1$ .

Since this is unacceptable, we could try the definition:

$\int_{-\infty}^\infty f(t)\,dt = \lim_{a\to-\infty}\lim_{b\to\infty}\int_a^b f(t)\,dt.$

Then if we attempt to integrate the function $f$ above, we get $+\infty$ , because we take the limit as $b$ tends to $\infty$ first. If we reverse the order of the limits, then we get $-\infty$ .

This is also unacceptable, so we could require that the integral exists and gives the same value regardless of the order. Even this does not give us what we want, because the Riemann integral no longer commutes with uniform limits. For example, let $f n (x) = 1 / n$ on $[0, n]$ and 0 everywhere else. For all $n$ , $\int f_n\,dx = 1$ . But $f n$ converges uniformly to zero, so the integral of $\lim f_n$ is zero. Consequently $\int f\,dx \not= \lim\int f_n\,dx$ . Even though this is the correct value, it shows that the most important criterion for exchanging limits and (proper) integrals is false for improper integrals. This makes the Riemann integral unworkable in applications.

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.

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 $x i - x i + 1$ 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.

[编辑] 参见

Antiderivative
Riemann-Stieltjes integral
Henstock-Kurzweil integral
Lebesgue integral
Darboux integral

[编辑] 参考文献

Shilov, G. E., and Gurevich, B. L., 1978. Integral, Measure, and Derivative: A Unified Approach, Richard A. Silverman, trans. Dover Publications. ISBN 0486635198.

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页面分类: 微积分

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