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	CATEGORIES ABOUT RIEMANN INTEGRAL
	definitions of mathematical integration

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OVERVIEW
	:<math>\left\sum {i	0}^{n-1} f(t_i) (x_{i+1}-x_i) - s ight < \epsilon\,</math>
	:<math>\left\sum {i	0}^{m-1} f(s_i) (y_{i+1}-y_i) - s ight < \epsilon\,</math>

To start, let $x\_0,\; \backslash ldots,\; x\_n$ and $t\_0,\; \backslash ldots,\; t\_\{n-1\}$ be a tagged partition (each $t\_i$ is between $x\_i$ and $x\_\{i+1\}$). Choose $\backslash epsilon\; >\; 0$. 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 $\backslash epsilon$ of either zero or one—our choice!

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 $\backslash epsilon$. If we confine each of them to an interval of length less than , then the contribution of each $t\_i$ to the Riemann sum will be at least and at most . This makes the total sum at least zero and at most $\backslash epsilon$. So let $\backslash delta$be a positive number less than . If it happens that two of the $t\_i$ are within $\backslash delta$of each other, choose $\backslash delta$smaller. If it happens that some $t\_i$ is within $\backslash delta$of some $x\_j$, and $t\_i$ is not equal to $x\_j$, choose $\backslash delta$smaller. Since there are only finitely many $t\_i$ and $x\_j$, we can always choose $\backslash delta$sufficiently small.

Now we add two cuts to the partition for each $t\_i$. One of the cuts will be at , and the other will be at . If one of these leaves the interval . If $t\_i$ is directly on top of one of the $x\_j$, then we let $t\_i$ be the tag for both and . 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 $1-\backslash epsilon$. 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 $\backslash epsilon$.

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.


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 $n\text{'}$th regular subdivision of $1$ 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 $I\_\{\backslash mathbb\{Q\}\}$ will appear to be integrable on $1$ 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 $I\_\{\backslash mathbb\{Q\}\}$. The Lebesgue integral is defined in such a way that all these integrals are 0.


FACTS ABOUT THE RIEMANN INTEGRAL


The Riemann integral is a linear transformation; that is, if $f$ and $g$ are Riemann-integrable on $${Link without Title} and $\backslash alpha$ and $\backslash beta$ are constants, then

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

A real-valued function $f$ on $${Link without Title} is Riemann-integrable if and only if it is Bounded and Continuous Almost Everywhere .

If a real-valued function on $${Link without Title} is Riemann-integrable, it is Lebesgue-integrable .

If $\{f\_n\}$ is a Uniformly Convergent sequence on $${Link without Title} with limit $f$, then

:$\backslash int\_\{a\}^\{b\}\; f\backslash ,\; dx\; =\; \backslash int\_a^b\{\backslash lim\_\{n\; o\; \backslash infty\}\{f\_n\}\backslash ,\; dx\}\; =\; \backslash lim\_\{n\; o\; \backslash infty\}\; \backslash int\_\{a\}^\{b\}\; f\_n\backslash ,\; dx.$

If a real-valued function is Monotone on the interval $${Link without Title} , it is Riemann-integrable.


GENERALIZATIONS OF THE RIEMANN INTEGRAL

It is easy to extend the Riemann integral to functions with values in the Euclidean vector space $\backslash mathbb\{R\}^n$ for any $n$. The integral is defined by linearity; in other words, if $\backslash mathbf\{f\}\; =\; (f\_1,\; \backslash dots,\; f\_n)$, then $\backslash int\backslash mathbf\{f\}\; =\; (\backslash int\; f\_1,\backslash ,\backslash dots,\; \backslash int\; f\_n)$. 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:

:$\backslash int\_\{-\backslash infty\}^\backslash infty\; f(t)\backslash ,dt\; =\; \backslash lim\_\{x\; o\backslash infty\}\backslash int\_\{-x\}^x\; f(t)\backslash ,dt.$

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 . Then,

:$\backslash int\_\{-x\}^x\; f(t)\backslash ,dt\; =\; \backslash int\_\{-x\}^0\; f(t)\backslash ,dt\; +\; \backslash int\_0^x\; f(t)\backslash ,dt\; =\; -x\; +\; x\; =\; 0$

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

:$\backslash int\_\{-x\}^x\; f(t-1)\backslash ,dt\; =\; \backslash int\_\{-x\}^1\; f(t-1)\backslash ,dt\; +\; \backslash int\_1^x\; f(t-1)\backslash ,dt\; =\; -(x+1)\; +\; (x-1)\; =\; -2$

for all $x\; >\; 1$.

Since this is unacceptable, we could try the definition:

:$\backslash int\_\{-\backslash infty\}^\backslash infty\; f(t)\backslash ,dt\; =\; \backslash lim\_\{a\; o-\backslash infty\}\backslash lim\_\{b\; o\backslash infty\}\backslash int\_a^b\; f(t)\backslash ,dt.$

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

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 $${Link without Title} and 0 everywhere else. For all $n$, $\backslash int\; f\_n\backslash ,dx\; =\; 1$. But $f\_n$ converges uniformly to zero, so the integral of $\backslash lim\; f\_n$ is zero. Consequently $\backslash int\; f\backslash ,dx$
ot= \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 .


SEE ALSO

• Antiderivative


• Riemann-Stieltjes Integral


• Henstock-Kurzweil Integral


• Lebesgue Integral


• Darboux Integral