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I need to come up with a precise mathematical definition of what a riemann integrable function is. I know what the riemann integral is but when I look for definitions all I find are proofs of how to prove that a function is riemann integrable. I need help create a definition of what it means for a function to be riemann integrable that does not include any notation, just a couple of mathematical sentences that defines riemann integrals.

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    $\begingroup$ For a nonnegative function $f$, it means that the area between the curve and the $x$ axis can be approximated arbitrarily closely using step functions. The advantage of step functions is that we know how to find their area, since they consist of finitely many rectangles. How can $f$ fail to be Riemann integrable? In one of three ways: either it is unbounded, so finitely many rectangles can't "reach" enough of the area, or its tails are too heavy, so its area is effectively infinite, or it has too many discontinuities, so approximation by rectangles just doesn't work well. $\endgroup$ – Bungo Dec 19 '15 at 1:35
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    $\begingroup$ people.math.sc.edu/schep/riemann.pdf $\endgroup$ – Will Jagy Dec 19 '15 at 1:46
  • $\begingroup$ Understanding the definition of the Riemann Integral $\endgroup$ – MJD Dec 20 '15 at 20:10
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The Riemann integral of a function on $[a,b]$ is the limit of Riemann sums whose partitions $[a,b]$ get finer and finer (i.e. the norm of the partition goes to zero). If this limit exists, then the function is said to be Riemann integrable and the value of the Riemann integral is the limit the sums approach.

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    $\begingroup$ I am not sure. The way I learned it, the norm of a partition is the width of the widest sub interval in the partition. I don't know what mesh refers to or how it is defined. $\endgroup$ – Laars Helenius Dec 19 '15 at 1:51
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    $\begingroup$ Great! Thanks for the reference. $\endgroup$ – Laars Helenius Dec 19 '15 at 1:59
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    $\begingroup$ While you're at it, it might be nice to contrast this with the same thing for Lebesgue integrability. $\endgroup$ – Mehrdad Dec 19 '15 at 4:08
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    $\begingroup$ Basically Riemann integration relies on summing areas of vertical strips derived from the partition of the domain of $f(x)$. On the other hand, Lebesgue integration relies on summing areas of horizontal strips derived from the partition of the range of $f(x)$ over $[a,b]$. Very loosely, each partition of the domain determines a step function that approximates $f(x)$ while each partition of the range determines a simple function of $f(x)$. All step functions are simple but not vice versa, so simple approximations yield a larger class of integrable functions. $\endgroup$ – Laars Helenius Dec 19 '15 at 5:42
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    $\begingroup$ To be clear, saying "the sums converge to $I$ as the mesh gets finer and finer" means "for any $\epsilon > 0$, there is a $\delta$ such that as long as the mesh is less than $\delta$, the sum is within $\epsilon$ of $I$". $\endgroup$ – Jack M Dec 19 '15 at 11:00
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The Riemann integral is defined in terms of Riemann sums. Consider this image from the Wikipedia page:

enter image description here

We approximate the area under the function as a sum of rectangles. We can see that in this case, the approximation gets better and better as the width of the rectangles gets smaller. In fact, the sum of the areas of the rectangles converges to a number, this number is defined to be the Riemann integral of the function.

Note however that we can draw these rectangles in a number of ways, as shown below (from this webpage)

enter image description here

If, no matter how we draw the rectangles, the sum of their area converges to some number $F$ as the width of the rectangles approaches zero, we say that the function is Riemann integrable and define $F$ as the Riemann integral of the function. For some functions the area will not converge, the canonical example being the indicator function for the rationals $\mathbb{1}_{\mathbb{Q}}(x)$, which is $1$ if $x$ is a rational and $0$ otherwise.

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Maybe an historical perspective will help. It never hurts to have many points of view on what a mathematical definition is doing.

For most of the 18th century the integral was considered just an antiderivative, much the way that many calculus students still consider it. If you want to compute $$ \int_a^b f(x)\,dx $$ you really must find an antiderivative $F$ for the function $f$ and then write or compute $$ \int_a^b f(x)\,dx =F(b)-F(a).$$ Cauchy in the early 19th century felt that this needed to be placed on a more rigorous footing.

If you assume that the function $f$ is continuous and that continuous functions have antiderivatives then take any points $a=x_0<x_1<x_2<x_3< \dots < x_n =b$ and use the mean-value theorem to select points $\hat x_i\in (x_{x-1},x_i)$ for which $$f(\hat x_i)\cdot(x_i-x_{i-1})=F(x_i)-F(x_{i-1}).$$ If you carefully check the very simple arithmetic you will get that $$ \int_a^b f(x)\,dx =F(b)-F(a) = \sum_{i=1}^n f(\hat x_i)\cdot(x_i-x_{i-1}).$$ Cauchy's bright idea was to notice that, while it would be hard to figure out the exact points $\hat x_i$ that make this work you can use, instead any other point $\xi_i\in [x_{i-1},x_i]$ as long as $f(\hat x_i)$ and $f(\xi_i)$ are close together. For continuous functions this is easy to arrange. You don't get an exact formula for the integral, you get an approximate formula: $$ \int_a^b f(x)\,dx =F(b)-F(a) \approx \sum_{i=1}^n f(\xi_i)\cdot(x_i-x_{i-1}).$$

The end result is that Cauchy proved that the integral of every continuous function could be approximated by these Riemann sums.

Riemann just asked the obvious question:

Are there other functions (not just continuous) that also have an integral using Cauchy's same method.

So the class of Riemann integrable functions is the class of functions for which Cauchy's method works. This is somewhat larger than the class of continuous functions, large enough a class that nineteenth century mathematicians thought that they had a pretty good theory of integration. (They didn't.)

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    $\begingroup$ While the mean value theorem in its modern form is attributed to Cauchy, you seem to imply that what you wrote was not obvious to Newton and Leibnitz; I seriously doubt that's the case. $\endgroup$ – Martin Argerami Dec 19 '15 at 11:16
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    $\begingroup$ @MartinArgerami Cauchy wasn't a nice guy and you are right he deserves no credit for some of this. Even the connection between Riemann sums and integration goes back at least to Euler. But I'll grant him two things: (i) the formalization of "continuity" is his, and (ii) the shift from defining an integral as an antiderivative to a more constructive definition is his. So, for better or worse, we need to date this more modern approach to the integral to the 1820s and Cauchy, however much of jerk he was and however much of what he did can be traced to earlier mathematicians. $\endgroup$ – B. S. Thomson Dec 19 '15 at 17:35
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    $\begingroup$ Can one say that Cauchy used "his" integral to prove that a continuous function is a derivative ? $\endgroup$ – Tony Piccolo Dec 19 '15 at 19:08
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    $\begingroup$ @TonyPiccolo I would say that is fair. Before Cauchy "continuous" meant something else and there was probably no similar sufficient condition that could have been stated for a function to be a derivative. $\endgroup$ – B. S. Thomson Dec 19 '15 at 21:45
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A positive function is Riemann integrable over the interval $[a,b]$ if the infimum of the upper sums equals the supremum of the lower sums. (You'll have to look up what an upper sum and lower sum are. An upper sum intuitively is an approximation of the area of the curve from above, while a lower sum is an approximation of the area of the curve from below.)

In other words, the best approximation of the area of the function from below equals the best approximation of the area of the function from above.

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    $\begingroup$ (+1) No need to assume positive, just "bounded". :) $\endgroup$ – Andrew D. Hwang Dec 19 '15 at 1:50
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    $\begingroup$ @AndrewD.Hwang: Maybe it was to avoid talking about area cancellation... $\endgroup$ – Mehrdad Dec 19 '15 at 4:09
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To answer this question, I will assume you already know what the integral of a step function is. First I will answer with notation, and then give an equivalent formulation without notation.

Let $f \colon [a,b] \to \mathbb{R}$ be a function. We say that $f$ is Riemann-integrable if for every $\epsilon > 0$, there are step functions $\varphi$ and $\psi$ on $[a,b]$ such that

  1. $\varphi \leq f \leq \psi$ and

  2. $\int (\psi - \varphi) \leq \epsilon$.

Without notation: A function on a closed bounded interval is called Riemann-integrable if there are two step functions respectively above and below it, and these two step functions can always be chosen so as to make the area between their graphs less than any specified quantity.

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Here is the way that I first encountered it (somewhat nonstandard): Let $f:[a,b] \rightarrow \mathbb{R}$ be a function, and $[a,b]$ be a non-degenerate closed bounded interval. Let $P$ is a partition of $[a,b]$ if $P=\{x_{0},x_{1},...,x_{n}\}$ is a set such that $a=x_{0}<x_{1}<...<x_{n}=b$. Then let the mesh of $P$ be defined by $||P||=max\{x_{1}-x_{0},x_{2}-x_{1}...,x_{n}-x_{n-1}\}$.

A representative set of $P$ is $t=\{t_{1},...t_{n}\}$ such that $t_{i} \in [x_{i},x_{i-1}]$ for all $i \in \{1,...,n\}$.

This representative set is necessary because riemann integrality is going to ultimately depend (intuitively) on a bunch of little rectangles. In computation, we pick either the left side of each rectangle or the right side in a Riemann sum. But, actually to prove integrability, we need to go further and say that the function's integral converges to a value, independently of how we choose these rectangles. Further, the mesh is important, because the rectangles need not be of equal width, all we need is that all of them are sufficiently "skinny." This means that the widest rectangle is still really "skinny." How skinny?

Let $P$ be a partition of $[a,b]$ and $t$ be a representative set. $f$ is Riemann-integrable if there exists some $K \in \mathbb{R}$ and $\delta>0$ such that $||p||<\delta$ imply that $| \sum_{i=1}^{n} f(t_{i})(x_{i}-x_{i-1})-K|<\epsilon$ for each $\epsilon>0$.

Please note that $K=\int_{a}^{b}f(x)dx$.

This is the maximum amount of generality (that I know of) to show that something is Riemann integrable without invoking upper and lower sums. It does not depend on our choice of "rectangles", since the representative set is arbitrary, and it only requires that the "widest" rectangle is smaller than some real number.

The difficulty with this definition is that you need to have a guess for what the integral actually is before proving that the function is integrable. This is why upper and lower sums are actually so important, and why most definitions require them. It can be proven that Darboux integrability implies Riemann-integrability.

For further references or further study, you can refer to Ethan Bloch's "The Real Numbers and Real Analysis."

Without additional definitions, it is probably easier to invoke some vague notions of the Darboux integrable. Assuming some familiarity with the concept: there exists a partition of $[a,b]$ such that $|U(p,f)-L(p,f)|<\epsilon$ for any $\epsilon>0$. even more loosely stated: the riemann sums of $f$ on a non-degenerate closed-bounded interval converge to a value as you take the sum over a sufficiently large number of subdivisions of the interval.

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