# I Definitions of the Riemann integral

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1. Oct 5, 2016

### Frank Castle

In some elementary introductions to integration I have seen the Riemann integral defined in terms of the limit of the following sum $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\Delta x$$ where the interval $[a,b]$ has been partitioned such that $a=x_{1}<x_{2}<\cdots <x_{n-1}<x_{n}=b$, with $x^{\ast}_{i}\in [x_{i-1},x_{i}]$ (and arbitrary point in the sub-interval $[x_{i-1},x_{i}]$), and $\Delta x=\frac{b-a}{n}$ the width of each sub-interval.

However, I have also seen it defined in terms of a similar sum, differing in the fact that the width of each sub-interval does not have to be equal. Indeed, given the same partition, we have that $\text{Max}\,\lvert\Delta x_{i}\rvert=\lbrace \lvert x_{2}-x_{1}\rvert,\ldots , \lvert x_{i}-x_{i-1}\rvert,\ldots , \lvert x_{n}-x_{n-1}\rvert\rbrace$, and then $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\lim_{\text{Max}\,\lvert\Delta x_{i}\rvert\rightarrow 0}\sum_{i=1}^{n}f(\zeta_{i})\Delta x_{i}$$ where $\Delta x_{i}=x_{i}-x_{i-1}$ is the width of the $i^{th}$ sub-interval and $\zeta_{i}\in [x_{i},x_{i-1}]$ is an arbitrary point within this interval.

I have a couple of questions about these definitions:

1. Is the second definition I gave in some sense "better" than the first (in the sense that it is more general - it allows for each of the sub-intervals to have a different length)?

2. It is my understanding that $\text{Max}\,\lvert\Delta x_{i}\rvert$ denotes the maximum width of a given sub-interval. Given this, why do we take the limit as this maximum width tends to zero (especially as the width of each interval in the sum isn't necessarily the maximum)? Is the intuition that we wish for the width of each interval to become infinitesimally small as the number of sub-intervals becomes infinitesimally large (hence taking the limit $n\rightarrow\infty$), and thus, by taking the limit as $\text{Max}\,\lvert\Delta x_{i}\rvert\rightarrow 0$ this ensures that the width of each interval will become infinitesimally small (regardless of whether $x_{i}-x_{i-1}=\text{Max}\,\lvert\Delta x_{i}\rvert$ or whether it is smaller than the maximum width)?!

Last edited: Oct 5, 2016
2. Oct 5, 2016

### pwsnafu

It turns out they are equivalent provided $x_i^*$ is arbritrary.

The requirement is there to ensure all sub-intervals become small in the limit. If you don't have it, you can fix, say, $x_2 - x_1 = 0.5$ but make the other sub-intervals small as $n\to\infty$. In this situation you never converge to the area under the curve.

Actually the $\lim_{n\to\infty}$ is misleading as you don't have a sequence. The limit is through refinements of the partition.

3. Oct 5, 2016

### Frank Castle

How does one show that they are equivalent?

Ah ok, so is the intuition that $\text{Max}\,\lvert\Delta x_{i}\rvert$ denotes that maximum possible width that any of the sub-intervals can have and so by making this become infinitesimally small this forces all of the sub-intervals to become infinitesimally small in the limit?!

So is the limit $\lim_{n\rightarrow\infty}$ symbolising the case in which the number of points considered between the end-points $a$ and $b$ becomes infinite, in the sense the at the interval is partition into an infinite number of points, and then the second limit $\text{Max}\,\lvert\Delta x_{i}\rvert\rightarrow 0$ ensures that the width between each successive point becomes infinitesimally small?

4. Oct 5, 2016

5. Oct 6, 2016

### Frank Castle

Would what I wrote in post #3 in response to pwsnafu's comments be correct at all?!

6. Oct 6, 2016

### mathman

Your first comment was a question. The others seem correct.

7. Oct 6, 2016

### pwsnafu

Usually by showing both are equivalent to the Darboux integral.

There is no such thing as "infinitesimally small". But the overall idea is intuition is correct.

By definition a partition is finite. Always. So no.

I have a feeling you don't know the definition of limit of sequence, let alone that of a net. Until you do, you will forever be confused.

Huh? H-K integrals you a gauge. There is no mention of that anywhere in the OP.

8. Oct 7, 2016

### Frank Castle

True, I just couldn't think of a better way to phrase it heuristically.

Isn't it the case that a sequence $(a_{n})$ has a limit if $\lim_{n\rightarrow}a_{n}=L$ where $L$ is some finite number?! My intuitive understanding of this is that, if this limit exists, then as we consider terms in the sequence for larger and larger values of $n$, their values will all be within a distance $\epsilon$ from the value $L$, and as such we say that the sequence tends towards a limiting value $L$.

Is the limit in this case then meant to symbolise the limit of a sequence of partial sums $$\Bigg(\lim_{\lvert\Delta x_{i}\rvert\rightarrow 0}\sum_{i =1}^{n}f(\zeta_{i})\left(x_{i}-x_{i-1}\right)\Bigg)_{n\in\mathbb{N}}$$ where each value of $n$ corresponds to a refinement of the partition of the interval $[a,b]$ (in the sense that as $n$ increases in value, the partition becomes more and more refined, for example, for $n=2$, the interval has the partition $P_{2}$ which would be a refinement of the $n=1$ partition, $P_{1}$). If the partial sums in the sequence tend towards a limiting value as $n$ becomes large in value (that is, as the partition becomes more and more refined), then we say that the sequence has a limiting value, equal to the integral $\int_{a}^{b}f(x)dx$.

Last edited: Oct 7, 2016
9. Oct 7, 2016

### pwsnafu

You do have a good intuition. But intuition is not a definition.
For all $\epsilon > 0$ there exists $N > 0$ such that whenever $n > N$ one has $|a_n - L| < \epsilon$.

Have you seen the above definition before?

Yes, you have the general idea. The problem with writing $\lim_{n\to\infty}$ is that there are many refinements of a partition. So you have to choose $P_1, P_2, \ldots$. However $Q_2, Q_3, \ldots$ may be different refinements of $P_1$. The definition needs to be written so that no matter how you take your partitions, you get the same limit.

Go back to the first definition you wrote in the OP, the one with fixed partition widths. Notice how you wrote $x_i^*$ is arbitrary. But it's not actually, Different choices of $x_i^*$ (these are called "tags" or "tag points" by the way) will result in a different Riemann sum. That's why writing it as a sequence is misleading.
An explicit example: $x=0$ if $x$ is rational, $x=1$ if $x$ is irrational. What happens if you choose tags to always be rational? What about always irrational?

Last edited: Oct 7, 2016
10. Oct 7, 2016

### Frank Castle

Good point, but you have to start somewhere right

Yes, it looks familiar. So, for values of $n>N$ every remaining term in the sequence will have a value within $\epsilon$ distance $L$, such that the sequence converges towards this unique limiting value $L$, right?!

Ah, ok. So larger values of $n$ correspond to further refinements of partition, but there will in general be many ways to partition the interval, and subsequently many ways to continually refine these partitions, but the point is that whichever way you choose to partition the interval, in the limit of increasing refinements, each choice should lead to the same limiting value?!

Is there a way to make this definition more rigorous? It seems to be used in quite a few introductory texts (at least those aimed at physicists)?!

11. Oct 7, 2016

### pwsnafu

Correct.

Yes.

I'm going to write the Riemann sum as $R(f, P)$.
Restrict our attention to partitions such that the tag point satisfies $x_i^* \in [x_i, x_{i+1}]$.

Suppose there exists a number $L$ with the property: for all $\epsilon > 0$ there exists a partition $P$ such that
$| R(f,Q) - L |<\epsilon$ whenever $Q$ is a refinement of $P$.

Edit: the definition I gave is called the refinement integral, which leads to the same integral but is not the same definition as in OP. To make that rigorous:

Suppose there exists a number $L$ with the property: for all $\epsilon > 0$ there exists $\delta > 0$ such that
$|R(f,P) - L| < \epsilon$ whenever $\max{\Delta x_i} < \delta$.

Last edited: Oct 7, 2016
12. Oct 9, 2016

### Frank Castle

Ok, cool. I'm glad I'm starting to understand things a little better.

Can one also define the definite integral in terms of upper and lower sums ($U(f;P)$ and $L(f;P)$ respectively), as follows:

A function $f$, bounded on the interval $[a,b]$, is said to be Riemann integrable on an interval $[a,b]$ if there exists a sequence of partitions $\left(P_{n}\right)$ such that $$\lim_{n\rightarrow\infty}\left[U(f;P_{n})-L(f;P_{n})\right]=0$$ in which case, we have that $$\int_{a}^{b}f(x)dx =\lim_{n\rightarrow\infty}U(f;P_{n})=\lim_{n\rightarrow\infty}U(f;P_{n})$$ Or does this fall into the same problems as the definition I gave in my OP?!

With regards to the definition $$\int_{a}^{b}f(x)dx =\lim_{n\rightarrow\infty}\sum_{k =1}^{n}f(\zeta_{k})\frac{(b-a)}{n}$$ Is the reason why this is, in some sense, a valid definition, because for any given partition we choose the Riemann sum should converge to the same limit. Hence we can choose a partitioning in which the interval is evenly partitioned, $x_{k}= a+k \frac{(b-a)}{n}$, leading to the definition above?! (I appreciate that this definition has its problems, but I just want to understand the motivation for it in the first place).

Last edited: Oct 9, 2016
13. Oct 10, 2016

### pwsnafu

It falls into the exact same trap: it's not a sequence so $\lim_{n\to\infty}$ is wrong.
To define it using upper and lower sums we look at $\inf U(f;P)$ and $\sup L(f;P)$, where the infimum and supremum are taken over all possible partitions. Remember, the real numbers have the property that any set bounded above has a supremum (and similarly for bounded below).
Then if those two quantities are equal we say it is the integral.

It's more to do with the fact that given any partition P, there exists equal distance one which approximates it (by taking n large enough).

14. Oct 11, 2016

### Frank Castle

I thought as much. It makes the whole concept difficult to learn though when seemingly good lecture notes use such definitions (see page 10 of these notes for example: https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf).

In practice then, how does one actually calculate a fairly standard integral, say $\int_{0}^{1}x^{2}dx$ using the rigorous definition you gave? (It seems reasonably straight forward using the definitions I've come across in terms of limits of Riemann sums, but I'm not sure how one would go about it using the proper definition you gave).

15. Oct 12, 2016

### pwsnafu

There's actually a reason for it. You see, if f is continuous over $[a,b]$ then it is uniformly continuous as well. So given any $\epsilon > 0$ there exists $\delta >0$ such that $|f(x)-f(y)| < \epsilon/(b-a)$ whenever $|x-y|<\delta$. Now choose a partition such that $\Delta x_i < \delta$. Observe $\sum (x_{i+1}-x_i) = b-a$. Then
$0 \leq \bigg| \sum f(\zeta_i)(x_{i+1}-x_i) - \sum f(\eta_i)(x_{i+1}-x_i) \bigg| \leq \sum |f(\zeta_i)-f(\eta_i)||x_{i+1}-x_{i}| < \sum \frac{\epsilon}{b-a} |x_{i+1}-x_i| = \epsilon$.

What this shows is that provided (1) f is continuous and (2) you only consider closed intervals, it doesn't matter where the tag points are. That "arbitrary" choice in the OP, is really arbitrary. At high school level, the course is designed so that only continuous are covered. So it works. But it's not a good definition to rely on as you advance through mathematics.

There is a misconception. The purpose of the definitions is not allow for calculations at all, it to prove their properties. We then use their properties to calculate the integral (usually the fundamental theorem). The reason for this is that the definition contains L, so you can't use the definition until you have calculated the integral beforehand.

As an aside even in numerics, you don't use the definition in the OP. You use trapezoidal rule or Simpson's rule.
You can also probably choose your partition so that if the function is steep you use narrow rectangles and when it is flat use fat rectangles. This idea of fat-vs-thin leads to the Henstock integral linked earlier.