# Why do the infinitely many infinitesimal errors from each term of an infinite Riemann sum still add up to only an infinitesimal error?

Ok, so after extensive research on the topic of how we deal with the idea of an infinitesimal amount of error, I learned about the standard part function as a way to deal with discarding this infinitesimal difference $\Delta x$ by rounding off to the nearest real number, which is zero. I've never taken nonstandard analysis before, but here's my question.

When you take a Riemann sum, you are approximating an area by rectangles, and each of those rectangles has an error in approximating the actual area under the curve for the corresponding part of the graph. As $\Delta x$ becomes infinitesimal, the width of these rectangles becomes infinitesimal, so each error becomes infinitesimal. But since there are infinitely many rectangles in that case, why is it that the total error from all of them still infinitesimal? In other words, shouldn't an infinite amount of infinitesimals add up to a significant amount?

• I think there may be a fundamental misunderstanding in your question. You bring up the topic of nonstandard analysis, though the most common definition of the integral lives comfortably in the realm of standard analysis. Are you really looking for an answer which uses the concepts of nonstandard analysis? – Antonio Vargas Jun 13 '16 at 2:16
• @AntonioVargas maybe. I'm saying that, if I understand correctly, there's infinitesimal error that we round down to zero. But, it's the limit of a Reimann sum to an infinite term that defines the integral. The hyperreals are negligible for any finite summation, but I don't see how we can take the standard part for an infinite sum. So it's a question of how we justify real analysis knowing the non-standard analysis approach. – rb612 Jun 13 '16 at 2:18
• Numbers such as slope and integral are defined to be the standard part of their hypperreal values. It's sort of like saying that the "real" world is a subset of the "hyperreal" world. – steven gregory Jun 13 '16 at 2:28
• I'm going to ignore nonstandard analysis. By basic geometry, any finite Riemann sum, as an approximation for signed area between the graph of a (let's say continuous) function and the x-axis, has discrepancy bounded by a bunch of rectangles with error $\le\sum |\Delta f(x_i)|\Delta x_i\le \max_i |\Delta f(x_i)|\cdot (b-a)$. On a compact interval $[a,b]$ at least, $f$ is uniformly continuous, hence one can bound $\max_i |\Delta f(x_i)|$ below any chosen $\varepsilon>0$ for sufficiently small mesh. Hence the Riemann sum converges to the correct signed area as mesh tends to zero. – arctic tern Jun 13 '16 at 4:25
• I have rewritten the question to try to make it clearer. Feel free to revert or further edit if I haven't stated the question exactly the way you meant it. – Eric Wofsey Jun 13 '16 at 5:06

If I've understood the question correctly, here is a heuristic explanation. Note that this is not rigorous, since to make your question rigorous you have to give some precise definition of what you mean by "the exact area", which is not at all easy to define in general.

Let us assume we are integrating a continuous function $f(x)$ from $0$ to $1$ by using a Riemann sum with infinitesimal increment $\Delta x$. Let us also assume for simplicity that $f$ is increasing (the general case works out essentially the same way but is a little more complicated to talk about). So we are approximating "the area under $f$" by replacing the region under the graph of $f$ from $x=c$ to $x=c+\Delta x$ by a rectangle of height $f(c)$, for $1/\Delta x$ different values of $c$. Now since $f$ is increasing, the difference between our rectangle of height $f(c)$ and the actual area under the graph of $f$ from $c$ to $c+\Delta x$ is at most $\Delta x(f(c+\Delta x)-f(c))$. But since $f$ is (uniformly) continuous, $f(c+\Delta x)-f(c)$ is infinitesimal. So our error is an infinitesimal quantity times $\Delta x$.

So although we are adding up $1/\Delta x$ (an infinite number) different errors to get the total error, each individual error is not just infinitesimal but infinitesimally smaller than $\Delta x$. So it is reasonable to expect that the sum of all of the errors is still infinitesimal.

• This makes so much sense! Thank you! Why is this explanation not used often in talking about integration in real analysis? I've done much research and haven't seen it explained as such before. – rb612 Jun 13 '16 at 4:32
• I think that many people get it drilled into their head early on that Riemann sums can approximate areas, so it doesn't even occur to them to question this. – Eric Wofsey Jun 13 '16 at 4:35
• well thank you for making it easy to understand. I just have one question about what you said. You said that because $f$ is continuous, $f(c+\Deltax)-f(c)$ is infinitesimal. It does make intuitive sense, but could you explain it a bit more? Is it because for a given $\Deltax$ then it can only grow at most $c\Deltax$ where $c$ is a finite constant; and a finite constant times an infinitesimal is still infinitesimally small? I don't quite understand how we know it's infinitesimal just because it's continuous. Sorry if this is an obvious question. – rb612 Jun 13 '16 at 4:41
• In the context of nonstandard analysis, this is usually the definition of continuity (or more properly, uniform continuity, but it doesn't make a difference in this context). A function $f$ is (uniformly) continuous iff whenever $x$ is infinitely close to $y$, $f(x)$ is infinitely close to $f(y)$. – Eric Wofsey Jun 13 '16 at 4:43
• now if a function is not uniformly continuous like $x^2$ as I have been reading, doesn't this change things a bit? You said it doesnt matter in this context though, so maybe I'm misunderstanding. Because as x increases by an infinitesimal we can't guarantee that the function will increase by an infinitesimal? Or am I misunderstanding uniform continuity – rb612 Jun 13 '16 at 7:53

Let's compute!

Suppose we integrate $\int_0^1 x \, \mathrm{d}x$ by pick a Riemann sum with $H$ equally sized intervals.

The smallest and largest value of $x$ in the $i$-th interval is $\frac{i-1}{H}$ and $\frac{i}{H}$ respectively, and that interval would contribute $\frac{i-1}{H} \cdot \frac{1}{H}$ and $\frac{i}{H} \cdot \frac{1}{H}$ respectively to the total sum, so the maximum possible error from ths interval is $\frac{1}{H^2}$.

However, there are only $H$ intervals — the error therefore cannot be any larger than $\frac{1}{H^2} \cdot H = \frac{1}{H}$, an infinitesimal.

And for fun,we compute the integral using the upper bounds: \begin{align*} \int_0^1 x \, \mathrm{d}x &= \text{std} \sum_{i=1}^H \left( \frac{i}{H} \right) \cdot \frac{1}{H} \\&= \text{std} \left( \frac{1}{H^2} \sum_{i=1}^H i \right) \\&= \text{std} \left( \frac{1}{H^2} \frac{H(H+1)}{2} \right) \\&= \text{std} \left( \frac{1}{2} + \frac{1}{2H} \right) \\&= \frac{1}{2} \end{align*}

You can see how the errors of