# The Unapologetic Mathematician

## Higher-Dimensional Riemann Integrals

Our coverage of multiple integrals will actually parallel our earlier coverage of Riemann integrals pretty closely. Only now we have to change our notion of “interval” to a higher-dimensional version.

For the moment, the only shapes we’ll integrate over will be closed rectangular $n$-dimensional parallelepipeds with sides parallel to the coordinate axes in $\mathbb{R}^n$. In one dimension, these are just coordinate intervals, but there aren’t many other obvious shapes to use in that case. In two dimensions, we are (for the moment) throwing out circles, ellipses, triangles, and anything else that’s not a rectangle; and even rectangles that are tilted with respect to the coordinate axes! Don’t worry, we’ll get them back later.

Anyway, such a shape in $n$-dimensional space is the product of $n$ closed intervals:

$\displaystyle[a^1,b^1]\times\dots[a^n,b^n]=\{(x^1,\dots,x^n)\in\mathbb{R}^n\vert a^1\leq x^1\leq b^1,\dots,a^n\leq x^n\leq b^n\}$

In this context, we’ll write this as $[a,b]$ for $a=(a^1,\dots,a^n)$ and $b=(b^1,\dots,b^n)$. This looks identical to, but is not to be confused with, the closed line segment from $a$ to $b$ we used in the mean value theorem. I’ll try to keep them separate by always referring to this rectangular parallelepiped as an “interval” and never using that term for a line segment. Of course, in one dimension the two are exactly the same.

Now we define a partition of an interval. We’ll do this just by partitioning each side of the interval. That is, we pick points

\displaystyle\begin{aligned}a^1={x^1}_0<&\dots<{x^1}_{m_1}=b^1\\&\vdots\\a^n={x^n}_0<&\dots<{x^n}_{m_n}=b^n\end{aligned}

where we cut the $i$th side into $m_i$ pieces, and each of the $m_i$ could be different. If we index the subintervals of the partition by $1\leq i_1\leq m_1$ through $1\leq i_n\leq m_n$, we can write

$\displaystyle I_{i_1\dots i_n}=[{x^1}_{i_1},{x^1}_{i_1+1}]\times\dots\times[{x^n}_{i_n},{x^n}_{i_n+1}]$

picking out the $i_k$th slice of the $k$th side. This cuts the whole interval up into a bunch of smaller pieces, which are themselves rectangular parallelepipeds. Again, we have the notion of a tagged partition, which picks out a sample point $t_{i_1\dots i_n}\in I_{i_1\dots i_n}$ for each subinterval. And again we say that one tagged partition is a “refinement” of another one if every partition point of the coarser partition is also one in the finer partition on the same side, and if every sample point in the coarser partition is one in the finer partition as well.

There’s a lot of numbers here and a lot of notation, but relax: most of it is going to go away very quickly. Go back and look at how partitions, tagged partitions, and refinements were defined in one variable, and just think of partitioning (and refining) the one-dimensional interval that makes up each side of the $n$-dimensional interval. Of course, like I said before, the number of parts on each side might have nothing to do with each other. Further, there’s no reason for the tags to have anything to do with each other. One might think of tagging the partition on each side of the interval, and then letting the sample points in the subintervals have those tags as coordinates. This is a perfectly valid way to come up with a collection of sample points, but the sample points don’t have to arise in this manner at all, so long as there’s exactly one within each subinterval.

So, just like in the one-dimensional case, the collection $\mathcal{P}[a,b]$ of all tagged partitions of an interval $[a,b]$ form a directed set, where we say that $P\preceq P'$ if $P'$ is a refinement of $P$. And again we define nets on this directed set; given a function $f$ defined on the interval $[a,b]$ and a partition $P\in\mathcal{P}[a,b]$, we define the Riemann sum

$\displaystyle f_P=\sum\limits_{i_1=1}^{m_1}\dots\sum\limits_{i_n=1}^{m_n}f(t_{i_1\dots i_n})\mathrm{vol}(I_{i_1\dots i_n})$

by sampling the function $f$ at the specified point $t_{i_1\dots i_n}$ in the subinterval $I_{i_1\dots i_n}$, multiplying by the $n$-dimensional volume of the subinterval (which is just the product of its side-lengths), and summing this over all the subintervals in $[a,b]$.

And again, as before, if this net converges to a limiting value $s$, we say that $f$ is Riemann integrable on the interval $[a,b]$ and we write

$\displaystyle\int\limits_{[a,b]}f\,dx=\int\limits_{[a,b]}f(x)\,dx=\int\limits_{[a,b]}f(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=s$

in three different notations, depending on what we want to emphasize. The first emphasizes the function as an object in and of itself; the second, the function’s dependence on the vector variable $x$; and the third, the function’s dependence on each individual real variable. In two and three dimensions, we often write the “double integral” and “triple integral” as

\displaystyle\begin{aligned}\iint\limits_{[a,b]}&f(x)\,dx\\\iiint\limits_{[a,b]}&f(x)\,dx\end{aligned}

as visual reminders that these are multiple integrals. This gets unwieldy very quickly, and so we usually just write one integral sign for each integration.

December 1, 2009 - Posted by | Analysis, Calculus

## 10 Comments »

1. […] and Lower Integrals and Riemann’s Condition Yesterday we defined the Riemann integral of a multivariable function defined on an interval . We want to move towards […]

Pingback by Upper and Lower Integrals and Riemann’s Condition « The Unapologetic Mathematician | December 2, 2009 | Reply

2. […] an -dimensional box — an interval . Then partition this interval with some partition like we did when we set up higher-dimensional Riemann sums. Then we just count up all the subintervals in the […]

Pingback by Jordan Content « The Unapologetic Mathematician | December 3, 2009 | Reply

3. […] It asserts, simply enough, that a bounded function defined on an -dimensional interval is Riemann integrable on that interval if and only if the set of discontinuities of has measure zero. Our proof will go […]

Pingback by Lebesgue’s Condition « The Unapologetic Mathematician | December 15, 2009 | Reply

4. […] Integrals I We may remember from a multivariable calculus class that we can evaluate multiple integrals by using iterated integrals. For example, if is a continuous, nonnegative function on a […]

Pingback by Iterated Integrals I « The Unapologetic Mathematician | December 16, 2009 | Reply

5. […] this case, we’re considering integrating an integrable function over some -dimensional interval . We want something like iterated integrals […]

Pingback by Iterated Integrals IV « The Unapologetic Mathematician | December 21, 2009 | Reply

6. […] Over More General Sets To this point we’ve only discussed multiple integrals over -dimensional intervals. But often we’re interested in more general regions, like circles […]

Pingback by Integrals Over More General Sets « The Unapologetic Mathematician | December 22, 2009 | Reply

7. […] is a simple consequence of the definition of a multiple integral as the limit of Riemann sums, since every Riemann sum for will be smaller than the corresponding sum for […]

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8. […] this is not the differential of a variable . When it shows up in a multiple integral, it’s a tiny little bit of -dimensional volume. And we measure the scaling of -dimensional […]

Pingback by Change of Variables in Multiple Integrals I « The Unapologetic Mathematician | January 5, 2010 | Reply

9. […] a particular integrator in our Riemann-Stieltjes integral. This could also be useful in setting up multiple integrals over -dimensional intervals that extend infinitely far in some […]

Pingback by Improper Integrals II « The Unapologetic Mathematician | January 15, 2010 | Reply

10. […] together the basic -forms in order. And then on the right we suddenly switch to a -dimensional Riemann integral over the standard -cube. The canonical basis -form corresponds to the volume element , and top […]

Pingback by Integration on the Standard Cube « The Unapologetic Mathematician | August 2, 2011 | Reply