Relating integration of forms to Riemann integration

[Eclair_de_XII]

TL;DR

Let $A:=[a_1,b_1]\times\cdots\times[a_n,b_n]$. Let $f:A\rightarrow\mathbb{R}$. Let $g_i:[a_i,b_i]\rightarrow \mathbb{R}$. Assume that $g_i$ defined as $g_i(x)=f(c_1,\ldots,c_{i-1},x,\ldots,c_n)$ is integrable on its domain. Then we can define integration of $f$ as such:

Partition each closed interval $[a_i,b_i]$ in the Cartesian product, $A$.
Denote the partition for the i-th closed interval as $\{x_i^1,\ldots,x_i^{k_i}\}$.
The Cartesian product of the partitions forms a partition of $A$ (think: a lattice of points that coincide with the points of each partition).
Let $a_0$ be a point in the partition.
Let $a'$ be the point in the partition s.t.:
1. Each component of $a'$ is greater than that of $a_0$.
2. The distance from $a'$ and $a_0$ is the infimum of the set of distances from $a_0$ to another point in the partition whose components exceed those of $a_0$
The vectors $a_0$ and $a'$ define hyper-rectangle that constitutes the volume element of the volume to be estimated.
Evaluate $f$ at some point $x_0$ in this hyper-rectangle in order to get a proper weight for this region of the domain.
In order to get the volume of that hyper-rectangle, we evaluate the volume element $dx^1\wedge\ldots\wedge dx^n$ at $(a'-a_0,a'-a_0,\ldots,a'-a_0)$.
This process yields a number $f(x_0) dx^1\wedge\ldots\wedge dx^n(a'-a_0,\ldots,a'-a_0)$ where the latter term is the "volume" of the element of the partition on which $f$ is to be evaluated.

Anyway, I was trying to relate integration of forms to Riemann integration earlier. I was confused by the relationship between the notation $\int_A f\,dx^1\wedge\,dx^2$ and $\int_A f(x)\,dx^1\,dx^2$, and then I tried to come up with my own explanation for why these two are equivalent, using my rather rusty knowledge of Riemann integration.

Forgive me if this is poorly-explained.

 

[FactChecker]

Maybe I am missing something. I'm struggling to understand this. Is this your wording or is this how a reference defines it? If it is a reference, please give it.

This process yields a number $f(x_0) dx^1\wedge\ldots\wedge dx^n(a'-a_0,\ldots,a'-a_0)$ where the latter term is the "volume" of the element of the partition on which $f$ is to be evaluated.

Is this a "number" or a vector? I am guessing that $(a'-a_0, a'-a_0, ...,a'-a_0)$ is a vector without the indices properly indicated. Exactly what term is the volume?

 

[Eclair_de_XII]

Is this your wording

Unfortunately, yes. I realize now that this rudimentary dump of ideas is a jumbled mess that even the most patient academic would have trouble deciphering.

Is this a "number" or a vector?

The $dx^1\wedge\ldots\wedge dx^n(a'-a_0,\ldots,a'-a_0)$ term would be the volume. The $dx^1\wedge\ldots\wedge dx^n$ maps the matrix $(a'-a_0,\ldots,a'-a_0)$ to the product of the "side-lengths" of the hyperrectangle.

I should probably have just explained that this Cartesian product of partitions forms a lattice of points in n-space. And each pair of nodes in the lattice defines a hyperrectangle. These hyperrectangles are analogous to the intervals that partition of given closed interval, which serves as the domain for a Riemann-integrable function.

 

[FactChecker]

Anyway, I was trying to relate integration of forms to Riemann integration earlier. I was confused by the relationship between the notation $\int_A f\,dx^1\wedge\,dx^2$ and $\int_A f(x)\,dx^1\,dx^2$, and then I tried to come up with my own explanation for why these two are equivalent, using my rather rusty knowledge of Riemann integration.

I'm not sure they are exactly equivalent. My understanding is that the wedge product, $dx^1\wedge\,dx^2$, is a generalized oriented area represented by the counterclockwise rotation from the vector $dx^1$ to $dx^2$. I'm not sure that $dx^1 dx^2$ represents the same thing unless the bounds on the integrand are a match. I will have to review the subject to give a more authoritative answer.
UPDATE: For $dx^1\wedge\,dx^2$, order is very important; $dx^1\wedge\,dx^2 = - dx^2\wedge\,dx^1$. That helps greatly in book-keeping and makes many results methodical. The same is not true of $dx^1 dx^2$, where $dx^1 dx^2 = dx^2 dx^1$ and you are left to your own devices to keep things straight.