Are differential forms lebesgue measures

Question

In my analysis class integral of differential form $w=fdx_1 \wedge dx_2 \wedge ... \wedge dx_k$ on some open subset of $R^n$ was defined as follows:

$$\int_Uw=\int_Ufd\lambda$$

So as lebesgue integral.

Consider set bounded by $\left|x \right|<1$ and $\left|y \right|<1$ (square), call this set $U$. I want to compute following integral:

$$\int_Udx \wedge dy$$

From definition it is equal to:

$$\int_Ud\lambda_2$$

Which is equal to:

$$\int_{-1}^1\int_{-1}^1dxdy$$

Now, from this post: what measure is dx ,if $dx$ is lebesgue measure $1$, is $dxdy$ lebesgue measure $2$? Is it true that $dx \wedge dy$ is lebesgue measure $2$?

So the question basically is whether differential forms are lebesgue measures.

Answer 1

A differential form is not a measure. But, you can use a differential form to induce a measure (see first paragraph here), and you can use the integral over Lebesgue measure to define integrals of forms over oriented manifolds (see second part of my answer here). It’s like lemons vs limes; they’re not the same thing, but they taste pretty similar.

So, on $\Bbb{R}^2$, it is not correct to say $dx\wedge dy=d\lambda_2$, but (with the usual conventions for orientation), we can say that by definition, for any Lebesgue-measurable set $E$, $\int_Edx\wedge dy=\int_E1\,d\lambda_2$. The thing is the symbol $\int_E$ on the left stands for the integral of a differential form, while the integral symbol on the right is $\int_E(\cdot)\,d\lambda_2$, the Lebesgue integral of a function over the set $E$. So, I’m using an already defined object (Lebesgue integral of a function on a measure space) to define a new object (integral of a differential form on an oriented manifold).

Having said this, by abuse of language, it is not uncommon to find people saying things like $d\lambda_2=dx\wedge dy$ (among other things).

Answer 2

The algebraic concept of alternating products of linear forms as linear maps from tangent vectors to the number field is differential geometry plus exterior tensor algebra to build up the local complex of all volume forms from k=0 to k=n over an n-dimensional manifold.

Under an integral, differential forms are maps on the boundaries written under the integral sign without any descriptin of the order of summation. This is making sense only, if the forms represent Lebesgue measures- which the fundamental formulas hold

$$ \int_a^b \ \mathbb d x = b-a, \ \ \mathbb d ( a x + b) = a \mathbb d x $$

An integral is the replacement of a volume sum by a surface sum. No Riemann-like integrals will work in all generality, except for a small class of smooth integrands and smooth surfaces.

In everyday speech at the blackboard, the term "Lebesgue measure" is often used as short form for a translation invariant measure as opposed to Stieltjes measures with position dependend weight functions, eg in curvilinear coordinates with a volume density.