# Integral of differential form and integral of measure

I am trying to understand the relations and differences between integral of differential form and integral of measure. From Wikipedia:

On a general differentiable manifold (without additional structure), differential forms cannot be integrated over subsets of the manifold; this distinction is key to the distinction between differential forms, which are integrated over chains, and measures, which are integrated over subsets.

1. Isn't a chain a manifold, and therefore a subset of a manifold? Why is that "differential forms cannot be integrated over subsets of the manifold"?
2. Is integral of a differential form defined in terms of Lebesgue integral via parameterization of the chain, as like here

$$\int_S \omega =\int_D \sum a_{i_1,\dots,i_k}(S({\mathbf u})) \frac{\partial(x^{i_1},\dots,x^{i_k})}{\partial(u^{1},\dots,u^{k})}\,du^1\ldots du^k$$

So can one say that integral of a differential form is not a different integration method from Lebesgue integral?

Thanks and regards!

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Related but not identical: math.stackexchange.com/q/49641/1543 –  Willie Wong Aug 15 '11 at 15:48
@Tim: A chain is neither a manifold nor a submanifold. Rather, it is a formal sum of oriented submanifolds, possibly mutually intersecting (or with other pathologies). –  Zhen Lin Aug 15 '11 at 16:04
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## 1 Answer

1. Differential forms cannot be integrated over all subsets of a manifold. You cannot integrate a 2-form over a curve, or a 1-form over a surface. More precisely: integration in measure space is defined over a $\sigma$-algebra. So if integration is defined on two sets $A$ and $B$, the integral is also defined on $A\cap B$. But the same is not true for integration of differential forms. Given a two form $\omega$ in $\mathbb{R}^3$, and take $S$ and $T$ be two smooth surfaces that intersects transversally. $\int_S\omega$ and $\int_T\omega$ make sense, but not $\int_{S\cap T}\omega$.

2. Generally a differential form, by definition, when taken in a local coordinate representation, is given by functions that are at least continuous. So in fact you don't need the machinery of Lebesgue integration: all the functions you are going to deal with can be treated with Riemann integration.

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