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This is probably a very naive question but I am trying to connect two pieces of information in my head regarding integration of differential forms and integration with respect to a measure.

The first piece is that Stokes' theorem implies the fundamental theorem of calculus in the following way:

$\int_{[a,b]}f(x)dx=\int_{[a,b]}dF(x)=\int_{\{a\}^-\cup\{b\}^+}F(x)=F(b)-F(a)$

Where $f(x)dx$ is the 1-form and $F(x)$ is the 0-form.

The second piece is that (Lebesgue) integration on a measure zero set would be equal to zero.

Since $\{a\}\cup\{b\}$ is a measure zero set, how would $\int_{\{a\}^-\cup\{b\}^+}F(x)$ being non-zero would fit into the Lebesgue framework?

Thanks

firemind
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  • Note, you really should write it as $\int_{{a}\cup{b}}$ since there is an orientation to the region - otherwise, how would you know that it is $F(b)-F(a)$ and not $F(a)-F(b)$. It's close, but there is a trickiness involving orientation. – Thomas Andrews May 20 '13 at 18:35
  • You are right, thanks. – firemind May 20 '13 at 18:37
  • The neasure on the boundary is different from the measure on the body. For example, the boundary of a disk is a circle, which has $2$-dimensional measure $0$, but there is a natural $1$-dimensional measure on the circle. – Thomas Andrews May 20 '13 at 18:38
  • Hmm, so the correct measure to think of for the above example in the case of boundary, would be a discrete measure, in which case ${a}\cup{b}$ would no longer be measure-zero. Am i getting this right? – firemind May 20 '13 at 18:44
  • I guess another way of asking the question is "how would one determine the measurability of ${a}$". – Lucas May 20 '13 at 18:44

1 Answers1

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As suggested by Thomas Andrews in the comments, the resolution of the apparent discrepancy is that there are different measures.

Length, area, and volume are different measures. If you take a disk of positive radius in the $xy$-plane in $\mathbb{R}^3$ and try to compute its volume (3-dimensional Lebesgue measure), you get zero. But it has a nonzero area, and if you use the 2-dimensional Lebesgue measure on $\mathbb{R}^2$, you will get the nonzero area.

If you compute the "area" of a circle, you get zero, but if you compute the length, you get $2\pi r$.

So what measure are we using when we compute the arc length of a curve? Is it Lebesgue measure on $\mathbb{R}^1$?

No, it the measure defined by the arc length formula $$s = \int \lvert r'(t)\rvert dt$$

I go into more detail about this measure in my answer in Surface measure=Lebesgue measure on $\mathbb{R}^{N-1}$?

So for your question, the measure we use on 0-dimensional sets is (signed) counting measure. This is also the measure induced on boundaries by arc-length or Lebesgue measure of intervals on $\mathbb{R}^1.$

In order to make Stokes' theorem and the fundamental theorem of calculus, it is a signed measure, which takes into account orientation. With that in mind, we have that $\{a\}^{-}\cup\{b\}^{+}$ has nonzero measure, and $$\int_{\{a\}^{-}\cup\{b\}^{+}}F(x) = F(b) - F(a),$$ as required.

ziggurism
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