Stokes' Theorem and Measure Zero Sets 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
 A: 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.
