Gauss tells us that $$ \int_V \text{div}\vec{F}\, d^{(n)}V=\int_{S}\vec{F}\cdot\vec{\nu}\, d^{(n-1)}S, $$ where $S=\partial V$ and $\vec{\nu}$ is the outer normal.
For $n=1$, $V=[a,b]$ and $f\colon V\to\mathbb{R}$, this gives $$ \int_a^b\frac{df}{dx}dx = \int_{S}f\cdot \nu d^{(0)}S, $$ $S=\partial V=\left\{a,b\right\}$ and $\nu(a)=-1, \nu(b)=1$.
In nearly all literature it is said without any reasoning that $$ \int_{S}f\cdot \nu d^{(0)}S=f(b)-f(a), $$ i.e. we get the fundamental theorem of calculus.
I really wonder how we really can compute this integral, that is, how can we parametrize $S$ and what the meaning of the $0$-dimensional Lebesgue-measure could be.
To be more concrete:
Assume $S=\gamma(A)$ for some parametrization $\gamma$ (here $A$ has to be $0$-dim.), then we compute the integral over S by $$ \int_A (f\cdot\nu)(\gamma(x))\sqrt{\text{det}(D\gamma(x)^T D\gamma(x))}\lambda^0 x $$
But...
(1) What is a good parametrization $\gamma$? Maybe just the identity map $\text{id}\colon S\to S$?
(2) What is the 0-dimensional Lebesgue-measure $\lambda^0$?!