surface measure and Gauss-Green theorem proof

Solution 1:

I'm not going to prove the divergence theorem here, because you can find proofs of this in books which cover integration on manifolds. Particularly, Amann and Escher's Analysis III covers all the necessary measure-theory and Lebesgue-integration background. In the end they also have a chapter dedicated to integration on manifolds (where Stokes theorem is proved in great generality, allowing for slightly rougher boundaries than what is usually proven, for example in Spivak's Calculus on manifolds. Of course, more analysis-related technicalities are present to make all the approximations work out).

What I will mention is the very basics of measures on manifolds. Throughout, $M$ shall denote a $C^r$ ($r\geq 1$), $m$-dimensional manifold, which we assume to be second-countable, which means we have a countable atlas (this is automatically true for submanifolds of $\Bbb{R}^n$, like $\partial \Omega$ in your case).

1. What is 'the' $\sigma$-algebra?

Well, $M$ is a $C^r$ manifold, so it is in particular a topological space, hence we can equip it with the Borel $\sigma$-algebra.

Alternatively, we can also equip it with the Lebesgue $\sigma$-algebra $\mathcal{L}_M$. The definition is that a subset $A\subset M$ belongs to $\mathcal{L}_M$ if for every chart $(U,\alpha)$ of $M$, we have that $\alpha[A\cap U]$ is a Lebesgue-measurable subset of $\Bbb{R}^m$. You can easily verify that the collection $\mathcal{L}_M$ is indeed a $\sigma$-algebra, which contains the Borel $\sigma$-algebra.

2. Volume Measures on Pseudo-Riemannian Manifolds.

Let $(M,g)$ be a pseudo-Riemannian manifold (of class $C^r$, $r\geq 1$, of dimension $m$, and again, second-countable). There is a very important measure $\lambda_g$ which can be defined on the Lebesgue-$\sigma$-algebra of $M$, and it is constructed from the metric tensor field $g$. This $\lambda_g$ captures our intuitive understanding of volumes of manifolds. The precise theorem is as follows:

Let $(M,g)$ be a pseudo-Riemannian manifold. There is a unique, positive measure $\lambda_g$ defined on the Lebesgue $\sigma$-algebra $\mathcal{L}_M$, such that for every chart $(U,\alpha)$ of $M$ and for any $A\in\mathcal{L}_M$, we have \begin{align} \lambda_g(A\cap U)&=\int_{\alpha[A\cap U]}\sqrt{|\det [g_{(\alpha)}]|}\,d\lambda_m,\tag{$*$} \end{align} where $\lambda_m$ is the (restriction of) Lebesgue measure on $\Bbb{R}^m$, and the integrand is the square root of the absolute value of the determinant of the matrix of components of the metric tensor relative to the chart $(U,\alpha)$.

In fact, $\lambda_g$ is complete and $\sigma$-finite (and we refer to this as the Riemann-Lebesgue volume measure on $M$).

The proof of uniqueness is clear. Fix a countable atlas $\{(U_j,\alpha_j)\}_{j=1}^{\infty}$ of $M$, and take any $A\in\mathcal{L}_M$. Then, $\lambda_g(A)$ is uniquely determined as \begin{align} \lambda_g(A)&=\lambda_g\left(\bigcup_{j=1}^{\infty}U_j\cap A\right)\\ &=\sum_{j=1}^{\infty}\lambda_g(U_j\cap A)\\ &=\sum_{j=1}^{\infty}\int_{\alpha_j[A\cap U_j]}\sqrt{|\det [g_{(\alpha_j)}]|}\,d\lambda_m \end{align}

To show the existence, you run this arugment in reverse: consider the set function $A\mapsto \sum_{j=1}^{\infty}\int_{\alpha_j[A\cap U_j]}\sqrt{|\det [g_{(\alpha_j)}]|}\,d\lambda_m$, and show it satisfies the properties of a measure, and that it satisfies the property $(*)$ stated in the theorem (in proving countable additivity, you'll have to swap a double series of non-negative terms, and for proving $(*)$, you'll have to use the usual change of variables in $\Bbb{R}^m$).

The $\sigma$-finiteness is clear, and completeness of $\lambda_g$ follows from completeness of $\lambda_m$.

(btw this result is also given in the book, though there it is proven using a partition of unity argument... but I prefer this more elementary approach, hence I presented this).

3. Usual Applications.

Usually, what happens is that we're given a (say Riemannian) manifold $(M,g)$, along with a submanifold $S$ of $M$. Then, the standard way of inheriting a metric tensor on $S$ is by taking the pullback $\iota^*g$, where $\iota:S\to M$ is the canonical injection. Then, by the theorem above, we have a corresponding measures $\lambda_g$ on $M$ and $\lambda_{\iota^*g}$ on $S$. In the case when $S$ has codimension $1$ in $M$, we refer to $\lambda_{\iota^*g}$ as the induced surface measure on $S$, and very often it is denoted as $\sigma$. So, volume integrals over $M$ are written $\int_M f\,d\lambda_g$ or just $\int_Mf \,dV$, while integrals over the submanifold $S$ are written $\int_Sf\,d\lambda_{\iota^*g}$ or just $\int_Sf\,d\sigma$, or maybe even $dS$ is written.

As a hyper special case, consider $M=\Bbb{R}^n$, with the usual Riemannian metric $g=\sum_{i=1}^ndx^i\otimes dx^i$, where $(x^1,\dots, x^n)$ is the usual coordinate system (i.e corresponding to the identity chart, or simply what we call cartesian coordinates). Then, we can consider any submanifold $S$ of $\Bbb{R}^n$, for example a sphere of certain radius, or for example in $\Bbb{R}^3$, we can consider a certain sheet of the hyperboloid $x^2+y^2-z^2=1$ etc etc. I your case, we can consider $M=\Omega$ to be an open set with nice boundary in $\Bbb{R}^m$, and $S=\partial\Omega$. Then, all the above machinery applies here.

If you know about differential forms, then you may know that on any oriented Riemannian manifold, one can define a volume form (i.e a top-degree, nowhere vanishing differential form) $\mu_g$, which has the property that on every positively oriented, $g$-orthonormal basis of each tangent space, $\mu_g$ evaluates to $1$. In this case, the integration of a function $f:M\to\Bbb{R}$ with respect to the measure $\lambda_g$ is the same thing as the integration of the top degree differential form $f\,\mu_g$ (so all the notions are consistent; though using $\lambda_g$ is more general since it doesn't require orientability).