# surface measure and Gauss-Green theorem proof

The famous Gauss-Green theorem states the follows. Let $$\Omega$$ be a bounded open set of $$\mathbb{R}^n$$and $$\partial \Omega$$ be $$C^1$$ boundary. Then for $$f,g \in C^1(\overline{\Omega}),$$ the following holds: $$\begin{eqnarray} \int\limits_{\Omega}f_{x_i}g dx =-\int\limits_{\Omega}fg_{x_i} dx + \int\limits_{\partial \Omega} fg dS \quad \text{for } i=1,2,\ldots,n. \end{eqnarray}$$ where $$dS$$ is the surface measure on $$\partial \Omega$$.

How to prove this result in a rigorous measure-theoretic setup? What is the underlying $$\sigma-$$algebra and measure on $$\partial \Omega$$?

P.S.: Detailed answer or a reference that contains all these details is highly appreciated.

• I wrote a brief sketch on this here (See after "EDIT"). But you are right that this is one of those things that is rarely mentioned explicitly. Commented Jul 27, 2021 at 11:12

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 $$A\in\mathcal{L}_M$$ with $$A\subset U$$, we have \begin{align} \lambda_g(A)&=\int_{\alpha[A]}\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 $$M$$).

To prove this, note that for each chart $$(U,\alpha)$$ belonging to the maximal atlas, we can define $$\lambda_{g,(U,\alpha)}(A):= \int_{\alpha[A]}\sqrt{|\det [g_{(\alpha)}]|}\,d\lambda_m$$. This defines a positive measure on $$\mathcal{L}_M|_{U}$$ (by monotone convergence) which is complete (since the standard Lebesgue measure $$\lambda_m$$ is, and since the integrand is strictly positive) and $$\sigma$$-finite (since the standard Lebesgue measure $$\lambda_m$$ is). Also, if we consider two different charts $$(U,\alpha)$$ and $$(V,\beta)$$, then in the overlap, these measures agree by a simple application of the change-of-variables theorem. Therefore, applying theorems 1 and 2 stated below gives us the desired $$\lambda_g$$.

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. In fact, we can generalize this and get a measure $$\lambda_{\rho}$$ from any non-negative, measurable, scalar density $$\rho$$. As a brief summary: a scalar density is essentially an object $$\rho$$ which to each chart $$(U,\alpha)$$ gives a function $$\rho_{\alpha}:\alpha[U]\to\Bbb{R}$$, such that for two different charts $$\rho_{\beta}$$ and $$\rho_{\alpha}$$ differ by a Jacobian determinant. This determinant is exactly the thing needed to cancel the Jacobian from the change-of-variables theorem, thereby giving us a whole bunch of compatible measures $$\lambda_{\rho,(U,\alpha)}$$ to which we can apply theorem 1 below (and by imposing more conditions on the scalar density, we can apply theorem 2).

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).

Measure-Theoretic Facts

Here are the necessary measure-theoretic facts used above.

Theorem 1 (Patching measures together)

Let $$(X,\mathfrak{M})$$ be a measurable space, let $$\{X_i\}_{i\in I}\subset\mathfrak{M}$$ be an arbitrary collection of measurable subsets which covers $$X$$, and suppose for each $$i\in I$$ we have a positive measure $$\mu_i$$ on $$X_i$$ (i.e defined on the $$\sigma$$-algebra $$\mathfrak{M}_i:=\{A\cap X_i\,:\, A\in\mathfrak{M}\}$$). If

• there is an at-most countable set of indices $$I_0\subset I$$ such that $$\{X_i\}_{i\in I_0}$$ covers $$X$$,

• for all $$i,j\in I$$, the measures $$\mu_i$$ and $$\mu_j$$ agree on the overlap $$\mathfrak{M}_i\cap\mathfrak{M}_j$$,

then there is a unique positive measure $$\mu$$ defined on $$(X,\mathfrak{M})$$ such that for each $$j\in I$$, we have that $$\mu$$ and $$\mu_j$$ agree on $$\mathfrak{M}_j$$.

To prove this we essentially follow our noses.

• Uniqueness: Suppose such a $$\mu$$ exists. Enumerate the index set $$I_0=\{1,2,\dots\}$$, and let us define $$E_1=X_1$$, and for each $$i\in I_0$$, recursively define $$E_{i+1}:=X_{i+1}\setminus(X_1\cup\cdots\cup X_i)$$. Then, $$\{E_i\}_{i\in I_0}$$ is an at-most countable family of measurable sets, which is pairwise disjoint and such that for each $$i\in I_0$$ we have $$E_i\subset X_i$$. So, for any $$A\in\mathfrak{M}$$, we have \begin{align} \mu(A)&=\sum_{i\in I_0}\mu(A\cap E_i)=\sum_{i\in I_0}\mu_i(A\cap E_i)\tag{*} \end{align} The first equal sign used the fact that $$A=\bigcup\limits_{i\in I_0}(A\cap E_i)$$ is a disjoint union of measurable sets and $$\mu$$ is a measure, and the second equal sign used the fact that $$\mu$$ agrees with each $$\mu_i$$. This shows $$\mu$$ is uniquely determined by $$\{\mu_i\}_{i\in I_0}$$ (and hence by $$\{\mu_i\}_{i\in I}$$).
• Existence: We simply flip the logic around by defining a set function $$\mu$$ for each $$A\in\mathfrak{M}$$ by the equality $$(*)$$. Then, each $$\mu_i$$ being a measure implies first of all that \begin{align} \mu(\emptyset):=\sum_{i\in I_0}\mu_i(\emptyset)=\sum_{i\in I_0}0=0. \end{align} Second, if $$\{A_n\}_{n=1}^{\infty}\subset\mathfrak{M}$$ is any collection of pairwise disjoint measurable sets, and $$A:=\bigcup\limits_{n=1}^{\infty}A_n$$ their union, then \begin{align} \mu(A)&:=\sum_{i\in I_0}\mu_i(A\cap E_i) =\sum_{i\in I_0}\sum_{n=1}^{\infty}\mu_i(A_n\cap E_i) =\sum_{n=1}^{\infty}\sum_{i\in I_0} \mu_i(A_n\cap E_i) =:\sum_{n=1}^{\infty}\mu(A_n). \end{align} The first equal sign is a definition, the second is because $$\mu_i$$ is a measure, the third we can swap the double series since these are countably many non-negative terms, and the last is definition again. Hence, the set-function $$\mu$$ is indeed a positive measure. Finally, we need to show compatibility, so let $$j\in I$$ be any index and let $$A\in \mathfrak{M}_j$$ be any measurable set. Then, \begin{align} \mu(A):=\sum_{i\in I_0}\mu_i(A\cap E_i) =\sum_{i\in I_0}\mu_j(A\cap E_i) =\mu_j(A), \end{align} where the second equal sign used that $$A\cap E_i\subset X_j\cap X_i$$ and hence $$\mu_i(A\cap E_i)=\mu_j(A\cap E_i)$$ due to compatibility of $$\mu_i$$ and $$\mu_j$$, and the third equal final equal sign used that $$\mu_j$$ is a measure.

This completes the proof of the theorem.

Now, we can ask what properties the measure $$\mu$$ inherits from the $$\{\mu_i\}_{i\in I}$$. Here’s an indication of results in this direction

Theorem 2 (Properties of patched measure)

Keep notation as in theorem 1.

• If $$\{\mu_i\}_{i\in I_0}$$ consists of complete measures, then $$\mu$$ is complete.

• If $$\{\mu_i\}_{i\in I_0}$$ consists of $$\sigma$$-finite measures, then $$\mu$$ is $$\sigma$$-finite.

• Proof of completeness: Suppose $$A\in\mathfrak{M}$$ and $$\mu(A)=0$$ and let $$Z\subset A$$ be any subset. Then, we have $$0=\mu(A)=\sum_{i\in I_0}\mu_i(A\cap E_i)$$, and since these are positive measures, we have each $$\mu_i(A\cap E_i)=0$$, and thus by completeness of $$\mu_i$$, we have $$Z\cap E_i\in \mathfrak{M}_i\subset\mathfrak{M}$$. So, $$Z=\bigcup\limits_{i\in I_0}Z\cap E_i$$ belongs to $$\mathfrak{M}$$, thereby proving completeness of $$\mu$$.
• Proof of $$\sigma$$-finiteness: Suppose each $$X_i$$ can be covered by countably many sets $$\{A_{i,j}\}_{j=1}^{\infty}$$, each having $$\mu_i$$-finite measure. But since $$\mu$$ agrees with $$\mu_i$$ on each of these sets, it follows these are also have $$\mu$$-finite measure. Thus, $$\{A_{i,j}\,:\, i\in I_0,\,j\in\Bbb{N}\}$$ is an atmost countable (since countable union of countable sets is countable) collection of sets which cover $$X$$, and each of which has finite $$\mu$$-measure. Thus, $$\mu$$ is $$\sigma$$-finite.

Note that we could investigate many other properties as well, such as how regularity (i.e approximability) properties of each $$\mu_i$$ implies that for $$\mu$$. Another thing to investigate would be what happens if we have another collection of measures $$\{\nu_i\}_{i\in I}$$ (same indexing set) which also satisfies theorem 1, and such that for each $$i\in I$$ (or simply $$i\in I_0$$) we have that $$\nu_i$$ and $$\mu_i$$ are $$\sigma$$-finite and $$\nu_i\ll \mu_i$$. Then, we could investigate whether or not the patched measures are also absolutely continuous $$\nu\ll\mu$$, and in this case we can ask how the Radon-Nikodym derivatives are related. Another thing to ask would be to consider a completely different family of $$\sigma$$-finite measures $$\{(Y_j,\nu_j)\}_{j\in J}$$, and ask how the product measure $$\mu\times \nu$$ is related to each of the $$\mu_i\times \nu_j$$ etc. The point is that once you have the ‘big theorems’ of Fubini and Radon-Nikodym, you can easily patch these results together and investigate what happens.

• I think there is a slight error in the proof of Theorem 1, in particular the compatibility part. It should read: $A \cap E_i \subset X_j \cap X_i$. Commented Apr 4 at 11:55
• @Algebro1000 yes thank you. When I originally wrote this down on paper, I used $E_i$ and $F_i$ but then I decided to change it to $X_i$ and $E_i$ in this answer. Obviously some things got overlooked in transcription. If you think there are more typos (I hope not), let me know. Commented Apr 4 at 12:00