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A version of Gauss's theorem says that

If $U$ is a bounded, open subset of $\mathbb{R}^n$ with $C^1$ smooth $\partial U$, then for $u \in C^1 (\bar U)$ $$ \int_U u_{x_i} \, dx = \int_{\partial U} u v^i \, dS $$ where $v^i$ denotes the $i$th component of outward normal to $\partial U$.

(1) Are these integrals Lebesgue integrals, or are they Riemann integrals?

(2) If the integrals can be interpreted in both Lebesgue and Riemann sense, are they the same?

(3) In Lebesgue sense, $dx$ is the Lebesgue meausure in $\mathbb{R}^n$ but which Lebesgue measure is used for the "surface element $dS$?

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Note that the conditions require that $u$ is continuous and $\partial U$ is $C^1$ smooth (i.e., it can be represented locally by continuous maps). Under those conditions, Riemann integration is well-defined. And anywhere Riemann integration is well-defined, so is Lebesgue integration and the two are equal in value.

So if you want to consider it Lebesgue integration, you are free to do so. If you want to consider it Riemann integration, you are also free to do so. If you prefer a different integration, it will almost certainly work as well, as any worthwhile itegration theory is going to be able to handle continuous functions.

The surface element $dS$ is the area measure/differential on the surface.

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  • $\begingroup$ Is the area measure on the surface same as the Lebesgue measure on $\mathbb{R}^{n-1}$? $\endgroup$ – nan May 12 '18 at 18:08
  • $\begingroup$ No, because the surface is not $\Bbb R^{n-1}$. There are various ways to define it. One way is even to use Riemann integration of the constant function $1$ on nice subsets of the surface to define a set function, which is then extended to the sigma algebra those subsets generate. $\endgroup$ – Paul Sinclair May 12 '18 at 21:24
  • $\begingroup$ Why is the surface not in $\mathbb{R}^{n-1}$? The boundary of a volume $\mathbb{R}^3$ is a surface in $\mathbb{R}^2$ $\endgroup$ – nan May 12 '18 at 21:36
  • $\begingroup$ No it isn't. The sphere is the boundary of a ball in $\Bbb R^3$, but it is most certainly not a subset of the plane $\Bbb R^2$. There is a big difference between being 2-dimensional and being in $\Bbb R^2$. $\endgroup$ – Paul Sinclair May 13 '18 at 0:18
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The question(s) has been already answered by Paul Sinclair, but I have nevertheless decided to post my answer(s), to which I started to work a few days ago, since I think it could be interesting as a few more details are given.

For the first question, recall Lebesgue's criterion for Riemann integrability ([1], §1.7 p.20): if $f:[a,b]\to \mathbb{R}$ a real function of one variable, continuous outside a set $D\subset[a,b]$ $$ f\text{ is Riemann integrable }\iff \mu(D)=0 $$

where $\mu$ is the Lebesgue measure in $\mathbb{R}$. The result is a consequence of the general properties of Riemann's integral, also true for functions $u:U\to \mathbb{R}$ with $U\subset\mathbb{R}^n$: in the OP it is stated that $\partial U\in C^1$ and $u\in C^1(\bar{U})$, thus $u$ and its gradient $\nabla u$ are at least continuous on the whole $\bar{U}$, therefore under such kind of hypotheses, the integrals in Gauss's theorem can be understood as Riemann's integrals. However, Riemann's integral can also be iterpreted as a particular case of Lebesgue's integral, where the simple functions necessary to define it are chosen to be the characteristic functions of finer and finer partitions in cubes of the domain $U$ (see [2], pp. 29-33 for all the details for $n=1$). Therefore, when Riemann's integral is defined, also Lebesgue's integral is and the two integrals have the same value (this is also true for every generalization of the concept of integral, as shown extensively and very clearly in reference [2]): this answers the OP second question.

Concerning the third question, $ds$ is the hypersurface measure/differential form defined on $\partial U$ as clearly stated by Paul Sinclair. It is important to emphasize this fact since Gauss theorem for Lebesgue integrals holds also for quite bad sets: for example it holds for Caccioppoli sets (i.e. sets of "finite perimeter"), for whic it takes the form ([1],ch. 3, §3.3, pp. 143-144, theorem 3.36) $$ \int\limits_U\!\nabla\cdot u\, \mathrm{d}x =\int\limits_{\mathbb{R}^n} u\cdot\nu_U\, \mathrm{d}|D\chi_U| $$ where $|D\chi_U|$ is in general a Radon measure supported by $\partial{U}$, that can be quite different from a standard Lebesgue measure.

[1] Shilov, G. E. and Gurevich, B. L. (1977), Integral, Measure, and Derivative: A Unified Approach, revised edition, Dover books on advanced mathematics, New York: Dover Publications, pp. xiv+233, ISBN 0-486-63519-8, MR 0466463, Zbl 0391.28007.

[2] Gordon, R. A. (1994), The integrals of Lebesgue, Denjoy, Perron, and Henstock, Graduate Studies in Mathematics, 4, Providence, RI: American Mathematical Society, pp. xi+395, ISBN 0-8218-3805-9, MR1288751, Zbl 0807.26004.

[3] Ambrosio, Luigi; Fusco, Nicola; Pallara, Diego (2000). Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434, ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001.

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