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Let $\Omega$ be a bounded $C^1$ domain with bounded boundary $\partial\Omega$. Can someone point me to a reference where the surface integral of a measurable function $f\colon \partial\Omega \to \mathbb{R}$ is defined: $$\int_{\partial\Omega} fdS = ?$$ without the use of a transformation of coordinates in the sense that for Lipschitz domains $\Omega$, the surface integral involves a transformation of coordinates via a rotation and translation, which I believe is unnecessary for a $C^1$ domain. I want to avoid this transformation which causes problems for something I am doing.

I have not seen such a definition yet which does not involve a transformation.

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    $\begingroup$ What about integration with respect to the $(d-1)$-dimensional Hausdorff measure (restricted to $\partial \Omega$? $\endgroup$
    – Jochen Wengenroth
    Commented Jan 13, 2014 at 13:20
  • $\begingroup$ Yeah, but I wanted a way to compute it for an application. $\endgroup$
    – michael_carbon
    Commented Jan 13, 2014 at 18:09

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Take any piecewise continuous function $f$ defined on the boundary $\partial \Omega$, and extend $f$ to be constant along all lines perpendicular to the boundary, at least until those lines collide with one another. If $\partial \Omega$ is compact, this extends $f$ to some $\varepsilon$-neighborhood, as long as $\partial \Omega$ is $C^1$. Then use usual Lebesgue measure on that neighborhood to integrate the extended $f$. Then divide by $2 \varepsilon$ and take the limit as $\varepsilon \to 0$. It is easy to check that this recovers the usual Lebesgue integral on $\partial \Omega$, by taking charts. In particular, this definition extends uniquely to integrable functions $f$, essentially by Fubini's theorem. However, proving that there is a limit as $\varepsilon \to 0$ probably requires charts.

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  • $\begingroup$ Thanks for answering. "..recovers the usual Lebesgue integral on $\partial\Omega$, by taking charts." -- this usual integral by taking charts does not require an affine transformation of coordinates, right? The chart function is a diffeomorphism which suffices. $\endgroup$
    – michael_carbon
    Commented Jan 12, 2014 at 11:47
  • $\begingroup$ The chart function is a diffeomorphism, which suffices. You can take any oriented orthonormal basis of the tangent space of $\partial \Omega$ at a point, and wedge it together to give a volume form. That is the volume form whose associated measure we want to integrate, and that is the volume form whose integrals agree with the integrals I have described by extending functions to be constant in perpendicular directions. $\endgroup$
    – Ben McKay
    Commented Jan 13, 2014 at 13:13
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A general way to define this is to use rectifiable sets. For example, if $\Omega$ is a set of finite perimeter, measurable functions and their integrations are well defined. Moreover, the divergence theorem holds in this setting. You can check the book of Giusti on BV functions or Washek F. Pfeffer's "The Divergence Theorem and Sets of Finite Perimeter".

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