Can divergence be thought of as a Radon-Nikodym derivative?

by Jack M   Last Updated May 16, 2018 14:20 PM - source

The divergence theorem states roughly

$$\int_\Omega \operatorname{div}U\ dV=\int_{\partial \Omega}U\cdot n\ dS$$

where $U$ is a vector field, $\Omega$ is a region of space with a smooth boundary, $n$ is a smooth normal vector field on its boundary.

This looks alot like a Radon-Nikodym derivative: a certain function on sets can be obtained by integrating something on that set. In this interpretation, we would think of the right hand side of the equation, the flux of $U$ out of $\Omega$, as a (signed) measure. It seems like it should be additive like a measure should be, and perhaps it can even be extended to the Borel sigma algebra via something like Carathéodory's extension theorem.

The other question is if this "flux measure" is absolutely continuous with respect to the Lebesgue measure. This is plausible since the flux out of a flat surface should be zero: remember that we're thinking of closed surfaces here, so we should be integrating on "both sides" of the flat surface, which will cancel out.

Can this interpretation in any way be made rigorous?

Related Questions

Lebesgue-Radon-Nikodym representation

Updated May 08, 2018 16:20 PM

Radon-Nikodym property

Updated June 19, 2018 13:20 PM

Calculating Explicit Radon-Nikodym Derivatives

Updated June 28, 2018 23:20 PM