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Let us define a measure defined on the sphere $\mathbb{S}^d$ as follows:

$\nu (B) := \int_B f d \sigma$ where $f$ is an arbitrary fixed continuous function and $\sigma$ is the surface measure

Is there a obvious interpretation of this measure?

One further question: when I am asking whether this defined measure is a Borel or resp. a Radon measure, does this statement only depend on $\sigma$?

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  • $\begingroup$ No, there is no canonical thing for any given function in an aesthetically pleasing way $\endgroup$ – Adam Hughes Dec 6 '16 at 15:54
  • $\begingroup$ when I am asking whether this defined measure is a Borel or resp. a Radon measure, does this statement only depend on $\sigma$? $\endgroup$ – tubmaster Dec 6 '16 at 18:28
  • $\begingroup$ You should put that in your original question. $\endgroup$ – Adam Hughes Dec 6 '16 at 18:41
  • $\begingroup$ ok sorry, done. $\endgroup$ – tubmaster Dec 6 '16 at 18:48
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If I am reading your question correctly, it would depend on $f$ and $\sigma$ to satisfy the properties of a Borel or Radon measure since you are basically saying that $f$ is the Radon-Nikodym derivative of $\nu$ w.r.t. $\sigma$. For example, let's take the Lebesgue measure on $\mathbb{R}$ (for simplicity) to be $\lambda$ and then take $\nu$ to be a measure defined by $$ \nu(A)=\int_{A} f \, d\lambda $$ and $f$ has compact support then $\nu$ is a finite measure while $\lambda$ is not. So yes, the interpretation of your measure $\nu$ is that it is absolutely continuous to the surface measure $\sigma$ with $f$ being the Radon-Nikodym derivative. You simply need to check for what conditions on $f$ you need for $\nu$ to satisfy the properties of a Borel or Radon measure.

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  • $\begingroup$ thanks! But the only trivial condition I see is that $f$ must be function definied on $S^d$... would that be enough? $\endgroup$ – tubmaster Dec 7 '16 at 10:40
  • $\begingroup$ Yes, since you previously stated that $f$ is continuous and since $S^{d}$ is compact then this is enough to guarantee that $\nu$ is a Radon measure. However, I'd encourage you to go through the details of verifying local finiteness (trivial since $f$ is continuous) and inner regularity (i.e. $\nu(A)=\{ \sup \nu(K) : K \subseteq A, \text{K is compact}\}$) so that you can convince yourself. $\endgroup$ – TheLaughingAlgebrist Dec 7 '16 at 13:42

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