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Let $ B $ be the set of all measures $ \phi $ of $ \mathbf{R}^{n} $ such that every open set is $ \phi $-measurable (sometimes these measures are called Borel measures). Note the measures in $ B $ are not required to be locally finite.

Let $ 1 \leq m \leq n $ be an integer. The upper density of a measure $ \phi $ at $ a \in \mathbf{R}^{n} $ is defined as

$ \Theta^{*m}(\phi,a) = \limsup_{r \to 0}\frac{ \phi(B(a,r))}{r^{m}}. $

My question is: is there a standard way to topologize $ B $ in order to have that $ \Theta^{*m} :B \times \mathbf{R}^{n} \rightarrow \mathbf{R} \cup \{\infty\} $ is a Borel map? Of course I am not looking for trivial topologies, but something that has reasonable properties (such as not too many open sets etc...).

In case I am looking for references about this problem.

Thank you for your answers.

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  • $\begingroup$ At a guess, I think the usual weak topology is likely to work. The map $(\phi, a, r) \mapsto \phi(B(a,r))$ seems likely to be Borel, and then I think you can take the limsup through the rationals. $\endgroup$ Jun 8, 2016 at 1:18
  • $\begingroup$ Yes, that'll work for sure, in fact $(a,\mu)\mapsto \mu(B(a))$ is semicontinuous with respect to the weak $*$ topology. $\endgroup$ Jun 8, 2016 at 2:01
  • $\begingroup$ Thank you for your comments! if some of you wants to write it as an answer I ll be glad to accept it! $\endgroup$ Jun 8, 2016 at 9:08
  • $\begingroup$ If you want to write out the details as your own answer, I'll be glad to upvote it... ;-) $\endgroup$ Jun 8, 2016 at 22:17

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