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Has there been study of a generalization of measure spaces along the following or similar lines ?

Given a measure space $(X, \Sigma, \mu)$, define for $U\in\Sigma$ a $\mu$-ball of radius $r$ of $U$ by

$ B(U;r) =$ { $ V\in\Sigma : \mu(U \cup V) < r $ }

So a $\mu$-ball is a set of sets. Then finding suitable properties of the collection of all $\mu$-balls on $(X, \Sigma, \mu)$ to use as axioms for a measure-space analogy of topological spaces, where such spaces that didn't arise from a measure-space would be called non-measurizable.

The above definition of a $\mu$-ball is just a simple analogy from an open ball in a metric space, however perhaps a totally different way of generalizing a measure space would more naturally relate to measure spaces as topological spaces relate to metric spaces.

If connectedness in topological spaces is similar to connectedness in graphs then perhaps connectedness in measure-space analogy of topological spaces is similar to connectedness in hypergraphs.

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Forgive me for being blunt, but why would you want to do this? –  Qiaochu Yuan Jun 29 '12 at 22:24
    
Note that your definition of $\mu$-balls is problematic, since it implies that for all $U$ of sufficiently small measure, the empty set belongs to the $B(U,r)$. This suggests to me that you really need to think more carefully about what you are trying to generalize. –  Yemon Choi Jun 29 '12 at 22:57
    
A better measure of "distance" between elements of the sigma-algebra might be to take the measure of the symmetric difference of sets. I have a vague recollection that this idea is pursued further in volume 1 of Dunford-Schwartz, but unfortunately cannot remember the details right now. –  Yemon Choi Jun 29 '12 at 23:24
    
Probably not what you want, but $d(U,V):=\mu(U\Delta V)=\|\chi_U-\chi_V\|_1$ makes the family of all "finite measure sets up to 0-measure perturbations" a complete metric space. –  Pietro Majer Jun 29 '12 at 23:25
    
What I see here is a hierarchy: metric -> topological -> measure space. You could even go to the left by riemannian manifold or so. –  Fernando Muro Jun 30 '12 at 0:12

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