## Properties of Metric Spaces of Measure Rings

Today we collect a few facts about the metric space associated to a measure ring .

First of all, the metric on is translation-invariant. That is, if , , and are sets in with finite measure, then . Indeed, we calculate

Next we need to define an “atom” of a measure ring to be a minimal nonzero element. That is, it’s an element so that if then either or . We also define a measure ring to be “non-atomic” if it has no atoms. As a quick result: a totally -finite measure ring can have at most countably many atoms, since must contain all of them, and no two of them can have a nontrivial intersection — if there were uncountably many of them, any decomposition of could only cover countably many of them.

On the other hand, we define a metric space to be “convex” if for any two distinct elements and there is an element between them. That is, is neither nor , and it satisfies the equation . We assert that the metric space of a -finite measure ring is convex if and only if the measure ring is non-atomic.

Let and be elements of . Without loss of generality we can assume that by using the translation-invariance of . Indeed, we can replace with and with . There will be an element between and if and only if is between and .

So for any is there an element between and ? Such a will satisfy

This is only possible if , which means that , and so . But for to be between and it cannot be equal to either of them, which means that cannot be an atom for any with . Since we can decompose any into a countable union of elements of finite measure, no element of infinite measure can be an atom either.

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Sir,

What else can be said of the metric space (X,d_m) associated to a measure ring (S,m) if we assume more structure for S?

For example I was wondering about this: if S is the Borel sigma algebra on a compact group G with Haar measure, must the associated metric space be compact?

Thank you

Comment by Saul | January 23, 2011 |

I’m not really sure offhand what can be said. Though it’s certainly an interesting question worth investigating.

Comment by John Armstrong | January 23, 2011 |