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.