Let be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that be positive — that only if — we don’t need to worry about negligible elements.
And so we write for the metric space consisting of the elements with . This has a distance function defined by . We also write for the metric associated with the measure algebra associated with the measure space . We say that a measure space or a measure algebra is “separable” if the associated metric space is separable.
Now, if we set
then , , and itself are all uniformly continuous.
Indeed, if we take two pairs of sets , , , and , we calculate
Similarly, we find that . And thus
And so if we have control over the distance between and , and the distance between and , then we have control over the distance between and . The bounds we need on the inputs uniform, and so is uniformly continuous. The proof for proceeds similarly.
To see that is uniformly continuous, we calculate
Now if is a -finite measure space so that the -ring has a countable set of generators, then is separable. Indeed, if is a countable sequence of sets that generate , then we may assume (by -finiteness) that for all . The ring generated by the is itself countable, and so we may assume that is itself a ring. But then we know that for every and for every positive we can find some ring element so that . Thus is a countable dense set in , which is thus separable.