Let be a totally finite measure algebra, and write for the maximal element. Without loss of generality, we can assume that is normalized so that .
We define a “partition” of an element to be a finite set of “disjoint” elements of whose “union” is . Remember, of course, that the elements of are not (necessarily) sets, so the set language is suggestive, but not necessarily literal. That is, if then and
The “norm” of a partition is the maximum of the numbers . If is a partition of and if is any element of below , then is a partition of .
If and are partitions, then we write if each element in is contained in an element of . We say that a sequence of partitions is “decreasing” if for each . A sequence of partitions is “dense” if for every and every positive number there is some and an element so that , and is exactly the union of some elements in . That is, we can use the elements in a fine enough partition in the sequence to approximate any element of as closely as we want.
Now, if is a totally finite, non-atomic measure algebra, and if is a dense, decreasing sequence of partitions of , then . Indeed, the sequence of norms is monotonic and bounded in the interval , and so it must have a limit. We will assume that this limit is some positive number , and find a contradiction.
So if then at least one of the must be big enough that for all . Otherwise the sequence of norms would descend below and that couldn’t be the limit. Let be just such an element, and consider the sequence of partitions of . The same argument is just as true, and we find another element from the partition , and so on.
Now, let be the intersection of the sequence . By assumption, each of the has , and so as well. Since is non-atomic, can’t be an atom, and so there must be an with . This element must be either contained in or disjoint from each element of each partition .
We can take smaller than either or . Now no set made up of the union of any elements of any partition can have a distance less than from . This shows that the sequence of partitions cannot be dense, which is the contradiction we were looking for. Thus the limit of the sequence of norms is zero.