Given a signed measure on a measurable space , we can use it to break up the space into two pieces. One of them will contribute positive measure, while the other will contribute negative measure. First: some preliminary definitions.
We call a set “positive” (with respect to ) if for every measurable the intersection is measurable, and . That is, it’s not just that has positive measure, but every measurable part of has positive measure. Similarly, we say that is “negative” if for every measurable the intersection is measurable, and . For example, the empty set is both positive and negative. It should be clear from these definitions that the difference of two negative sets is negative, and any disjoint countable union of negative sets is negative, and (thus) any countable union at all of negative sets is negative.
Now, for every signed measure there is a “Hahn decomposition” of . That is, there are two disjoint sets and , with positive and negative with respect to , and whose union is all of . We’ll assume that , but if takes the value (and not ) the modifications aren’t difficult.
We write , taking the infimum over all measurable negative sets . We must be able to find a sequence of measurable negative sets so that the limit of the is — just pick so that — and we can pick the sequence to be monotonic, with . If we define as the union — the limit — of this sequence, then we must have . The measurable negative set has minimal measure .
Now we pick , and we must show that is positive. If it wasn’t, there would be a measurable subset with . This cannot itself be negative, or else would be negative and we’d have , contradicting the minimality of .
So must contain some subsets of positive measure. We let be the smallest positive integer so that contains a subset with . Then observe that
So everything we just said about holds as well for . We let be the smallest positive integer so that contains a subset with . And so on we go until in the limit we’re left with
after taking out all the sets .
Since , the measure of is finite, and so the measure of any subset of must be finite as well. Thus the limits of the must be zero, so that the measure of the countable disjoint union of all the can converge. And so any remaining measurable set that can fit into must have . That is, must be a measurable negative set disjoint from . But we must have
which contradicts the minimality of just like would have if it had been a negative set. And thus the assumption that is untenable, and so every measurable subset of has positive measure.