It’s not too hard to construct examples showing that Hahn decompositions for a signed measure, though they exist, are not unique. But if we have two of them — — there’s something we can show to be unique. For every measurable set we have and .
Indeed, it’s easy to see that , so (since is positive) . But we can also see that , so (since is negative) . And so , and similarly . We then see that
The proof that is similar.
We can thus unambiguously define two set functions on the class of all measurable sets
for any Hahn decomposition . We call these the “upper variation” and “lower variation”, respectively, of . We can also define a set function
called the “total variation” of . It should be noted that and have almost nothing to do with each other.
Now, I say that each of these set functions — , , and — is a measure, and that . If is (totally) finite or -finite, then so are and , and at least one of them will always be finite.
Each of these variations is clearly non-negative, and countable additivity is also clear from the definitions. For example, given a pairwise-disjoint sequence we find
and similarly for and . Thus each one is a measure. The equation is clear from the definitions. The fact that takes at most one of implies that one of is finite. Finally, if every measurable set (say, ) is a countable union of -finite sets (say, ), then is the countable union of the , and so as well, and similarly for . Thus and are -finite.
We see that every signed measure can be written as the difference of two measures, one of which is finite. The representation of a signed measure as the difference between its upper and lower variations is called the “Jordan decomposition” of .