Assume that this function satisfies the stronger sigma additivity assumption
for any disjoint family of elements of such that . (Functions obeying these two properties are known as pre-measures.) Then, extends to a measure defined on the sigma-algebra generated by ; i.e., there exists a measure
If is not -finite then the extension need not be unique, even if the extension itself is -finite.
Here is an example:
We call rational closed-open interval, any subset of of the form , where .
Let be and let be the algebra of all finite union of rational closed-open intervals contained in . It is easy to prove that is, in fact, an algebra. It is also easy to see that the cardinal of every non-empty set in is .
Let be the counting set function () defined in . It is clear that is finitely additive and -additive in . Since every non-empty set in is infinite, we have, for every non-empty set ,
Now, let be the -algebra generated by . It is easy to see that is the Borel -algebra of subsets of , and both and are measures defined on and both are extensions of .
This theorem is remarkable for it allows one to construct a measure by first defining it on a small algebra of sets, where its sigma additivity could be easy to verify, and then this theorem guarantees its extension to a sigma-algebra. The proof of this theorem is not trivial, since it requires extending from an algebra of sets to a potentially much bigger sigma-algebra, guaranteeing that the extension is unique (if is -finite), and moreover that it does not fail to satisfy the sigma-additivity of the original function.