Robert Bartoszynski - Probability and Statistical Inference

It is important to realize that closure under countable operations is stronger than closure under any finite operations. This means that there exist classes of sets that are fields but not ‐fields. This is illustrated by the following example:

Let and let be the class of all subsets of such that either or is finite. Then is a field but not a ‐field. First, if then because the definition of is symmetric with respect to complementation. Next, if and are both in , so is their union. If and are both finite, then is finite and hence belongs to . On the other hand, if either or (or both) are finite, then is also finite because it is contained in and also in .

Thus, is a field. However, is not a ‐field. Let be the set consisting only of the element (i.e., ). Clearly, . Take now . This is a countable union of sets in that is not in since the set of all even numbers is not finite, nor does it have a finite complement.

Typically, it is easy to determine that a class of sets is a field, while direct verification that it is a ‐field can be difficult. On the other hand, it is sometimes easy to verify that a class of sets is a monotone class.

Theorem 1.4.2 A ‐field is a monotone class. Conversely, a field that is a monotone class is a ‐field.

Proof : To prove this theorem, assume first that is a ‐field, and let be a monotone sequence of elements of . If then , whereas if then So is a monotone class. On the other hand, let be a monotone class and a field, and let be an arbitrary sequence of elements of . Put . Then since is a field, and also for every . Further, since is a monotone class, . However, , so is a ‐field, as asserted.