Measure, Integration & Real Analysis, 2021a

122 Chapter 5 Product Measures
Now we define a monotone class as a collection of sets that is closed under
countable increasing unions and under countable decreasing intersections.
5.15 Definition monotone class
Suppose W is a set and M is a set of subsets of W. Then M is called a monotone
class on W if the following two conditions are satisfied:
• If E 1 ⊂ E 2 ⊂···is an increasing sequence of sets in M, then
• If E 1 ⊃ E 2 ⊃··· is a decreasing sequence of sets in M, then
∞⋃
k=1
∞⋂
k=1
E k ∈M;
E k ∈M.
Clearly every σ-algebra is a monotone class. However, some monotone classes
are not closed under even finite unions, as shown by the next example.
5.16 Example a monotone class that is not an algebra
Suppose A is the collection of all intervals of R. Then A is closed under countable
increasing unions and countable decreasing intersections. Thus A is a monotone
class on R. However, A is not closed under finite unions, and A is not closed under
complementation. Thus A is neither an algebra nor a σ-algebra on R.
If A is a collection of subsets of some set W, then the intersection of all monotone
classes on W that contain A is a monotone class that contains A. Thus this
intersection is the smallest monotone class on W that contains A.
The next result provides a useful tool when the standard technique for showing
that every set in a σ-algebra has a certain property does not work.
5.17 Monotone Class Theorem
Suppose A is an algebra on a set W. Then the smallest σ-algebra containing A is
the smallest monotone class containing A.
Proof Let M denote the smallest monotone class containing A. Because every σ-
algebra is a monotone class, M is contained in the smallest σ-algebra containing A.
To prove the inclusion in the other direction, first suppose A ∈A. Let
E = {E ∈M: A ∪ E ∈M}.
Then A⊂E(because the union of two sets in A is in A). A moment’s thought
shows that E is a monotone class. Thus the smallest monotone class that contains A
is contained in E, meaning that M⊂E. Hence we have proved that A ∪ E ∈M
for every E ∈M.
Now let
D = {D ∈M: D ∪ E ∈Mfor all E ∈M}.
Measure, Integration & Real Analysis, by Sheldon Axler