THE MONOTONE CLASS THEOREM JACOPO DE SIMOI Abstract. The proof of the Monotone Class Theorem is given in the bookby Lieb–Loss but differs from the one I gave in class. Please find below the version that was given in class on Sept 13. Recall the relevant definitions; here 푋 is a non-empty set. • an algebra of sets on 푋 is a family of sets 풜 ⊂ 풫(푋) that is closed under complement and finite unions • an algebra is a σ-algebra if it is also closed under countable unions. • a monotone class is a family of sets 풞 ⊂ 풫(푋) with the property that the (countable) union of any increasing sequence of sets in 풞 is also in 풞 and the (countable) intersection of any decreasing sequence of sets in 풞 is also in 풞. Observe that any σ-algebra is a monotone class. Lemma 0.1. If a family ℱ is an algebra and a monotone class then it is indeed a σ-algebra. Proof. We need to show that countable union of any sequence of sets in ℱ belongs to ℱ. Let {퐸푘}푘∈N ⊂ ℱ be a sequence of sets; then construct the sequence 푘 퐹푘 = ⋃ 퐸푘 푖=1 Since ℱ is an algebra, 퐹푘 ∈ ℱ for any 푘 ∈ N; moreover the sets 퐹푘 form an increasing sequence, therefore ℱ ∋ ⋃ 퐹푘 = ⋃ 퐸푘. 푘∈N 푘∈N We conclude that ℱ is a σ-algebra. Given any family ℱ ⊂ 풫(푋) we can define the following objects • the σ-algebra generated by ℱ, denoted by ℳ(ℱ) is the smallest σ-algebra of sets of 푋 that contains ℱ; • the monotone class generated by ℱ, denoted by 풞(ℱ) is the smalles t monotone class that contains ℱ. Notice that if ℱ ⊂ ℳ(ℱ′) (resp. ℱ ⊂ 풞(ℱ′)) we have ℳ(ℱ) ⊂ ℳ(ℱ′). Moreover, since any σ-algebra is a monotone class we immediately conclude that for any family ℱ, 풞(ℱ) ⊂ ℳ(ℱ). The Monotone Class Theorem guarantees that if ℱ is an algebra, then this becomes an equality. 1 2 JACOPO DE SIMOI Theorem 0.2. Let 풜 be an algebra of sets on 푋; then 풞(풜) = ℳ(풜). Proof. To ease notation, since 풜 is fixed let us write 풞 = 풞(풜) and ℳ = ℳ(풜) without possibility of confusion. First of all observe that it suffices to show that 풞 is a σ-algebra; then since 풞 ⊃ 풜 and by the minimality of ℳ we can conclude that ℳ ⊂ 풞; this, together with the trivial inclusion mentioned above concludes the proof. Observe moreover that by the above lemma it suffices to show that 풞 is an algebra; in order to do so we need to show that the class is invariant under finite unions and complements. In order to show the closure by the complement operation, let us define 풞∗ = {퐸 ∈ 풞|푋 ∖ 퐸 ∈ 풞} Observe that by design 풞∗ ⊂ 풞; moreover since 풜 is an algebra we have that each element of 풜 will be in 풞∗, therefore we have 풜 ⊂ 풞∗. If we now show that 풞∗ is a monotone class, we can conclude –by the minimality of 풞– that 풞 ⊂ 풞∗, which would then imply that 풞∗ is the entire 풞, i.e. 풞 is closed by the complement operation. The proof that 풞∗ is a monotone class is a routine check and follows from the definition [it is left to the reader] Let us now show that the set 풞 is closed under finite unions. For any element 퐸 ∈ 풞 we define 풞퐸 = {퐹 ∈ 풞|퐸 ∪ 퐹 ∈ 풞} ⊂ 풞} Observe that, by the symmetry of the definition we have 퐹 ∈ 풞퐸 if and only if 퐸 ∈ 풞퐹. Also, a routine check shows that 풞퐸 is a monotone class for any choice of 퐸 [do it!]. If we manage to show that 풜 ⊂ 풞퐸 for any 퐸 ∈ 풞, we will conclude that 풞 ⊂ 풞퐸, and thus 풞 = 풞퐸 for any 퐸 which shows closure of 풞 under finite unions. In order to prove that 풜 ⊂ 풞퐸 for any 퐸 ∈ 풞, let us now choose 퐴 ∈ 풜; observe that for any 퐵 ∈ 풜 we have 퐴 ∪ 퐵 ∈ 풜 (by definition of algebra), thus 퐴 ∪ 퐵 ∈ 풞, which implies that 퐵 ∈ 풞퐴. Since 퐵 was arbitrary we conclude that 풜 ⊂ 풞퐴 for any 퐴 ∈ 풜. Since 풞퐴 is a monotone class, by minimality of 풞 among all minimal classes containing 풜, we conclude that 풞퐴 = 풞. Otherwise said: for any 퐸 ∈ 풞 and 퐴 ∈ 풜 we have 퐸 ∈ 풞퐴. But by symmetry this implies that 퐴 ∈ 풞퐸 for any 퐸 ∈ 풞, or 풜 ⊂ 풞, that was what we wanted to show. .