Exercise 10. Let X be a set, and let M be a σ-algebra on X. Show that M is stable under set differences.
Proof. Given two sets A and B in M, we want to show that their set difference A B also belongs to M. Start with the assumption that A and B are in M, meaning they are both in the σ-algebra. By the second property of a σ-algebra, if A is in M, then X A is also in M. Since X A is a set difference, it follows that X A is in M. Now, let's apply property 3 of the σ-algebra. Consider the countable sequence A_n = X (B ∩ A), where A_n represents the set difference between X and the intersection of B and A. Since X (B ∩ A) is in M (as shown in step 2), A_n is a sequence of sets in M. According to property 3, the union of the sets A_n is also in M. Therefore, we have:
So each of these A_n's are exactly the same? This proof is much simpler than all of this. Just rewrite A/B as A ∩ B^c, then go from there.
(AB) = A (XB) = A (X(B∩A)) = A (A∪A∪A∪..)
Since A B is the intersection of sets in M, it also belongs to M. So, we have shown that M is stable under set differences, as any two sets A and B in M imply that their set difference A B is also in M.