Misc. Analysis Results

Proof of Claim 1.1: Of course, we assume that {nj}j=1 are all distinct. Let n1 = 1. Then there must exist An2 such that An2 ∩An1 6= ∅. If not, then An1 ∩Aj = ∅ for all j 6= n1. Thus, An1 ⊆ Aj for all j 6= n1. Now, fix some j0 6= n1 such thatAj0 6= An1 . Then, sinceF is a σ-algebra, we have thatAk0 = A c j0 for some k0 6= n1. Thus, setting j = k0 6= n1, An1 ⊆ Ack0 = ( Acj0 )c = Aj0 . The only way An1 ∩Aj0 = ∅ and An1 ⊆ Aj0 is if An1 = ∅, which is a contradiction. Thus, there exists n2 such that An1 ∩An2 6= ∅.