Mark S. answered • 09/16/24
Middle School to College Level Mathematics Tutor
This is over a year late, but I like this sort of math, so I'll give it a go.
(i) Suppose that E is a countable set in Θ, then either E is countable or E^c is countable. It thus follows that E^c is also in Θ as E^c is countable or (E^c)^c = E is countable. Now consider a sequence {E_n} ⊂ Θ, either all E_n are countable, so that their countable union is countable and thus in Θ, or there is at least one E_j that is uncountable so that E_j^c is countable, then (∪E_n)^c = ∩(E_n^c) ⊂ E_j^c so that (∪E_n)^c is countable making ∪E_n an element of Θ.
(ii) I'm unclear on what this question is asking.
(iii) It is clear that μ(∅) = 0. Let E_n be a sequence of disjoint sets in Θ, then either all E_n are countable, in which case μ(∪E_n) = 0 = ∑μ(E_n), or some E_j has E_j^c countable. Since all E_n are disjoint, we have E_n ⊂ E_j^c for n ≠ j, so that all E_n (n ≠ j) are countable as subsets of a countable set. Then μ(∪E_n) = 1 = ∑μ(E_n).
