Question 10.T.9: Let A be a σ-algebra on Ω, μ a measure on A, and (Ei) a sequ...
Let \mathcal{A} be a σ-algebra on Ω, μ a measure on \mathcal{A}, and (E_{i}) a sequence in \mathcal{A}.
(i) If (E_{i}) is increasing, then μ(\lim E_{i}) = \lim μ(E_{i}), and μ is said to be continuous from below on \mathcal{A}.
(ii) If (E_{i}) is decreasing and there is an N such that μ(E_{N} ) < ∞, then μ(\lim E_{i}) = \lim μ(E_{i}), and μ is said to be continuous from above on \mathcal{A}.
(i) Let E = \lim E_{i}. From Example 10.5 we know that E = \cup_{i=1}^{∞} E_{i}. As we did in the proof of Theorem 10.8, we first express E as a disjoint union by defining
F_{1} = E_{1}, F_{i} = E_{i} \setminus \overset{i−1}{\underset{j=1}{\cup}} E_{j}, i ≥ 2.
F_{i} ∈ \mathcal{A} for all i ∈ \mathbb{N} and the unions
E_{i} = \overset{i}{\underset{j=1}{\cup}} F_{j}, E = \overset{∞}{\underset{j=1}{\cup}} F_{j}
are disjoint. By countable additivity,
μ(E) = \sum\limits_{i=1}^{∞}{μ(F_{i})}
= \underset{n→∞}{\lim} \sum\limits_{i=1}^{n}{μ(F_{i})} = \underset{n→∞}{\lim} μ(E_{n}).
(ii) In this case, set E = \lim E_{i} = \cap_{i=1}^{∞} E_{i}. Since (E_{i}) is decreasing, G_{i} = E_{N}\setminus E_{i} is an increasing sequence in \mathcal{A} with limit E_{N}\setminus E. From part (i) we therefore have
μ(E_{N}\setminus E) = \lim μ(E_{N}\setminus E_{i}).
The monotonicity of μ implies that μ(E) < ∞ and that μ(E_{i}) < ∞ for all i ≥ N. Now the subtractive property gives
μ(E_{N} ) − μ(E) = \lim[μ(E_{N} ) − μ(E_{i})] = μ(E_{N} ) − \lim μ(E_{i})
and the desired result.