Probability Foundations – Problem (38/365)

Let \mu be a finitely additive measure on an algebra \mathcal{A}, and let A_1, A_2, \dots \in \mathcal{A} be pairwise disjoint and satisfy A = \cup_{i=1}^\infty A_i \in \mathcal{A}. Then show that \mu(A) \ge \sum_{i=1}^\infty \mu(A_i).

\displaystyle  \text{Since } A \in \mathcal{A}, A - A_1 \in \mathcal{A} \text{, and } A - \cup_{i=1}^k A_i \in \mathcal{A} \\ \text{Let } B_i = \cup_{k=i}^\infty A_k \\ \sum_{i=1}^\infty \mu(A_i) \\ = \sum_{i=1}^\infty \mu(B_i - B_{i+1}) \\ = \sum_{i=1}^\infty \mu(B_i) - \mu(B_{i+1}) \\ = \mu(B_1) - \mu(B_2) + \mu(B_2) - \mu(B_3) + \mu(B_4) - \mu(B_5) + \dots \\ = \mu(B_1) - \lim_n B_n \\ = \mu(A) - \lim_n B_n \\ \le \mu(A)

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