## 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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