> [!tldr] > Given a set $\Omega$ and a collection of its subsets $\mathcal{A} \subseteq \mathcal{P}(\Omega), \emptyset \in \mathcal{A}$, a **set function** $\mu$ is a function $\mathcal{A} \to [0,\infty]$ such that $\mu(\emptyset)=0$. A set function is **additive** if for all finite collections of disjoint sets $(A_{i})_{i=1}^{n} \subseteq \mathcal{A}$, $\mu\left( \bigcup_{i=1}^{n}A_{i}\right)=\sum_{i=1}^{n}\mu(A_{i})$Furthermore, it is **countably additive** or **σ-additive** if the same holds for all countable collection of disjoint sets $(A_{i})$. For an additive and finite set function $\mu$ on an algebra $\mathcal{A}$, it is countably additive (hence a [[Measures|measure]]) $\iff$ for any $(A_{i}) \subseteq \mathcal{A}$ where $A_{i} \downarrow \emptyset$, there is $\mu(A_{i}) \to 0$. > [!proof] > $[\Rightarrow]$ Consider the "differences" $D_{i}:= A_{i}-A_{i+1}$. They are disjoint, so countable additivity and finiteness guarantees $\begin{align*} \mu(A_{i})&= \mu(\cup_{j \ge i}D_{j})\\[0.4em] &= \sum_{j \ge i} \mu(D_{j}) & [\text{countable additivity}]\\ &\to 0. &[\text{finiteness}] \end{align*}$ > $[\Leftarrow]$ take any disjoint $(B_{i}) \subseteq \mathcal{A}$, and denote $C:= \bigcup_{i}B_{i}$ and $A_{n}:= C-\bigcup_{i=1}^{n}B_{i}$. Then $(A_{n}) \downarrow \emptyset$, so $\mu(A_{n}) \to 0$. Then (finite) additivity gives $\mu(\bigcup_{i=1}^{n}B_{i}) =\mu(C)-\mu(A_{n})\to \mu(C)$.