Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞]

In the setting of Measure Spaces, Borel Sets and Lebesgue Measure on the Real Line: Standing Notation, sums over a finite set are as in Sums and Products over a Finite Set and over an Interval (i), Sums and Products over a Finite Set and over an Interval (ii), Sums and Products over a Finite Set and over an Interval (iii) and Sums and Products over a Finite Set and over an Interval (v). Sums over a finite set of maps into $[0,\infty]$ are formed with the sum $+$ of $[0,\infty]$, which is associative and commutative with neutral element $0$ by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i).

(i) Let $M$ be a set with an associative and commutative binary operation $+$ that has a neutral element $0$. Let $A$ be a finite set, $E\subseteq A$, and $f:A\to M$ a map with $f(a)=0$ for every $a\in A\setminus E$. Then $\sum_{a\in A}f(a)=\sum_{a\in E}f(a)$.

(ii) Let $M$, $+$ and $0$ be as in (i), let $A$ be a finite set and $f:A\to M$ a map with $f(a)=0$ for every $a\in A$. Then $\sum_{a\in A}f(a)=0$.

(iii) Let $M$, $+$ and $0$ be as in (i), let $A$ be a finite set, $a_{0}\in A$, and $f:A\to M$ a map with $f(a)=0$ for every $a\in A$ with $a\neq a_{0}$. Then $\sum_{a\in A}f(a)=f(a_{0})$.

(iv) Let $(X,\mathcal{F},\mu)$ be a measure space, let $J$ be a finite set, and let $j\mapsto B_{j}$ be a map from $J$ to $\mathcal{F}$ with $B_{i}\cap B_{j}=\emptyset$ for all $i,j\in J$ with $i\neq j$. Let $\bigcup_{j\in J}B_{j}$ be the union of the family $(B_{j})_{j\in J}$. Then $\bigcup_{j\in J}B_{j}\in\mathcal{F}$ and

$$\mu\Bigl(\bigcup_{j\in J}B_{j}\Bigr)=\sum_{j\in J}\mu(B_{j}).$$

(v) Let $C$ be a finite set and $b:C\times\mathbb{N}\to[0,\infty]$ a map. For $c\in C$, $\sum_{m\in\mathbb{N}}b(c,m)$ is the sum of the sequence $m\mapsto b(c,m)$ in $[0,\infty]$, and $c\mapsto\sum_{m\in\mathbb{N}}b(c,m)$ is a map from $C$ to $[0,\infty]$; for $m\in\mathbb{N}$, $\sum_{c\in C}b(c,m)\in[0,\infty]$, and $m\mapsto\sum_{c\in C}b(c,m)$ is a sequence in $[0,\infty]$. Then

$$\sum_{c\in C}\sum_{m\in\mathbb{N}}b(c,m)=\sum_{m\in\mathbb{N}}\sum_{c\in C}b(c,m).$$