First, it is proven that $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ converges. Then, the inequality $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$ is well-defined, and hence can be proven. === Proof of Convergence === For every $n \in \N$, let $F_n \subseteq I$ be finite such that: :$\displaystyle \sum_{i \mathop \in G} b_i > \sum \left\{{ b_i: i \in I }\right\} - 2^{-n}$ for all finite $G$ with $F_n \subseteq G \subseteq I$ By passing over to $\displaystyle F'_n = \bigcup_{i \mathop = 1}^n F_i$ if necessary, it may be arranged that $F_n \subseteq F_m$ for $n \le m$. Next, define the sequence $\left({a_n}\right)_{n \in \N}$ by $a_n := \displaystyle \sum_{i \mathop \in F_n} a_i$. To show $\left({a_n}\right)_{n \in \N}$ is a Cauchy sequence, let $\epsilon > 0$. Subsequently let $N \in \N$ be such that $2^{-N} < \epsilon$, and let $m \ge n \ge N$. Then: {{begin-eqn}} {{eqn|l = d \left({a_m, a_n}\right)      |r = \left\vert{\left({\sum_{i \mathop \in F_m} a_i}\right) - \left({\sum_{i \mathop \in F_n} a_i}\right)}\right\vert      |c = Definition of Induced Metric }} {{eqn|r = \left\vert{\sum_{i \mathop \in F_m \setminus F_n} a_i}\right\vert      |c = $F_n \subseteq F_m$ }} {{eqn|o = \le      |r = \sum_{i \mathop \in F_m \setminus F_n} b_i      |c = $b_i \ge a_i \ge 0$ for all $i \in I$ }} {{end-eqn}} Now to estimate this last quantity, observe: {{begin-eqn}} {{eqn|l = \sum \left\{ {b_i : i \in I}\right\} - 2^{-n} + \sum_{i \mathop \in F_m \setminus F_n} b_i      |o = <      |r = \sum_{i \mathop \in F_n} b_i + \sum_{i \mathop \in F_m \setminus F_n} b_i      |c = Definition of $F_n$ }} {{eqn|r = \sum_{i \mathop \in F_m} b_i      |c = Union with Relative Complement, $F_n \subseteq F_m$ }} {{eqn|o = \le      |r = \sum \left\{ {b_i : i \in I}\right\}      |c = Generalized Sum is Monotone }} {{eqn|ll= \implies      |l = \sum_{i \mathop \in F_m \setminus F_n} b_i      |o = <      |r = 2^{-n} }} {{end-eqn}} Finally, as $n \ge N, 2^{-n} < 2^{-N} < \epsilon$ (by defining property of $N$). Combining all of these estimates leads to the conclusion that $d \left({a_m, a_n}\right) < \epsilon$. It follows that $\left({a_n}\right)_{n \in \N}$ is a Cauchy sequence. By Real Number Line is Complete Metric Space, this implies there exists an $a \in \R$ such that $\displaystyle \lim_{n \to \infty} a_n = a$. Having identified a candidate $a$ for the sum $\displaystyle \sum \left\{{a_i: i \in I}\right\}$ to converge to, it remains to verify that this is indeed the case. According to the definition of considered sum, the convergence is convergence of a net. Next, Metric Induces Topology ensures that we can limit the choice of opens $U$ containing $a$ to open balls centered at $a$. Now let $\epsilon > 0$. We want to find a finite $F \subseteq I$ such that: :$d \left({\displaystyle \sum_{i \mathop \in G} a_i, a}\right) < \epsilon$, for all finite $G$ with $F \subseteq G \subseteq I$ Now let $N \in \N$ such that for all $n \ge N$, $d \left({v_n, v}\right) < \dfrac \epsilon 2$ (with the $v_n$ as above). By taking a larger $N$ if necessary, ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well. Let us verify that the set $F_N$ defined above has the sought properties. So let $G$ be finite with $F_N \subseteq G \subseteq I$. Then: {{begin-eqn}} {{eqn|l = d \left({\sum_{i \mathop \in G} a_i, a}\right)      |r = \left\vert{\left({\sum_{i \mathop \in G} a_i}\right) - a}\right\vert      |c = Definition of Induced Metric }} {{eqn|o = \le      |r = \left\vert{\left({\sum_{i \mathop \in G} a_i}\right) - \left({\sum_{i \mathop \in F_N} a_i}\right)}\right\vert + \left\vert{\left({\sum_{i \mathop \in F_N} a_i}\right) - a}\right\vert      |c = Triangle Inequality }} {{eqn|o = <      |r = \left\vert{\sum_{i \mathop \in G \setminus F_N} a_i}\right\vert + \frac \epsilon 2      |c = $F_N \subseteq G$, defining property of $N$ }} {{eqn|o = \le      |r = \sum_{i \mathop \in G \setminus F_N} b_i + \frac \epsilon 2      |c = $b_i \ge a_i \ge 0$ for all $i \in I$ }} {{end-eqn}} For the first of these terms, observe: {{begin-eqn}} {{eqn|l = \sum \left\{ {b_i: i \in I}\right\} - 2^{-N} + \sum_{i \mathop \in G \setminus F_N} b_i      |o = <      |r = \sum_{i \mathop \in F_N} b_i + \sum_{i \mathop \in G \setminus F_N} b_i      |c = Defining property of $F_N$ }} {{eqn|r = \sum_{i \mathop \in G} b_i      |c = Union with Relative Complement, $F_N \subseteq G$ }} {{eqn|o = \le      |r = \sum \left\{ {b_i: i \in I}\right\}      |c = Generalized Sum is Monotone }} {{eqn|ll= \implies      |l = \sum_{i \mathop \in G \setminus F_N} b_i      |o = <      |r = 2^{-N} }} {{end-eqn}} Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain: :$\displaystyle d \left({\sum_{i \mathop \in G} a_i, a}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$ By definition of convergence of a net, it follows that: :$\displaystyle \sum \left\{{a_i: i \in I}\right\} = a$ {{qed|lemma}} === Proof of Inequality === Suppose that $\displaystyle \sum \left\{{ a_i: i \in I }\right\} > \sum \left\{{ b_i: i \in I }\right\}$. Then, as the sums converge, there exists a finite $F \subseteq I$ such that: :$\displaystyle \sum_{i \mathop \in F} a_i > \sum \left\{{ a_i: i \in I }\right\} - \epsilon$ for every $\epsilon > 0$.  So by picking a suitable $\epsilon$, it may be arranged that: {{begin-eqn}} {{eqn|l = \sum \left\{ { b_i: i \in I }\right\}      |o = <      |r = \sum \left\{ { a_i: i \in I }\right\} - \epsilon }} {{eqn|o = <      |r = \sum_{i \mathop \in F} a_i      |c = Defining property of $F$ }} {{eqn|o = \le      |r = \sum_{i \mathop \in F} b_i      |c = $a_i \le b_i$ for all $i$ }} {{eqn|o = \le      |r = \sum \left\{ { b_i: i \in I }\right\}      |c = Generalized Sum is Monotone }} {{end-eqn}} These inequalities together constitute a contradiction, and therefore: :$\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$ {{qed}} Category:Generalized Sums rgumdbhfc82we1qtta77khjt2o7gn5n