Let $(M,+,0)$ be a naturally ordered commutative monoid (i.e. such that the natural preorder is antisymmetric) such that $(M,\sqsubseteq)$ is a complete lattice. Then $(M,\sum^*)$ is a summation structure with $$\sum^* \alpha=\sup_{( {\mathbb{M}},\sqsubseteq)}\left(\sum(\alpha|J) | J\in 2^{I}_{\text{fin}}\right),$$
That is $M$ is a set and $\sum^*$ is a map that assigns to $\alpha:I\to M$, $\sum^*\alpha$ with
- If $\alpha:\{i\} \to M$, $\sum^* \alpha= \alpha i$.
- Partitioning $I$ doesn't change the sum: For $\eta: I \to A$ and $\beta : A\to M$ with $a\mapsto \sum^*(\alpha | \eta^{-1}a)$ always: $\sum^* \alpha = \sum^* \beta$.
Now
Let $\alpha:{i}\to M$, then $\sum^*\alpha=\sup\{\alpha i\}=\alpha i$.
For $\alpha:I\to M$, $\eta: I\to A$ and $\beta:A\to M$, with $\beta a=\sum^*(\alpha|\eta^{-1}a)$.
We have to show $\sum^*\alpha=\sum^* \beta$.
The inequality $\sum^* \beta \leq \sum^*\alpha $: We have $$\sum^*\beta=\sup_{A_0\subseteq A}\sum_{a\in A_0}\beta a=\sup_{A_0}\sum_{a} \sum^* (\alpha|\eta^{-1}a)=\sup_{A_0}\sum_a\sup_{J\subset \eta^{-1}a}\sum_{i\in J} \alpha i$$ Now the last supremum is less than the supremum taken over all finite subsets of $I$, that is $$ \leq \sup_{A_0}\sum_{a\in A_0}\sup_{J\subseteq I}\sum_{i\in J}\alpha i=\sum^*\alpha$$
For the other inequality: we want to show that $\sum^*\alpha=\sup_{J\in \mathfrak{I}}\sum_{i\in J}\alpha i \leq \sum^* \beta$. That is for every $J\in \mathfrak{I}$ it is to show $$\sum_{i\in J}\alpha i\leq\sup_{A_0\in \mathfrak{A}}\sum_{a\in A_0} \sup_{J_a\subseteq_{\text{fin}}\eta^{-1}a}\sum_{i\in J_a}\alpha i.$$ For each $J$ it suffies to find one $A_0$ with $$\sum_{i\in J}\alpha i\leq \sum_{a\in A_0} \sup_{J_a \subseteq_{\text{fin}}\eta^{-1}a} \sum_{i\in J_a}\alpha i.$$ Pick $A_0=\eta J$ and for each $a \in \eta J$ the set $J_a=J\cap \eta^{-1}a$. Then we have $J=\cup_{a\in A_0}J_a=\cup_{a\in A_0} J \cap \eta^{-1}a$ and $$\sum_{i\in J}\alpha i=\sum_{a\in \eta J}\sum_{i\in J_a} \alpha i.$$