Let $A$ be a Boolean algebra, let $p$ be a probability measure an $A$.
Let $N = \{1,2,3,…,n\}$ for a fixed $n \in \mathbb{N}$, and let $\epsilon$ be a function from the set of all subsets of $N$ (the power set $2^N$ of $N$) into real numbers $\epsilon: 2^N \rightarrow \mathbb{R}$ with the assumption that $\epsilon(\emptyset) = 0$.
For each such function define a function $p_{\epsilon}$ from the cartesian product $A^n$ into the real numbers: $p_{\epsilon}(A_1, \dotsc, A_n):= \sum\limits_{S \subseteq N} \epsilon (S) p \left(\cap_{i \in S}A_i\right)$ for all $A_1, \dotsc, A_n \in A$. ($\cap$ denotes the Boolean operation of the meet in $A$)
Now choose for $A$ the Boolean algebra of all the subsets of the set $2^N$ (this algebra has $2^{2^n}$ elements). The elements $S \subseteq N$ are the atoms of $A$.
Take $A_i = \cup_{\{i\}\subseteq S \subseteq N}S, \, i = 1, 2, …, n$.
Then the following equation holds:
$p_{\epsilon}\left(A_1, \dotsc, A_n\right) = \sum\limits_{T \subseteq N}\left(\sum\limits_{S \subseteq T} \epsilon(S)\right)p(X_T)$, $\quad $where $X_T = T$ by the definition of $A$.
My question now is how you can see that the equality does hold?
Besides, how does the Boolean algebra $A$ look like? I mean, it seems to be the power set of the power set of $N$. What do its elements and the operations of meet, join and complement look like? Moreover, if $S \subseteq N$, then why is $S$ an atom of $A$? Wouldn't it have to be of the form $\{S\}$? (I'm familiar with the power set as a Boolean algebra, but not with the power set of the power set.)
Best Answer
There is nothing unusual about the Boolean algebra $A$; it is just the usual power set Boolean algebra of the set $2^N$, with union and intersection and set complement as its operations. The fact that $2^N$ happens to itself be a power set is irrelevant.
It appears that there is either a typo or a rather unfortunate abuse of notation that is confusing you: when it refers to $S$ as being an atom of $A$, it should indeed be $\{S\}$ instead. Similarly, the definition of $A_i$ should be $$A_i = \bigcup_{\{i\}\subseteq S \subseteq N}\{S\}.$$ In other words, $A_i$ is just the set of all subsets of $N$ that contain $i$.
So, this means that for $S\subseteq N$, $\bigcap_{i\in S}A_i$ is just the set of all $T\subseteq N$ such that $S\subseteq T$. We can write this as the union of the singletons $\{T\}$ for each such $T$. So, since $p$ is additive, $$p\left(\bigcap_{i\in S}A_S\right)=\sum_{T\supseteq S}p(\{T\})$$ and thus $$p_\epsilon(A_1,\dots,A_n)=\sum_{S\subseteq N}\epsilon(S)\sum_{T\supseteq S}p(\{T\})=\sum_{T\subseteq N}\sum_{S\subseteq T}\epsilon(S)p(\{T\}).$$