= Finite additivity of a set function
{title2=$F(A\cup B)=F(A)+F(B)\quad(A\cap B=\varnothing)$}
A real-valued <set function> on an algebra of sets is finitely additive if it has the displayed property for disjoint pairs. It follows that $F(\varnothing)=0$. On a finite ambient <set>, induction gives $F(A)=\sum_{x\in A}F(\{x\})$, so arbitrary signed singleton weights describe all such functions. Summing the pointwise <indicator function> identity for the <inclusion-exclusion principle> against these weights proves the same formula for $F$.
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