Given a semiring.
Consider a premeasure.
Regard the following constructions: $$\inf_{A\subseteq S_1\sqcup\ldots\sqcup S_n}\{\mu(S_1)+\ldots+\mu(S_n)\}\quad\inf_{A\subseteq\bigsqcup_kS_k}\sum_k\mu(S_k)\quad\inf_{A\subseteq\bigcup_kS_k}\sum_k\mu(S_k)$$ Do these differ and which are still subadditive?
1 Answer
This answer is not up-to-date...
Consider the finite integral measure: $$\mu(E):=\int_Ee^{-x^2}\mathrm{d}x:\quad\mu(\mathbb{R})=\sqrt{\pi}$$ Take the ring of bounded Borel sets: $$E\in\mathcal{B}(\mathbb{R}):\quad|E|<\infty$$
Then the first differs from the latter both: $$\mu_0(\mathbb{R})=\infty\neq\sqrt{\pi}=\mu_{12}(\mathbb{R})$$
It also fails to be subadditive: $$\mu_0(\mathbb{R})=\infty\nleq\sqrt{\pi}=\sum_{k=-\infty}^\infty\mu_0(k,k+1]$$ (This even prohibits it from becoming a measure at all!)
The latter both agree since: $$S=(S\setminus S')\sqcup(S\cap S')=(S_1\sqcup\ldots\sqcup S_n)\sqcup(S\cap S')$$ (That gives among other subadditivity.)
