Let $X$ be a set. $\mathcal{A}$ an algebra over $X,$ and $\mu : \mathcal{A} \rightarrow [0, \infty]$ a $\sigma-$additive pre-measure. Prove that $\mu$ is $\sigma-$subadditive.
First of all, I beleive that, as $\mu$ is $\sigma-$additive, for all finite sequences of disjoint measurable sets $A_1, A_2, …, A_n$ we have $$\mu(\bigcup_{i \in \mathbb{N}}A_i) = \sum\mu(A_i) ,$$ and I should try to prove that for all finite sequences of measurable sets $B_1, B_2,.., B_n$, not necessariliy disjoint, $$\mu(\bigcup_{i \in \mathbb{N}}B_i) \leq \sum\mu(B_i) $$ holds.
Let's suppose not. e.g. $$\mu(\bigcup_{i \in \mathbb{N}}B_i) > \sum\mu(B_i) ,$$ for all sequences. This would contradict the hypothesis, which says that a disjoint sequence of sets exists, for which the equality holds.
I beleive that my proof might be wrong, because my argument is too easy, and does not require something else than the $\sigma -$additivity.
Best Answer
I think your definitions of $\sigma$-additivity and $\sigma$-subadditivity are incorrect. You should replace “there exists” with “for all”.