Track / Stage 6 / 84 of 400 #1084 of 2000

Problem 1084

National olympiad, first round

Algebra Difficulty 6.1 Prove it

14. Using Zorn's Lemma, equivalent to the Axiom of Choice:

«if on a set $X$ there is a partial order $\prec$ such that every chain, i.e., a set of pairwise comparable elements, has a maximal element, then there exists an element $x_{\max} \in X$ such that $x \prec x_{\max}$ for all $x \in X$ », show that every finitely additive probability measure $\mathrm{P}$ defined on an algebra $\mathscr{A}$ of subsets of $\Omega$ can be extended to a finitely additive probability on all subsets of $\Omega$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. For any $A \notin \mathscr{A}$ , extend $\mathrm{P}$ to the algebra generated by $\mathscr{A}$ and $A$ in the same way as in the solution to problem II.3.13. Define a partial order $\prec$ on all such extensions, considering $\mathrm{P}_{1} \prec \mathrm{P}_{2}$ if $\mathrm{P}_{i}$ is an extension of $\mathrm{P}$ on the algebra $\mathscr{A}_{i} \supseteq \mathscr{A}$ and $\mathscr{A}_{1} \subseteq \mathscr{A}_{2}$ . Clearly, every chain of measures $\mathrm{P}_{\lambda}$ on algebras $\mathscr{A}_{\lambda}, \lambda \in \Lambda$ , has a maximal element $Q$ , correctly defined by the formula

$\left.\mathrm{Q}\right|_{\mathscr{A}_{\lambda}}=\mathrm{P}_{\lambda}$

on the algebra $\bigcup_{\lambda} \mathscr{A}_{\lambda}$ . Thus, by Zorn's lemma, there exists a maximal extension $\mathrm{P}_{\max}$ . It will be defined on all subsets of the set $\Omega$ (otherwise, it could be extended, leading to a contradiction with its maximality).

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.