Symmetry and measurability
In the first part of the paper, the authors show the following theorem: Let \(X\) be a set, \(G\) be a finite group, and \(\bullet\) be an action of \(G\) on \(X\). Let \(\mathcal{C}\) be an algebra of subsets of \(X\) such that \(g \bullet C \in\mathcal{C}\) whenever \(g \in G\) and \(C\in\mathcal{C}\). If \(E\) belongs to the \(\sigma\)-algebra of subsets generated by \(\mathcal{C}\) and \(g \bullet E = E\) for every \(g \in G\), then \(E\) belongs to the \(\sigma\)-algebra generated by the family \[ \{ C\in\mathcal{C}: g\bullet C=C \text{ for every } g\in G\}. \] The proof is by a series of propositions. In the second part, the authors apply this result to the \(n\)-fold product case. For \(H\subset X^n\) let \[ \text{sym}(H)=\{ (x_{\pi(1)},\dots,x_{\pi(n)}): (x_1,\dots,x_n)\in H,\; \pi\in S_n\}, \] where \(S_n\) is the \(n\)-th symmetric group. \(H\subset X^n\) is called fully symmetric if \(H = \text{sym}(H)\). Let \((X, \Sigma)\) be a measurable space and let \(\widehat{\bigotimes}_{i=1}^n\Sigma\) denote the \(n\)-fold product of \(\Sigma\) by itself, see [\textit{D. H. Fremlin}, Measure theory. Vol. 2. Broad foundations. Corrected second printing of the 2001 original. Colchester: Torres Fremlin (2003; Zbl 1165.28001)]. Then the \(\sigma\)-algebra of fully symmetric sets in \(\widehat{\bigotimes}_{i=1}^n\Sigma\) is generated by the sets \[ \text{sym}(A^k \times X^{n-k}), \] where \(A\in\Sigma\) and \(k \geq (n+1)/2\), and also by the sets \[ \text{sym}(A^k \times X^{n-k}), \] where \(A\in\Sigma\) and \(k \leq (n+1)/2\). In the case of the countable product case, the authors prove the following result. The sets \[ \text{sym}(\prod_{k=1}^\infty A_k), \] \(A_k\in\Sigma\), \(k\in\mathbb{N}\), generate the \(\sigma\)-algebra of symmetric sets in \(\widehat{\bigotimes}_{i=1}^\infty\Sigma\) if and only if \(\Sigma=\{ X,\emptyset\}\).
scientific article; zbMATH DE number 6850200