Nonmeasurable products of absolutely negligible sets in uncountable solvable groups (Q6111384)

Let \((G,{\cdot})\) be a~group and let \(\mu\) be a~measure whose domain \(\mathrm{dom}(\mu)\) is a~\(\sigma\)-algebra of subsets of~\(G\). Recall that the measure~\(\mu\) is called a~left \(G\)-quasi invariant measure (\(G\)-invariant measure, respectively) on~\(G\) if \(\mathrm{dom}(\mu)\) is a~left \(G\)-invariant family of sets and for each \(g\in G\) and each set \(X\in\mathrm{dom}(\mu)\), \(\mu(X)=0\) if and only if \(\mu(gX)=0\) (\(\mu(gX)=\mu(X)\), respectively). A~set \(X\subset G\) is called \(G\)-absolutely negligible in~\(G\), if for every \(\sigma\)-finite left \(G\)-quasi-invariant (left \(G\)-invariant) measure~\(\mu\) there exists a~left \(G\)-quasi-invariant (left \(G\)-invariant) measure~\(\mu'\) extending~\(\mu\) such that \(\mu'(X)=0\). The main result of the paper states that for any uncountable solvable group \((G,{\cdot})\) and for any nonzero \(\sigma\)-finite left \(G\)-quasi invariant measure~\(\mu\) there exist two \(G\)-absolutely negligible sets \(A,B\subset G\) such that \(A\cdot B=\{a\cdot b:a\in A\) and \(b\in B\}\) is not measurable with respect to~\(\mu\).