On the measurability of functions defined on the product of two topological spaces (Q1978854)

The author gives some sufficient conditions for the measurability of functions defined on the product \(X\times Y\) of topological spaces. Let \(\mu _1\), \(\mu _2\) be \(\sigma \)-finite measures on \(\sigma \)-fields \(M_1\) and \(M_2\), respectively, such that all open sets are measurable. Moreover, for every \(A\in M_1\), \(\mu _1 A>0\), the interior \(A^0\) has also positive measure and \(\mu _1 A>0\) for every non empty open set. Put \(M=M_1\times M_2\) and \(\mu =\mu _1\times \mu _2\). Assume further that for every \(A\in M\), with \(\mu A>0\) there is \(B\subset A\), \(\mu B>0\), such that all sections \(B^y\) and \(B_x\) are open. Let \(f:X\times Y\rightarrow R\) be such that all sections \(f^y\) are quasicontinuous. Suppose that there is a countable family \(M_0 \subset M_2\) containing only sets of positive \(\mu _2\) measure such that for every \(\varepsilon >0\), \(x\in X\), and every open set \(U\subset Y\) there is \(A\in M_0\) such that \(\mu _2(A\cap U)>0\) and osc\(_{A\cap U}f_x<\varepsilon \). Then \(f\) is \(\mu \)-measurable. Author also presents two other similar statements, where the quasicontinuity is replaced by \(\mu _1\)-measurability.