Condition for a regular set function in a topological space to be exhaustive (Q1181608)

Let \((T,\eta)\) be a \(\sigma\)-topological space in the sense of \textit{A. D. Alexandroff} [Mat. Sb., Nov. Ser. 9(51), 563-621 (1941; Zbl 0028.07201)], \(\Sigma\) an algebra of subsets of \(T\) with \(\eta\subseteq\Sigma\) and \(S\) a class of closed subsets of \(T\). Let \(X\) be a topological space and let \(\varphi:\Sigma\to X\) be a set function with \(\varphi(\emptyset)=\theta\), where \(\theta\) is an element of \(X\) and \(\emptyset\) the empty set. Let also \(H\) be a fundamental system of neighborhoods of the element \(\theta\). In particular, if \(X\) is an Abelian topological group, \(\theta\) may be taken as the neutral element of \(X\). Let \(L\) be a subclass of \(\Sigma\). \(\varphi\) is called \(L\)-exhaustive (resp. \(L\)-continuous from above at the \(\emptyset)\) if every pairwise disjoint (resp. decreasing) sequence \((E_ n)\) in \(L\) implies \(\lim\varphi(E_ n)=\theta\). Further, \(\varphi\) is called \(L\)-regular with respect to the class \(S\) if for every neighborhood \(U\) in \(H\) and for every element \(E\) in \(L\) there exists \(F\in S\) such that: \(\tilde\varphi(E-F)\subseteq U\), where \(\tilde\varphi(E-F):=\{\varphi(A):A\subseteq E-F,A\in\Sigma)\). On the other hand \(\varphi\) is absolutely continuous on \(\Sigma\), if for every neighborhood \(U\in H\) there exists a neighborhood \(V\in H\) such that for every two elements \(A,B\in\Sigma\) the following statements are true: (i) If \(\varphi(A)\in V\) and \(\varphi(B)\in V\) then \(\varphi(A\cap B)\in U\), (ii) If \(\varphi(A)\in V\) and \(\varphi(A\cap B)\in V\) then \(\varphi(B)\in U\). The main result of the paper is the following: Let \((T,\eta)\) be a \(\sigma\)-topological space, \(\Sigma\) an algebra of subsets of \(T\) with \(\eta\subseteq\Sigma\), \(S\) a class of closed subsets of \(T\) and \(L\) a class of subsets of \(T\) with \(L\subset\Sigma\). If \(\varphi:\Sigma\to X\) is continuous from above at the origin on \(\Sigma\), \(L\cup\eta\)-regular with respect to the class \(S\) and absolutely continuous on \(\Sigma\) then \(\varphi\) is \(L\cup\eta\)-exhaustive.