Not only the proofs but also the full statement of theorems in this paper are too technical to reproduce here but the following may give some idea of its contents and importance. A relatively easy way to solve functional equations is by reduction to differential equations. Since, as Hilbert pointed out in the second part of his 1900 fifth problem, differentiability (sometimes of higher order) and often even continuity is not a natural assumption, regularization theorems are useful. These should show, preferably for large classes of equations, that all measurable solutions are continuous, then differentiable, of any order, maybe even analytic.
The author had proved many regularization theorems in the last 25 years [for surveys see his booklet Regularity properties of functional equations, Leaflets in Mathematics, Janus Pannonius University press, Pécs (1996) and his joint paper with L. Székelyhidi, Aequationes Math. 52, 10–29 (1996; Zbl 0858.39010)]. They deal with functional equations in several variables; for such equations the abundance of variables (relative to the number of places in the unknown functions) is great help in applying this and other solution methods.
In the present paper, that contains remarkable results and difficult proofs, he shows, roughly speaking, that measurability implies continuity for \(f\) in functional equations of the form \(f(x)=h(x,y,f(y), f[g_1(x,y)], \dots,f[ g_n (x,y) ])\) and in their generalizations with different \(f\)’s. Here \(f : X\to Z\) is the unknown function, the functions \(g_j : D\to X\) \((j=1,\dots,n),\) continuously differentiable, and \(h : D\times Z^{n+1}\to Z,\) continuous, are given, \(X, Z,\) and \(D\) being open subsets of \(\mathbb{R}^s, \mathbb{R}^m,\) and of \(X\times X,\) respectively. The vector variable \(y\) represents the “free” variables with which one can work. The task is easier if \(m>s\) but the author gets significant results also when \(1\leq m\leq s.\) Actually, the values of \(f\) are considered to be in separable metric and in topological spaces. Examples and counterexamples illustrate the strength of the results.