Commutativity of integral quasi-arithmetic means on measure spaces
Abstract: Let $(X, \mathscr{L}, \lambda)$ and $(Y, \mathscr{M}, \mu)$ be finite measure spaces for which there exist $A \in \mathscr{L}$ and $B \in \mathscr{M}$ with $0 < \lambda(A) < \lambda(X)$ and $0 < \mu(B) < \mu(Y)$, and let $I\subseteq \mathbf{R}$ be a non-empty interval. We prove that, if $f$ and $g$ are continuous bijections $I \to \mathbf{R}+$, then the equation $$ f{-1}!\left(\int_X f!\left(g{-1}!\left(\int_Y g \circ h\;d\mu\right)\right)d \lambda\right)! = g{-1}!\left(\int_Y g!\left(f{-1}!\left(\int_X f \circ h\;d\lambda\right)\right)d \mu\right)$$ is satisfied by every $\mathscr{L} \otimes \mathscr{M}$-measurable simple function $h: X \times Y \to I$ if and only if $f=c g$ for some $c \in \mathbf{R}+$ (it is easy to see that the equation is well posed). An analogous, but essentially different, result, with $f$ and $g$ replaced by continuous injections $I \to \mathbf R$ and $\lambda(X)=\mu(Y)=1$, was recently obtained in [Indag. Math. 27 (2016), 945-953].
Paper Prompts
Sign up for free to create and run prompts on this paper using GPT-5.
Top Community Prompts
Collections
Sign up for free to add this paper to one or more collections.