# Commutativity of integral quasi-arithmetic means on measure spaces

Research paper by **D. Głazowska, P. Leonetti; J. Matkowski; S. Tringali**

Indexed on: **23 Dec '17**Published on: **01 Dec '17**Published in: **Acta Mathematica Hungarica**

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#### 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
$$f^{-1}\Big(\int_X f\Big(g^{-1}\Big(\int_Y g \circ h \,d\mu\Big)\Big)d \lambda\Big) = g^{-1}\Big(\int_Y g\Big(f^{-1}\Big(\int_X f \circ h \,d\lambda\Big)\Big)d \mu\Big)$$
is satisfied by every
\({\mathscr{L} \otimes \mathscr{M}}\)
-measurable simple function
\({h\colon X \times Y \to I}\)
if and only if f = cg 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 [7].