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\begin{document}$$I \subset {\mathbb{R}}$$\end{document} be an open interval and \documentclass[12pt]{minimal}
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\begin{document}$$M, N : I^2 \longrightarrow I$$\end{document} be means on I. Let \documentclass[12pt]{minimal}
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\begin{document}$$ \varphi : I \longrightarrow {\mathbb{R}}$$\end{document} be solution of the functional equation
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\begin{document}$$\varphi(M(x, y)) + \varphi(N(x, y)) = \varphi(x) + \varphi(y), \quad \quad x, y \in I$$\end{document}. We give sufficient conditions on \documentclass[12pt]{minimal}
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\begin{document}$$M, N$$\end{document} and the function \documentclass[12pt]{minimal}
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\begin{document}$$\varphi$$\end{document} such that for every Lebesgue measurable solution \documentclass[12pt]{minimal}
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\begin{document}$$f : I \longrightarrow \mathbb{R}$$\end{document} of the functional inequality
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\begin{document}$$f(M(x, y)) + f(N(x, y)) \leq f(x) + f(y), \quad \quad x, y \in I$$\end{document}, the function \documentclass[12pt]{minimal}
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\begin{document}$$f \circ \varphi^{-1} : \varphi(I) \longrightarrow {\mathbb{R}}$$\end{document} is convex.
