# 100€ bounty ended: Do invariant measures maximize the integral?

Update: The negative answer to the following question has been provided by Matthew Daws, who won, but also rejected, the bounty of 100 euro that I set over the question.

Let $\mathcal M(\mathbb Z)$ be the set of all finitely additive probability measures on the power set of $\mathbb Z$. Let $\phi:\mathbb Z\rightarrow\mathbb R$ be nonnegative and bounded. Observe that $\phi$ is integrable with respect to any $\mu\in\mathcal M(\mathbb Z)$. Let me say that $\mu$ is $\phi$-translation invariant if for all $y\in\mathbb Z$ one has

$$\int\phi(x+y)d\mu(x)=\int\phi(x)d\mu(x)$$

Let $I_\phi(\mathbb Z)$ be the class of $\phi$-invariant measures in $\mathcal M(\mathbb Z)$. Let $\mu\in I_\phi(\mathbb Z)$ be fixed.

Question: Is it true that the mapping $F:\nu\in\mathcal M(\mathbb Z)\rightarrow\int\int\phi(x+y)d\nu(x)d\mu(y)$ attains its maximum on a measure $\nu\in I_\phi(\mathbb Z)$?

A simple observation which might be silly because I don't know if Fubini's theorem holds for finitely additive probability measures: If we are in a situation where we can use Fubini's theorem the mapping in the question is constant since $$\int\int\phi(x+y)d\nu(x)d\mu(y) = \int\int\phi(x+y)d\mu(y)d\nu(x)$$ $$= \int\int\phi(y)d\mu(y)d\nu(x) = \int\phi(y)d\mu(y).$$ –  Tapio Rajala Sep 3 '11 at 8:34