Recently in my investigations I faced a problem related to amenable groups. 
I have no idea if my question is suitable for this site, but after ask some collegues in my department I decided to try to find some help here.


Suppose that $G$ is an amenable group having a mean 
$ m\in \left(L^{\infty}(G,\mathbb{R}) \right)^{\*} $. 
Is it true that there exist a mean $\tilde{m}\in \left(L^{\infty}(G\times G,\mathbb{R}) \right)^\*$ such that for all borelians 
$B$ we have 
$$ 
\tilde{m}(B\times G)=m(B).
$$


If the answer is negative could you point me out the counter-example ?


Remark: The answer is positive if $G$ is locally compact group because the mean in $G\times G$ is the product Haar measure of $G$.



Have a general mean of an amenable group a kind of Fubini's property ?
Can't you argue as follows: given $f\in L^\infty(G\times G)$ for any $a\in L^1(G)$ define $f\_a$ by $$ f\_a(s) = \int\_G f(s,t) a(t) \ dt. $$ Then $f\_a\in L^\infty(G,\mathbb R)$ with $\|f\_a\| \leq \|f\| \|a\|$. Thus the map $$L^1(G) \rightarrow \mathbb R, \quad a\mapsto m(f\_a) $$ is bounded and linear, and so there is some $g=m\_f\in L^\infty(G)$ with $$ \int\_G g(s) a(s) = m(f\_a) \qquad (a\in L^1(G)). $$ Now define $\tilde m$ by $$ \tilde m(f) = m(g). $$


Then, let $x,y\in G$ and define $\hat f(s,t) = f(xs,yt)$, say $\hat f = (x,y)\cdot f$. Then check, for the obvious notation, that $$ \hat f\_a = x\cdot f\_{y^{-1}\cdot a}, \quad \hat g = y^{-1} \cdot g, \quad \tilde m(\hat f) = m(y^{-1}\cdot g)=m(g) = \tilde m(f), $$ so $\tilde m$ is left invariant. Notice I use that $m$ is left invariant in the calculation of $\tilde m(\hat g)$, and in the calculation of $\hat g$.


Finally, clearly if $f=1$ then $g=1$ and so $\tilde m(f)=1$. Then set $f=\chi\_{B\times G}$ so that $g = m(B) 1$ and hence $\tilde m(f) = m(B)$ as you want.


