most recent 30 from http://mathoverflow.net 2013-05-22T00:57:28Z http://mathoverflow.net/feeds/question/84120 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84120/fubinis-theorem-and-unique-mean-value Valerio Capraro 2011-12-22T21:12:58Z 2011-12-22T23:09:09Z

<p>Following the terminology of Rosenblatt, I will say that a bounded function $f:\mathbb Z\rightarrow\mathbb R$ has a unique mean value if for every pair of finitely additive translation invariant probability measures $\lambda_1,\lambda_2$ on $\mathbb Z$, one has $\int fd\lambda_1=\int fd\lambda_2$.</p> <p>I will say that $f$ has Fubini's property if for all finitely additive probability measures $\mu,\nu$ on $\mathbb Z$, one has $\int\int f(x+y)d\mu d\nu=\int\int f(x+y)d\nu d\mu$.</p> <p><strong>Question:</strong> Is it true that the two properties above are equivalent? </p> <p>It is obvious that if $f$ has Fubini's property, then it has also a unique mean value, but the converse is not clear to me. I don't have a real evidence why the two properties should be the same.. let's say, that I am interested in studying the relation between these two properties and I was not able to find a function with a unique mean value which does not have Fubini's property.</p>

http://mathoverflow.net/questions/84120/fubinis-theorem-and-unique-mean-value/84126#84126 Jesse Peterson 2011-12-22T23:09:09Z 2011-12-22T23:09:09Z

<p>Let $g$ be the characteristic function on the positive odd numbers and let $f(x) = g(x) - g(x + 1)$, so that $f$ has a unique mean value. If $\mu$ is non-principle and supported on the positive odd numbers and $\nu$ is non-principle and supported on the negative even numbers then we have $\iint f(x + y) \; d\mu d\nu = 1$, while $\iint f(x + y) \; d\nu d\mu = 0$.</p>