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Given two atomic measures $\mu$ and $\nu$ on $\mathbb{Z}$, write $\mu \sim \nu$ iff there exist countable decompositions $\mu = \mu_1 + \mu_2 + \cdots$ and $\nu = \nu_1 + \nu_2 + \cdots$ along with some constant $A>0$ such that, for all $i$, $\mu_i$ and $\nu_i$ are finite, nonzero measures supported on some interval $[n_i-A,n_i+A]$, and $\mu_i$ and $\nu_i$ have the same total mass and the same center of mass.

Is $\sim$ transitive?

In Transitive closure of balanced mass transport in Z (move to close) I conjectured (and then wrongly asserted) that the answer was "no", but a reply by Christian Remling (refuting a purported counterexample of mine) has made me less sure.

The situation that interests me most is where $\mu$ and $\nu$ are uniformly bounded, in the sense that there exists $B$ such that $\mu(n)$ and $\nu(n)$ are less than $B$ for all $n$. But I'm not sure how such an assumption would affect my central question, so I omitted it as an hypothesis. Feel free to assume it if the assumption gives you some traction.

In the case where $\sum_{n \in \mathbb{Z}} \mu(n) = \sum_{n \in \mathbb{Z}} \nu(n) = 1$, this setup is reminiscent of the theory of martingales, so I'm tagging this question pr.probability as well as co.combinatorics. Feel free to add other tags if they seem appropriate; I'm finding this question difficult to classify.

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  • $\begingroup$ Do you have an example for two measures that are not equivalent, but have the same (possibly infinite) total mass and center of mass (which can be an interval for infinite mass measures)? $\endgroup$
    – domotorp
    Commented Apr 7, 2015 at 22:20
  • $\begingroup$ I believe that the measure that assigns mass 1 to every integer $n \geq 1$ is not equivalent under $\sim$ to the measure that assigns mass 1 to every integer $n \geq 0$, but I suspect that if I tried to write down a proof I'd need to use the uniform boundedness assumption. $\endgroup$ Commented Apr 8, 2015 at 2:16
  • $\begingroup$ Indeed not hard to find counterexamples to my quesiton, if you pick the second measure to be 1 for only primes, that will guarantee that A cannot be bounded. $\endgroup$
    – domotorp
    Commented Apr 8, 2015 at 11:15

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