Transitive closure of balanced mass transport in Z (move to close) 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.
I suspect that $\sim$ is not transitive, though I don't see how to prove it (note that the intervals $[n_i-A,n_i+A]$ can overlap with one another). But my real question is: What is the topological closure $\approx$ (relative to the total variation metric topology) of the transitive closure of $\sim$? Which is to say: How can one concretely characterize the resulting topologically closed, transitive relation?
The characterization of $\approx$ should be sufficiently concrete that one can immediately deduce (for instance) that the measure $\mu$ that assigns measure 1 to each integer $n \geq 1$ (and measure 0 to everything else) is not $\approx$-equivalent to the measure $\nu$ that assigns measure 1 to each integer $n \geq 0$ (and measure 0 to everything else).
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.
This is a discretized version of Transitive closure of balanced bounded mass transport , which I believe has not received sufficient attention.
[ADDED ON APRIL 3, MODIFIED LATER THAT DAY: I proposed a disproof of the assertion that $\sim$ is transitive, as follows: "Consider a mass distribution that puts two particles of mass 1 at each odd positive integer. Send the two particles at each odd positive integer in opposite directions to the adjacent even integers. Now we see a single particle at 0 and two particles at each even positive integer. Send the two particles at each even positive integer to the adjacent odd integers. The resulting distribution has lone particles at 0 and 1 and paired particles at 3, 5, 7, etc. The end result is that 1 unit of mass has gotten shifted from 1 to 0 while the rest of the mass distribution is unchanged. Finally, note that there is no way to achieve this redistribution with a single $\sim$-operation." The last sentence is incorrect, and as Christian Remling showed, the example is not a counterexample at all.]
Inasmuch as this thread has gotten difficult too follow, and entangles two questions (is $\sim$ transitive? if not, what is the topological closure of its transitive closure?), I request that some moderator close this thread. I have raised the first of the two questions in a new post: Transitivity of balanced mass transport in Z. If the answer turns out to be negative, then I will post a revised version of the second question.
 A: UPDATE: As pointed out by James, this attempt is not convincing. The procedure I suggested in the last paragraph is not really clearly defined. I still believe that $\sim$ is transitive, and that a more careful ``common refinement'' type argument should work. For now, I can only show that James's example is not a counterexample.
We have
$$
\mu = \sum_{n=0}^{\infty} 2\delta_{2n+1}, \quad \nu = \delta_0 + \delta_1 + \sum_{n=1}^{\infty} 2\delta_{2n+1} .
$$
Both measures are equivalent to a common third measure, so we want to show that $\mu\sim\nu$. This we can do by working our way from the left to the right.
First of all, split off $\delta_0+(1/2)\delta_3$ from $\nu$, and pair this with $(3/2)\delta_1$, and also split off $(1/2)\delta_1$ from each measure. Now the remaining part of $\nu$ starts with $(1/2)\delta_1 + (3/2)\delta_3 + 2(\delta_5+\delta_7+\ldots)$, while $\mu$ has been reduced to $2(\delta_3+\delta_5+\ldots)$.
Next, pair $(1/2)(\delta_1+\delta_5)$ (from $\nu$) with $\delta_3$, and also split off $\delta_3$ from both measures. I'm now in the same configuration as before this step, but everything has been shifted to the right by two units. So I can continue indefinitely in this style.
ORIGINAL (IMPRECISE) ANSWER STARTS HERE:
The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.
Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if
$$
a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b
$$
and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.
So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.
Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.
