Uniqueness of the limit sequence of discrete probability measures

Question

Let $N_n$ be a sequence of natural numbers increasing to infinity, and suppose we have a sequence of finite sets of distinct points $X_n = \{x_1^{n},x_2^{n},\ldots,x^{n}_{N_n}\} \subset[0,1] \subset \mathbb{R}$. Consider the discrete probability measure $$ \rho_n = \frac{1}{N_n}\sum_{i=1}^{N_n}\delta_{x^{n}_i}, $$ a normalized sum of delta functions centered at the points $x^{n}_i$. Being bounded as a linear operator on $C([0,1])$, there exists a vaguely convergent subsequence of the $\rho_n$ i.e. there exists a probability measure $\rho$ on [0,1] such that $$ \int_0^1fd\rho_{n_k} \to_{k\to \infty} \int_0^1 fd\rho $$ for all $f \in C([0,1])$. Let me further impose a spacing condition that if $$ r_n := \inf_{i\neq j} |x^n_i - x^n_j| $$ is the minimum distance between distinct pairs of the $x^n_i$, then $$ \inf_n N_n r_n > 0. $$ (in particular, this implies $x^n_i$ are distinct). This loosely can be interpreted as enforcing that the $X_n$ not accumulate too much on 0-dimensional sets (or perhaps I should say on sets of Hausdorff/Minkowski dimension < 1? I'm not sure and would be interested in answers to this as well, though it's not my main question).

As a simple example, if $X_n$ is regularly spaced on $[0,1]$, then $d\rho = dx = $ Lebesgue measure.

My question is: What further conditions can be imposed on the sets $X_n$ to guarantee that the original sequence $\rho_n$ converges (as opposed to a subsequence)?

Note that this is a rewrite of an earlier question of mine (my first ever), since closed (Uniqueness of the limit of a sequence of (discrete) probability measures). I understand if it gets closed again, because it's pretty specific yet open-ended at the same time. I imagine any nontrivial answer would be kind of creative, perhaps involving a rule or algorithm for how the points in the $X_n$ are distributed, and/or involving some nestedness property. Nestedness alone (i.e. $X_n \subset X_{n+1}$) does not guarantee uniqueness of the limit, as I have constructed counterexamples to demonstrate.

Answer 1

For this answer the assumption that the points $x_i^n$ are distinct is not necessary. The restriction to $[0,1]$ is also not necessary. The answer applies to probability measures with finite support on any complete metric space.

On the set of Borel probability measures on $[0,1]$, the topology in question is given by the Kantorovich-Rubinshtein metric $d_0$. See V.I. Bogachev, "Measure theory", Chapter 8, or the Wikipedia article "Wasserstein metric". In this metric the set of probability measures is complete. So a sequence converges if and only if it is Cauchy.

The Kantorovich-Rubinshtein metric is the optimal transport metric. The distance $d_0(\rho,\sigma)$ is the least cost of moving some divisible matter from a state distributed according to $\rho$ to one distributed according to $\sigma$, when the cost of moving a unit amount from $x\in[0,1]$ to $y\in[0,1]$ is the distance $|x-y|$.

If the distributions $\rho$ and $\sigma$ have finite support, every transport plan for moving from $\rho$ to $\sigma$ has a simple description: There are an integer $K>0$, elements $y_1,y_2,...,y_K$ in the support of $\rho$ and $z_1,z_2,...,z_K$ in the support of $\sigma$, and numbers $\alpha_1,\alpha_2,...,\alpha_K \in[0,1]$ such that $\rho = \sum_{i=1}^K \alpha_i \delta_{y_i}$, $\sigma = \sum_{i=1}^K \alpha_i \delta_{z_i}$, and the plan moves amount $\alpha_i$ from $y_i$ to $z_i$ for $i=1,2,...,K$. The cost of such a plan is $\sum_{i=1}^K \alpha_i |y_i - z_i |$.

The sequence of $\rho_n$ converges if and only if it is Cauchy. So this is a necessary and sufficient condition for it to converge:

$(1) \;\; \forall \varepsilon>0 \;\exists M \;\forall m,n > M \;\exists K=K(m,n) \;\exists y_1,y_2,...,y_K,z_1,z_2,...,z_K \in[0,1]$
$\exists \alpha_1,\alpha_2,...,\alpha_K\in[0,1] \;\;\rho_m = \sum_{i=1}^K \alpha_i \delta_{y_i} , \;\;\rho_n = \sum_{i=1}^K \alpha_i \delta_{z_i}, \;\; \sum_{i=1}^K \alpha_i |y_i - z_i | < \varepsilon$.

From this one can derive various sufficient conditions, depending on what one desires. For example, in the following condition for $n>m$ the amount $\frac{1}{N_m}$ at each point of $X_m$ is split into $\frac{N_n}{N_m}$ equal parts and each of those parts is transported to one point of $X_n$. (Of course this is only possible if $N_m$ divides $N_n$.)

$(2) \;\; \forall \varepsilon>0 \;\exists M \;\forall n>m > M \;\exists y_1,y_2,...,y_{N_n} \in[0,1]$ $\rho_m = \frac{1}{N_n}\sum_{i=1}^{N_n} \delta_{y_i}, \;\frac{1}{N_n}\sum_{i=1}^{N_n} |x_i^n - y_i | < \varepsilon$.

A variation of condition (2) may be useful when the sets $X_n$ form the levels of a tree in which each point in $X_{n+1}$ has exactly one parent in $X_n$ (for example, a nearest point in $X_n$) and each point in $X_n$ has the same number of children in $X_{n+1}$. The transport plan then follows the edges of the tree. For $n>m$ and $x\in X_n$, denote by $A_m(x)$ the unique ancestor of $x$ in $X_m$ (reached through a chain of parents). The following sufficient condition for the convergence of $\rho_n$ is a special case of (2):

$(3) \;\; \forall \varepsilon>0 \;\exists M \;\forall n>m > M \;\;\frac{1}{N_n}\sum_{i=1}^{N_n} |x_i^n - A_m(x_i^n) | < \varepsilon$.

Answer 2

Let $F_n$ be the distribution function of $\rho_n$. Then $F_n(0-) = 0$ and $F_n(1)$. Let $\rho$ be any limit point (w.r.to the weak topology) of $(\rho_n)_{n \in N}$ with distribution function $F$. Then there is a subsequence $(F_{n_k})_{k \in N}$ s.t. $F_{n_k}$ converges weakly to $F$, i.e. $\lim_{k \to \infty} F_{n_k}(t) = F(t)$ for each continuity point $t$ of $F$. If in particular $F$ is continuous (as implied by the spacing conditions, Remark of Christian Remling), then this convergence is uniform, i.e. \begin{eqnarray} (*) \lim_{k \to \infty} \|F_{n_k} - F\|_\infty = 0. \end{eqnarray} Thus if we know that any possible limit distribution $\rho$ has a continuous distribution function $F$, then this limit distribution function $F$ is uniquely defined if and only if \begin{eqnarray} (**) \lim_{m,n \to \infty} \|F_n - F_m\|_\infty = 0. \end{eqnarray} Note that here $F_n$ may be discountinuous. Of course this condition can be translated into conditions for the origninal $X_n$. Note that $N_n \cdot F_n(t)$ is the number of points $x_i^n$ in $[0,t]$.