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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 $\sum_{i=1}^K \alpha_i |y_i - z_i | < \varepsilon$.

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):

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

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):

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

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 $\sum_{i=1}^K \alpha_i |y_i - z_i | < \varepsilon$.

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:

$(2) \;\; \forall \varepsilon>0 \;\exists M \;\forall n>m > M \;\exists y_1,y_2,...,y_{N_n} \in[0,1] \;\; \exists \alpha_1,\alpha_2,...,\alpha_{N_n}\in[0,1]$
$\rho_m = \sum_{i=1}^{N_n} \alpha_i \delta_{y_i}, \;\sum_{i=1}^{N_n} \alpha_i |x_i - y_i | < \varepsilon$.

A variation of condition (2) may be useful when the sets $X_n$ form the levels of a tree, so that each point in $X_{n+1}$ is linked to exactly one point in $X_n$. The transport plan then follows the edges of the tree.

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 $\sum_{i=1}^K \alpha_i |y_i - z_i | < \varepsilon$.

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):

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

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 Radon probability measures 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. 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 $|x-y|$.

So for discrete measures one can describe the necessary and sufficient Cauchy condition in some more or less combinatorial ways. The following formulation describes a "transport plan" from $\rho_m$ to $\rho_n$; we split $\rho_m$ and $\rho_n$ into smaller pieces which are then paired one-to-one, describing what is transported where:

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

From this one can derive various sufficient conditions, depending on what one desires. For example:

$(2) \;\; \forall \varepsilon>0 \;\exists M \;\forall n>m > M \;\exists y_1,y_2,...,y_{N_n} \in[0,1] \;\; \exists \alpha_1,\alpha_2,...,\alpha_{N_n}\in[0,1]$
$\rho_m = \sum_{i=1}^{N_n} \alpha_i \delta_{y_i}, \;\sum_{i=1}^{N_n} \alpha_i |x_i - y_i | < \varepsilon$.

A variation of condition (2) may be useful when the sets $X_n$ form the levels of a tree, so that each point in $X_{n+1}$ is linked to exactly one point in $X_n$. The transport plan then follows the edges of the tree.

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 $\sum_{i=1}^K \alpha_i |y_i - z_i | < \varepsilon$.

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:

$(2) \;\; \forall \varepsilon>0 \;\exists M \;\forall n>m > M \;\exists y_1,y_2,...,y_{N_n} \in[0,1] \;\; \exists \alpha_1,\alpha_2,...,\alpha_{N_n}\in[0,1]$
$\rho_m = \sum_{i=1}^{N_n} \alpha_i \delta_{y_i}, \;\sum_{i=1}^{N_n} \alpha_i |x_i - y_i | < \varepsilon$.

A variation of condition (2) may be useful when the sets $X_n$ form the levels of a tree, so that each point in $X_{n+1}$ is linked to exactly one point in $X_n$. The transport plan then follows the edges of the tree.