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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 $[0,t]$.