Proof of Prokhorov's Theorem on a Metric Space: a Tight Sequence of Borel Probability Measures Has a Weakly Convergent Subsequence

Throughout, for $j\in\mathbb{N}$ the real number $1/j$ is formed from the image of $j$ in $\mathbb{R}$ (The Natural Numbers Read in ℝ: a Predicative Condition and a Function from ℕ₀ to ℝ (ii)), which is positive (Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vi)).

The route: tightness gives an increasing sequence of compact sets $K_{j}$, each carrying all but less than $1/j$ of the mass of every $\mu_{n}$; one recursion applies Helly's selection theorem on $K_{1},K_{2},\dots$ in turn, each time thinning the previous subsequence, and the diagonal subsequence converges on every $K_{j}$ to a measure $\lambda_{j}$; the $\lambda_{j}$ increase to a Borel probability measure $\mu$, and tightness turns convergence on each $K_{j}$ into weak convergence on $S$.

Step 1 (an increasing sequence of compact sets). The sequence $(\mu_{n})$ is a tight family indexed by $\mathbb{N}$ of Borel probability measures, which are finite (Measure, Measure Space, and Probability Measure (iii)). So A Tight Family of Finite Borel Measures Has an Increasing Sequence of Compact Sets Carrying All but 1/j of Every Mass (i) gives a sequence $(K_{j})_{j\in\mathbb{N}}$ of subsets of $S$, each compact in $(S,\mathcal{T}_{d})$, with $K_{j}\subseteq K_{j+1}$ and

$$\mu_{n}(S\setminus K_{j})<\frac{1}{j}\qquad\text{for all }n,j\in\mathbb{N}.$$

Let $d_{j}$ be the restriction of $d$ to $K_{j}\times K_{j}$, $\iota_{j}:K_{j}\to S$ the inclusion, and $\mu_{n}^{j}$ the restriction of $\mu_{n}$ to $\mathcal{B}(K_{j})$. By A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (ii) and A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (iii), $(K_{j},d_{j})$ is a locally compact separable metric space, and by A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (vi) each $\mu_{n}^{j}$ is a finite Borel measure on it with $\mu_{n}^{j}(K_{j})=\mu_{n}(K_{j})\le1$.

Step 2 (Helly's theorem on one compact set). Each $\mu_{n}^{j}$ is a finite signed measure (Finite Signed Measures from Finite Measures and from Integrals (ii)), so an element of the space $M(K_{j})$ of finite signed Borel measures on $(K_{j},d_{j})$ used in Helly Selection Theorem for Finite Signed Borel Measures on a Locally Compact Separable Metric Space, which is a set by The Finite Signed Measures on a Measurable Space Form a Set (i), with total variation norm $\|\mu_{n}^{j}\|=\mu_{n}^{j}(K_{j})\le1$ by Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (iv). Fix $j\in\mathbb{N}$ and a strictly increasing $\tau:\mathbb{N}\to\mathbb{N}$. By Helly Selection Theorem for Finite Signed Borel Measures on a Locally Compact Separable Metric Space (i) and Helly Selection Theorem for Finite Signed Borel Measures on a Locally Compact Separable Metric Space (ii), applied on $(K_{j},d_{j})$ to the sequence $(\mu_{\tau(i)}^{j})_{i}$ in $M(K_{j})$, there are a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$ and a $\nu\in M(K_{j})$ with $\|\nu\|\le1$ and $\nu\ge0$ setwise such that $(\mu_{\tau(\sigma(i))}^{j})_{i}$ converges vaguely to $\nu$; by Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (iv), $\nu$ is a finite measure on $(K_{j},\mathcal{B}(K_{j}))$ with $\nu(K_{j})=\|\nu\|\le1$. Let $\lambda=\nu\circ\iota_{j}^{-1}$, a finite Borel measure on $S$ with $\lambda(S\setminus K_{j})=\nu(\emptyset)=0$ by A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (vii). Let $f$ belong to the set $C_{b}(S)$ of bounded continuous functions. By A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (iv), $f|_{K_{j}}$ is continuous with compact support on $(K_{j},d_{j})$, so it lies in the space $C_{c}(K_{j})$ of The Space of Continuous Functions with Compact Support and the Uniform Norm (i). Moreover $f$ is bounded and $\mathcal{B}(S)$-measurable (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (i)), and $K_{j}\in\mathcal{B}(S)$ (A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (v)), as A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (vii) and A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (vi) require. Vague convergence, with the integrals against the nonnegative $\mu_{n}^{j}$ and $\nu$ read as Lebesgue integrals (Properties of the Integral with Respect to a Finite Signed Measure: Linearity, Bounds, Bounded Functions, Linearity in the Measure, Indicators and Finite Measures (vii)), and A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (vii) and A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures (vi) give (the integrals of $f$ against the finite Borel measures $\lambda$ and $\mu_{n}$ exist by Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (ii), and the real sequences of integrals by The Simple, Nonnegative, Integrable and Signed Integrals Are Predicative in the Measure and the Function, at Any Variable Positions (iii))

$$\int_{S}f\,d\lambda=\int_{K_{j}}f|_{K_{j}}\,d\nu=\lim_{i\to\infty}\int_{K_{j}}f|_{K_{j}}\,d\mu_{\tau(\sigma(i))}^{j}=\lim_{i\to\infty}\int_{S}f\,\mathbf{1}_{K_{j}}\,d\mu_{\tau(\sigma(i))} .$$

Thus, for each $j\in\mathbb{N}$ and each strictly increasing $\tau:\mathbb{N}\to\mathbb{N}$, there are a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$ and a finite Borel measure $\lambda$ on $S$ with $\lambda(S\setminus K_{j})=0$ such that $\int_{S}f\,\mathbf{1}_{K_{j}}\,d\mu_{\tau(\sigma(i))}\to\int_{S}f\,d\lambda$ as $i\to\infty$ for every $f\in C_{b}(S)$.

Step 3 (one recursion, a diagonal subsequence, and the limits on $S$). Call a triple $(j,\tau,\lambda)$ a stage if $j\in\mathbb{N}$, $\tau:\mathbb{N}\to\mathbb{N}$ is strictly increasing and $\lambda$ is a finite Borel measure on $S$, and let $\mathcal{S}$ be the set of stages (a set, as the strictly increasing functions $\mathbb{N}\to\mathbb{N}$ lie in $\mathbb{N}^{\mathbb{N}}$ by The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (ii) and the finite Borel measures on $S$ form a set by The Measures, the Finite Measures and the Probability Measures on a Measurable Space Form Sets (ii)). Let $Q$ be the relation on $\mathcal{S}$ (Maps and Relations Given by Formulas (iii)) with $((j,\tau,\lambda),(j',\tau',\lambda'))\in Q$ if and only if $j'=j+1$, $\tau'=\tau\circ\sigma$ for some strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$, $\lambda'(S\setminus K_{j'})=0$, and $\int_{S}f\,\mathbf{1}_{K_{j'}}\,d\mu_{\tau'(i)}\to\int_{S}f\,d\lambda'$ as $i\to\infty$ for every $f\in C_{b}(S)$. By Step 2 with $j+1$ for $j$, every stage $(j,\tau,\lambda)$ has a $Q$-successor $(j+1,\tau\circ\sigma,\lambda')$ ($\tau\circ\sigma$ is again strictly increasing by Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences (iii)), and Step 2 with $j=1$ and $\tau$ the identity gives a stage $(1,\tau_{1},\lambda_{1})$ with $\lambda_{1}(S\setminus K_{1})=0$ and $\int_{S}f\,\mathbf{1}_{K_{1}}\,d\mu_{\tau_{1}(i)}\to\int_{S}f\,d\lambda_{1}$ for every $f\in C_{b}(S)$. By Axiom of Dependent Choice (i) there is a sequence of stages starting at $(1,\tau_{1},\lambda_{1})$ whose consecutive terms are related by $Q$; by induction its $j$-th term is $(j,\tau_{j},\lambda_{j})$, where $\tau_{j+1}=\tau_{j}\circ\sigma_{j+1}$ with $\sigma_{j+1}$ strictly increasing, $\lambda_{j}(S\setminus K_{j})=0$, and the integrals of $f\mathbf{1}_{K_{j}}$ against $\mu_{\tau_{j}(i)}$ converge to $\int f\,d\lambda_{j}$. Put $n_{i}=\tau_{i}(i)$. A strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$ has $\sigma(i)\ge i$ (Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences (ii)), so $n_{i+1}=\tau_{i}(\sigma_{i+1}(i+1))\ge\tau_{i}(i+1)>n_{i}$ and $(n_{i})$ is strictly increasing. For the same reason, $\tau_{m+1}=\tau_{m}\circ\sigma_{m+1}$ means that for every $t\in\mathbb{N}$ there is $r\ge t$ with $\tau_{m+1}(t)=\tau_{m}(r)$. By induction on $i\ge j$ this passes to $\tau_{i}$ and $\tau_{j}$: for every $t\in\mathbb{N}$ there is $r\ge t$ with $\tau_{i}(t)=\tau_{j}(r)$ (for $i=j$ take $r=t$; if $\tau_{i+1}(t)=\tau_{i}(r')$ with $r'\ge t$ and $\tau_{i}(r')=\tau_{j}(r)$ with $r\ge r'$, then $\tau_{i+1}(t)=\tau_{j}(r)$ with $r\ge t$). Hence for $i\ge j$ we have $n_{i}=\tau_{i}(i)=\tau_{j}(r_{i})$ with $r_{i}\ge i$, and since $(n_{i})$ and $\tau_{j}$ are strictly increasing, so is $(r_{i})_{i\ge j}$; putting $r_{i}=i$ for $i<j$ keeps it strictly increasing, as $r_{j}\ge j$. So the sequence $(\tau_{j}(r_{i}))_{i}$, which agrees with $(n_{i})$ from the $j$-th term on, indexes a subsequence of $(\tau_{j}(i))_{i}$. A subsequence of a convergent sequence has the same limit (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (xiii)), and the first $j-1$ terms do not affect it (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (v)), so for every $j\in\mathbb{N}$ and every $f\in C_{b}(S)$

$$\int_{S}f\,d\lambda_{j}=\lim_{i\to\infty}\int_{S}f\,\mathbf{1}_{K_{j}}\,d\mu_{n_{i}} .$$

With $f=1$ (in $C_{b}(S)$ by Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iii)) this gives $\lambda_{j}(S)=\lim_{i}\mu_{n_{i}}(K_{j})$; as $1-1/j<\mu_{n_{i}}(K_{j})\le1$ for every $i$ by Step 1, comparing limits (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x)) gives $1-1/j\le\lambda_{j}(S)\le1$.

Step 4 ($\lambda_{j}\le\lambda_{j+1}$ setwise). Fix $j$, a closed $F\subseteq S$ and $\varepsilon>0$, and put $C=F\cap K_{j}$. Since $\lambda_{j}(S\setminus K_{j})=0$ by Step 3, $\lambda_{j}(F)=\lambda_{j}(C)$, so we may assume $C\neq\emptyset$. As $K_{j}$ is closed (Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (ii)), $C$ is closed in $(S,\mathcal{T}_{d})$ and contained in $K_{j}$, hence compact (Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (i)). First, Closed–Open Regularity of a Finite Borel Measure on a Metric Space (i), applied to $\lambda_{j+1}$ and the Borel set $C$ (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii)), gives a closed $F'\subseteq C$ and an open $V\supseteq C$ with $\lambda_{j+1}(V\setminus F')<\varepsilon$, so $\lambda_{j+1}(V)=\lambda_{j+1}(C)+\lambda_{j+1}(V\setminus C)<\lambda_{j+1}(C)+\varepsilon$, as $V\setminus C\subseteq V\setminus F'$. Next we choose $k\in\mathbb{N}$ such that $k\operatorname{dist}(x,C)\ge1$ for every $x\in S\setminus V$. If $V=S$, any $k$ will do. Otherwise $A=S\setminus V$ is nonempty and closed. Let $g(x)=\max\{0,\,1-\operatorname{dist}(x,A)\}$ for $x\in S$, the function of Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vii) for $A$ and $k=1$: it is Lipschitz with constant $1$, $0\le g\le1$, and $g(x)\ge1-\operatorname{dist}(x,A)$. For $m\in\mathbb{N}$ put

$$U_{m}=\Bigl\{x\in S:\ g(x)<1-\frac{1}{m}\Bigr\}.$$

Each $U_{m}$ is open: for $x\in U_{m}$ let $2r=(1-1/m)-g(x)>0$; every $y$ with $d(x,y)<r$ has $g(y)\le g(x)+d(x,y)<1-1/m$, so the ball of radius $r$ about $x$ lies in $U_{m}$. Also $U_{m}\subseteq U_{m+1}$, as $1-1/m<1-1/(m+1)$. The $U_{m}$ cover $C$: for $c\in C$ we have $\operatorname{dist}(c,A)>0$ (The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (v)), hence $g(c)<1$, and an $m$ with $1/m<1-g(c)$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii)) puts $c$ in $U_{m}$. So $C\subseteq U_{M}$ for some $M\in\mathbb{N}$ (Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (iii)); let $k=M$. If some $x\in S\setminus V$ had $M\operatorname{dist}(x,C)<1$, then $\operatorname{dist}(x,C)<1/M$ would give $c\in C$ with $d(c,x)<1/M$ (The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (i)); as $x\in A$, the same clause gives $\operatorname{dist}(c,A)\le d(c,x)<1/M$, so

$$g(c)\ge1-\operatorname{dist}(c,A)>1-\frac{1}{M},$$

contradicting $c\in U_{M}$. Finally let

$$f(x)=\max\bigl\{0,\ 1-k\operatorname{dist}(x,C)\bigr\}\qquad(x\in S),$$

the function of Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vii) for the nonempty closed set $C$ and this $k$. It is Lipschitz with constant $k$ and $0\le f\le1$, so $f\in C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vi)); moreover $f=1$ on $C$, and $f=0$ off $V$ by the choice of $k$, that is, $\mathbf{1}_{C}\le f\le\mathbf{1}_{V}$. As $f\mathbf{1}_{K_{j}}\le f\mathbf{1}_{K_{j+1}}$, monotonicity of the integral (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix)) gives $\int f\mathbf{1}_{K_{j}}\,d\mu_{n_{i}}\le\int f\mathbf{1}_{K_{j+1}}\,d\mu_{n_{i}}$ for every $i$, and comparing the limits of Step 3 (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x)) gives $\int f\,d\lambda_{j}\le\int f\,d\lambda_{j+1}$. By monotonicity of the integral again, therefore,

$$\lambda_{j}(F)=\lambda_{j}(C)\le\int_{S}f\,d\lambda_{j}\le\int_{S}f\,d\lambda_{j+1}\le\lambda_{j+1}(V)<\lambda_{j+1}(C)+\varepsilon\le\lambda_{j+1}(F)+\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $\lambda_{j}(F)\le\lambda_{j+1}(F)$ for every closed $F\subseteq S$. For $B\in\mathcal{B}(S)$ and $\varepsilon>0$, Closed–Open Regularity of a Finite Borel Measure on a Metric Space (i) applied to $\lambda_{j}$ gives a closed $F$ and an open $W$ with $F\subseteq B\subseteq W$ and $\lambda_{j}(W\setminus F)<\varepsilon$, so $\lambda_{j}(B\setminus F)<\varepsilon$ as $B\setminus F\subseteq W\setminus F$, whence

$$\lambda_{j}(B)<\lambda_{j}(F)+\varepsilon\le\lambda_{j+1}(F)+\varepsilon\le\lambda_{j+1}(B)+\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $\lambda_{j}(B)\le\lambda_{j+1}(B)$.

Step 5 (the limit measure). By Step 4 the measures $\lambda_{j}$ on $(S,\mathcal{B}(S))$ increase setwise, so The Setwise Supremum of an Increasing Sequence of Measures Is a Measure (i) (the measures on $(S,\mathcal{B}(S))$ form a set by The Measures, the Finite Measures and the Probability Measures on a Measurable Space Form Sets (i)) gives a measure $\mu$ on $(S,\mathcal{B}(S))$ with

$$\mu(B)=\sup_{j\in\mathbb{N}}\lambda_{j}(B)\qquad\text{for every }B\in\mathcal{B}(S).$$

By Step 3, $1-1/j\le\lambda_{j}(S)\le1$ for every $j$, so $\mu(S)=1$: $\mu$ is a Borel probability measure. For each $j$, $\mu$ and $\lambda_{j}$ are finite measures, as $\mu(S)=1$ and $\lambda_{j}(S)\le1$, so $\rho_{j}=\mu-\lambda_{j}$ is a finite signed measure (Finite Signed Measures from Finite Measures and from Integrals (i)). It is nonnegative, since $\mu(B)=\sup_{k}\lambda_{k}(B)\ge\lambda_{j}(B)$ for every $B\in\mathcal{B}(S)$, so a finite measure with $\|\rho_{j}\|=\rho_{j}(S)=1-\lambda_{j}(S)\le1/j$ (Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (iv)).

Step 6 (weak convergence; proves (i)). Let $f\in C_{b}(S)$ and let $M>0$ be a real number with $|f|\le M$; the integrals of $f$ against the finite measures $\mu_{n_{i}}$, $\mu$ and $\lambda_{j}$ exist by Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (ii). For all $i,j\in\mathbb{N}$, splitting $f=f\mathbf{1}_{K_{j}}+f\mathbf{1}_{S\setminus K_{j}}$ under $\mu_{n_{i}}$ by linearity of the integral (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)), and $\mu$ into $\lambda_{j}$ and $\rho_{j}$ (Properties of the Integral with Respect to a Finite Signed Measure: Linearity, Bounds, Bounded Functions, Linearity in the Measure, Indicators and Finite Measures (iv)),

$$\Bigl|\int_{S}f\,d\mu_{n_{i}}-\int_{S}f\,d\mu\Bigr|\le M\,\mu_{n_{i}}(S\setminus K_{j})+\Bigl|\int_{S}f\,\mathbf{1}_{K_{j}}\,d\mu_{n_{i}}-\int_{S}f\,d\lambda_{j}\Bigr|+M\,\|\rho_{j}\| ,$$

using, for the first term, Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x) and monotonicity of the integral (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix)),

$$\Bigl|\int_{S}f\,\mathbf{1}_{S\setminus K_{j}}\,d\mu_{n_{i}}\Bigr|\le\int_{S}|f|\,\mathbf{1}_{S\setminus K_{j}}\,d\mu_{n_{i}}\le M\,\mu_{n_{i}}(S\setminus K_{j}),$$

and Properties of the Integral with Respect to a Finite Signed Measure: Linearity, Bounds, Bounded Functions, Linearity in the Measure, Indicators and Finite Measures (iii) for the last. The first and last terms are at most $M/j$ by Steps 1 and 5. Now let $\varepsilon>0$. First choose $j\in\mathbb{N}$ with $1/j<\varepsilon/(4M)$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii)); then, as the middle term tends to $0$ by Step 3, choose $N\in\mathbb{N}$ such that it is less than $\varepsilon/2$ for every $i\ge N$. For $i\ge N$ the display gives $|\int f\,d\mu_{n_{i}}-\int f\,d\mu|<2M/j+\varepsilon/2<\varepsilon$, so $\int f\,d\mu_{n_{i}}\to\int f\,d\mu$ by Convergent Sequences of Real Numbers (i) (a real sequence by The Simple, Nonnegative, Integrable and Signed Integrals Are Predicative in the Measure and the Function, at Any Variable Positions (iii)). As $f$ was arbitrary, $\mu_{n_{i}}\Rightarrow\mu$ by Weak Convergence of Finite Borel Measures on a Metric Space (i) (the finite Borel measures form a set by The Measures, the Finite Measures and the Probability Measures on a Measurable Space Form Sets (ii), and probability measures are finite by Measure, Measure Space, and Probability Measure (iii)), and $(\mu_{n_{i}})$ is a subsequence of $(\mu_{n})$ since $(n_{i})$ is strictly increasing.