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

Each result cited below is universally quantified over the data in its own statement and is applied with the data named at the point of citation.

Throughout, $\mathcal{T}_{d}$ is the collection of subsets of $X$ open in $(X,d)$, $\mathcal{B}(X)$ is the Borel $\sigma$-algebra of $(X,d)$, $d_{\mathbb{R}}(s,t)=|s-t|$ is the absolute-value metric on $\mathbb{R}$, whose Borel $\sigma$-algebra is the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of the real line by claim 2 of Borel Measurability and Bounded Integration on a Metric Space, and "measurable" for a real function on a metric space means measurable with respect to its Borel $\sigma$-algebra and $\mathcal{B}(\mathbb{R})$. For a set $A\subseteq X$, $d_{A}$ is the restriction of $d$ to $A$, a metric on $A$ by claim 1 of The Restriction of a Metric to a Subset Induces the Subspace Topology, and $\mathcal{B}(A)$ is the Borel $\sigma$-algebra of $(A,d_{A})$. For $A\subseteq X$, $\mathbf{1}_{A}:X\to\mathbb{R}$ is its indicator. The space $X$ is nonempty, since $\mu_{1}(X)=1\neq0=\mu_{1}(\varnothing)$.

Step 0 (two tools). (a) Let $(Z,d_{Z})$ be a metric space, $\nu$ a Borel measure on it with $\nu(Z)<\infty$, $A$ a Borel subset of $Z$, $M$ a nonnegative real number, and $u:Z\to\mathbb{R}$ measurable with $|u|\le M$ on $Z$ and $u=0$ on $Z\setminus A$. Then $u$ is integrable with respect to $\nu$ and $\bigl|\int_{Z}u\,d\nu\bigr|\le M\,\nu(A)$. Indeed, $u$ is integrable by claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space. By The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure §integral, applied to $(Z,d_{Z})$, the set $A$, the measure $\nu$ and the function $u$, the restriction $u|_{A}$ is measurable on $(A,d_{A})$, integrable with respect to $\nu|_{A}$, and $\int_{Z}u\,d\nu=\int_{A}u|_{A}\,d(\nu|_{A})$. By The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure §borel-subset, $\nu|_{A}$ is a Borel measure on $(A,d_{A})$ with $\nu|_{A}(A)=\nu(A)$, which is finite since $\nu(A)\le\nu(Z)$ by Basic Properties of a Measure §monotone. As $|u|_{A}|\le M$, claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, applied on $(A,d_{A})$, gives $\bigl|\int_{A}u|_{A}\,d(\nu|_{A})\bigr|\le M\,\nu(A)$.

(b) If $f:X\to\mathbb{R}$ is measurable and $A\in\mathcal{B}(X)$, then $f\mathbf{1}_{A}$ is measurable. Indeed, for $C\in\mathcal{B}(\mathbb{R})$ the preimage of $C$ under $f\mathbf{1}_{A}$ is $f^{-1}(C)\cap A$ if $0\notin C$, and $\bigl(f^{-1}(C)\cap A\bigr)\cup(X\setminus A)$ if $0\in C$; both belong to the $\sigma$-algebra $\mathcal{B}(X)$. Likewise, the preimage of any set under $\mathbf{1}_{A}$ is one of $\varnothing$, $A$, $X\setminus A$, $X$, so $\mathbf{1}_{A}$ is measurable.

Step 1 (an increasing sequence of compact sets). For every $i\in\mathbb{N}$ the number $\tfrac{1}{i+1}$ is positive, so the tightness of $(\mu_{n})_{n\in\mathbb{N}}$, Tight Family of Borel Measures on a Metric Space §sequence together with Tight Family of Borel Measures on a Metric Space §tight, gives a set $L_{i}\subseteq X$, compact in $(X,\mathcal{T}_{d})$, with $\mu_{n}(X\setminus L_{i})\le\tfrac{1}{i+1}$ for every $n\in\mathbb{N}$. First, the union of two sets $A,B$ compact in $(X,\mathcal{T}_{d})$ is compact in $(X,\mathcal{T}_{d})$: an open cover $(U_{a})_{a\in I}$ of $A\cup B$ in $X$ is an open cover of $A$ and of $B$, so the implication from 1 to 2 of Compact Subset Criterion via Open Covers in the Ambient Space gives finite sets $J_{A},J_{B}\subseteq I$ indexing subcovers of $A$ and of $B$; the set $J_{A}\cup J_{B}$ is finite by claim 3 of Peeling an Element off a Finite Set, and Unions of Finite Sets and indexes a subcover of $A\cup B$, so $A\cup B$ is compact by the implication from 2 to 1 of the same theorem. Put $K_{1}=L_{1}$ and $K_{i+1}=K_{i}\cup L_{i+1}$. By induction every $K_{i}$ is compact in $(X,\mathcal{T}_{d})$; moreover $K_{i}\subseteq K_{i+1}$ and $L_{i}\subseteq K_{i}$. By Compact Subsets of a Metric Space are Closed and Borel §borel, $K_{i}$ and $X\setminus K_{i}$ belong to $\mathcal{B}(X)$, and, as $X\setminus L_{i}\in\mathcal{B}(X)$ by the same clause and $X\setminus K_{i}\subseteq X\setminus L_{i}$, Basic Properties of a Measure §monotone gives

$$\mu_{n}(X\setminus K_{i})\le\tfrac{1}{i+1}\qquad(n,i\in\mathbb{N}).$$

Write $d_{i}=d_{K_{i}}$. By claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology the subsets of $K_{i}$ open in $(K_{i},d_{i})$ are exactly the members of the subspace topology of $K_{i}$; as $K_{i}$ is compact in $(X,\mathcal{T}_{d})$, Compact Topological Space and Compact Subset shows that $K_{i}$ is compact for the topology of open subsets of the metric space $(K_{i},d_{i})$.

Step 2 (restricted and normalised measures). Fix $n,i\in\mathbb{N}$ and put $m_{n,i}=\mu_{n}(K_{i})$. By Basic Properties of a Measure §differences and Step 1, $m_{n,i}=1-\mu_{n}(X\setminus K_{i})\ge1-\tfrac{1}{i+1}\ge\tfrac{1}{2}$, and $m_{n,i}\le\mu_{n}(X)=1$ by Basic Properties of a Measure §monotone. Let $\nu_{n,i}=\mu_{n}|_{K_{i}}$; by The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure §borel-subset it is a Borel measure on $(K_{i},d_{i})$ with $\nu_{n,i}(K_{i})=m_{n,i}$. Let $\pi_{n,i}$ be the measure $\gamma$ of Nonnegative Combinations of Two Finite Measures, and the Average of Two Couplings §combination, a claim about an arbitrary measurable space (the dimension of that lemma plays no role in this claim and is taken to be $1$; it is unrelated to the metric $d$ of $X$), read for $(K_{i},\mathcal{B}(K_{i}))$, $\alpha=\beta=\nu_{n,i}$, $s=\tfrac{1}{m_{n,i}}$ and $t=0$. So $\pi_{n,i}(A)=\tfrac{1}{m_{n,i}}\nu_{n,i}(A)$ for $A\in\mathcal{B}(K_{i})$, $\pi_{n,i}$ is a Borel measure on $(K_{i},d_{i})$ with $\pi_{n,i}(K_{i})=1$, and for every bounded measurable $u:K_{i}\to\mathbb{R}$, which is integrable with respect to $\nu_{n,i}$ by claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, the same claim gives

$$\int_{K_{i}}u\,d\nu_{n,i}=m_{n,i}\int_{K_{i}}u\,d\pi_{n,i}. \qquad\text{(2.1)}$$

Moreover, if $f:X\to\mathbb{R}$ is measurable with $|f|\le M$, then $f\mathbf{1}_{K_{i}}$ is measurable by Step 0(b), integrable with respect to $\mu_{n}$ by claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, vanishes off $K_{i}$ and restricts to $f|_{K_{i}}$ on $K_{i}$; so The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure §integral gives

$$\int_{X}f\mathbf{1}_{K_{i}}\,d\mu_{n}=\int_{K_{i}}f|_{K_{i}}\,d\nu_{n,i}. \qquad\text{(2.2)}$$

Step 3 (a diagonal subsequence). (a) The real array $(m_{n,i})_{n,i\in\mathbb{N}}$ satisfies $|m_{n,i}|\le1$, so The Diagonal Subsequence Lemma for Bounded Real Arrays gives a strictly increasing sequence $s^{(0)}=(s^{(0)}_{l})_{l\in\mathbb{N}}$ in $\mathbb{N}$ such that for every $i$ the sequence $(m_{s^{(0)}_{l},i})_{l\in\mathbb{N}}$ converges; call its limit $m_{i}$.

(b) Recursively, for $i\in\mathbb{N}$, once the strictly increasing sequence $s^{(i-1)}$ is chosen, apply Weak Sequential Compactness of Borel Measures of Total Mass One on a Compact Metric Space to the compact metric space $(K_{i},d_{i})$ of Step 1 and the sequence $(\pi_{s^{(i-1)}_{l},i})_{l\in\mathbb{N}}$ of Borel measures of total mass one: it gives a strictly increasing sequence $(k^{(i)}_{l})_{l\in\mathbb{N}}$ in $\mathbb{N}$ and a Borel measure $\beta_{i}$ on $(K_{i},d_{i})$ with $\beta_{i}(K_{i})=1$ such that $(\pi_{s^{(i-1)}_{k^{(i)}_{l}},i})_{l\in\mathbb{N}}$ converges weakly to $\beta_{i}$. Put $s^{(i)}_{l}=s^{(i-1)}_{k^{(i)}_{l}}$; the sequence $s^{(i)}$ is strictly increasing by claim 2 of A Subsequence of a Subsequence is a Subsequence.

(c) For integers $0\le i\le i'$ and every $l\in\mathbb{N}$ there is $l'\in\mathbb{N}$ with $l\le l'$ and $s^{(i')}_{l}=s^{(i)}_{l'}$. This is clear for $i'=i$ (take $l'=l$). If it holds for $i'$, then $s^{(i'+1)}_{l}=s^{(i')}_{k}$ with $k=k^{(i'+1)}_{l}\ge l$ by Strictly Increasing Sequences of Natural Numbers Dominate Their Index, and the hypothesis for $i'$ applied to the index $k$ gives $l'\ge k\ge l$ with $s^{(i')}_{k}=s^{(i)}_{l'}$.

(d) Put $n_{j}=s^{(j)}_{j}$ for $j\in\mathbb{N}$. By (c) with $i=j$, $i'=j+1$, there is $l'\ge j+1$ with $n_{j+1}=s^{(j)}_{l'}$, and $s^{(j)}_{j}<s^{(j)}_{l'}$ by claim 1 of A Subsequence of a Subsequence is a Subsequence; so $(n_{j})_{j\in\mathbb{N}}$ is strictly increasing.

(e) Let $i\ge0$ be an integer and $(c_{n})_{n\in\mathbb{N}}$ a real sequence such that $(c_{s^{(i)}_{l}})_{l\in\mathbb{N}}$ converges to $L$. Then $(c_{n_{j}})_{j\in\mathbb{N}}$ converges to $L$. Given a positive real $\varepsilon$, Limit of a Sequence of Real Numbers gives $N$ with $|c_{s^{(i)}_{l}}-L|<\varepsilon$ for $l\ge N$. For $j\ge\max(i,N)$, (c) with $i'=j$ gives $l'\ge j\ge N$ with $n_{j}=s^{(i)}_{l'}$, so $|c_{n_{j}}-L|<\varepsilon$.

(f) Consequences. Fix $i\in\mathbb{N}$. By (a) and (e) with stage $0$, $(m_{n_{j},i})_{j\in\mathbb{N}}$ converges to $m_{i}$; since $1-\tfrac{1}{i+1}\le m_{n,i}\le1$ for all $n$, claim 1 of Order Properties of Limits of Real Sequences gives $1-\tfrac{1}{i+1}\le m_{i}\le1$. Let $u:K_{i}\to\mathbb{R}$ be bounded and continuous on $(K_{i},d_{i})$; it is measurable by claim 3 of Borel Measurability and Bounded Integration on a Metric Space applied to $(K_{i},d_{i})$, so (2.1) applies to it. By (b) and Weak Convergence of Finite Borel Measures on a Metric Space, $\bigl(\int_{K_{i}}u\,d\pi_{s^{(i)}_{l},i}\bigr)_{l\in\mathbb{N}}$ converges to $\int_{K_{i}}u\,d\beta_{i}$, so by (e) with stage $i$, $\bigl(\int_{K_{i}}u\,d\pi_{n_{j},i}\bigr)_{j\in\mathbb{N}}$ converges to the same limit. By (2.1) and Arithmetic of Limits of Real Sequences §products,

$$\int_{K_{i}}u\,d\nu_{n_{j},i}\longrightarrow m_{i}\int_{K_{i}}u\,d\beta_{i}\qquad(j\to\infty). \qquad\text{(3.1)}$$

The sequence $(n_{j})_{j\in\mathbb{N}}$ is fixed from now on.

Step 4 (the limit measures on the compact sets, and their extensions). Fix $i\in\mathbb{N}$. Let $\alpha_{i}$ be the measure $\gamma$ of Nonnegative Combinations of Two Finite Measures, and the Average of Two Couplings §combination read for $(K_{i},\mathcal{B}(K_{i}))$, $\alpha=\beta=\beta_{i}$, $s=m_{i}$, $t=0$. It is a Borel measure on $(K_{i},d_{i})$ with $\alpha_{i}(K_{i})=m_{i}$ and $\int_{K_{i}}u\,d\alpha_{i}=m_{i}\int_{K_{i}}u\,d\beta_{i}$ for bounded measurable $u$ (integrable with respect to $\beta_{i}$ by claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space). So (3.1) reads: $\int_{K_{i}}u\,d\nu_{n_{j},i}\to\int_{K_{i}}u\,d\alpha_{i}$ for every bounded $u$ continuous on $(K_{i},d_{i})$, such $u$ being measurable by claim 3 of Borel Measurability and Bounded Integration on a Metric Space.

Let $\iota_{i}:K_{i}\to X$ be the inclusion. Since $d(\iota_{i}(y),\iota_{i}(x))=d_{i}(x,y)$ by the symmetry condition 3 of Metric Space, Continuous Map Between Metric Spaces holds with $\delta=\varepsilon$, so $\iota_{i}$ is continuous on $K_{i}$ as a map from $(K_{i},d_{i})$ to $(X,d)$, hence measurable with respect to $\mathcal{B}(K_{i})$ and $\mathcal{B}(X)$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. Let $\bar{\alpha}_{i}$ be the image measure of $\alpha_{i}$ under $\iota_{i}$, formed by claim 1 of Image Measures, Measures with Densities, and Change of Variables: a Borel measure on $(X,d)$ with $\bar{\alpha}_{i}(B)=\alpha_{i}(B\cap K_{i})$ for $B\in\mathcal{B}(X)$ and $\bar{\alpha}_{i}(X)=\alpha_{i}(K_{i})=m_{i}\le1$. For bounded measurable $f:X\to\mathbb{R}$, integrable with respect to $\bar{\alpha}_{i}$ by claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, claim 2 of Image Measures, Measures with Densities, and Change of Variables gives, as $f\circ\iota_{i}=f|_{K_{i}}$,

$$\int_{X}f\,d\bar{\alpha}_{i}=\int_{K_{i}}f|_{K_{i}}\,d\alpha_{i}. \qquad\text{(4.1)}$$

Now let $f:X\to\mathbb{R}$ be bounded, with bound $M$, and continuous on $X$. Then $f$ is measurable by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. The map $f|_{K_{i}}$ is continuous on $K_{i}$ as a map from $(X,d)$, directly from Continuous Map Between Metric Spaces (the $\delta$ that works at a point for $X$ works for $K_{i}\subseteq X$), hence continuous on $(K_{i},d_{i})$ by The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure §continuity-map, and bounded by $M$. Combining (2.2), the reading of (3.1) above, and (4.1),

$$\int_{X}f\mathbf{1}_{K_{i}}\,d\mu_{n_{j}}\longrightarrow\int_{X}f\,d\bar{\alpha}_{i}\qquad(j\to\infty). \qquad\text{(4.2)}$$

Step 5 (the extensions increase with $i$). Fix $i\in\mathbb{N}$. (a) Let $h:X\to\mathbb{R}$ be bounded, continuous on $X$ and nonnegative. Since $K_{i}\subseteq K_{i+1}$ we have $h\mathbf{1}_{K_{i}}\le h\mathbf{1}_{K_{i+1}}$ pointwise, so claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $\int_{X}h\mathbf{1}_{K_{i}}\,d\mu_{n}\le\int_{X}h\mathbf{1}_{K_{i+1}}\,d\mu_{n}$ for every $n$. Along $(n_{j})$ both sides converge by (4.2), so claim 1 of Order Properties of Limits of Real Sequences gives $\int_{X}h\,d\bar{\alpha}_{i}\le\int_{X}h\,d\bar{\alpha}_{i+1}$.

(b) Let $B\in\mathcal{B}(X)$ and let $\varepsilon$ be a positive real number. Let $\lambda$ be the measure $\gamma$ of Nonnegative Combinations of Two Finite Measures, and the Average of Two Couplings §combination for $(X,\mathcal{B}(X))$, $\alpha=\bar{\alpha}_{i}$, $\beta=\bar{\alpha}_{i+1}$, $s=t=1$; it is a Borel measure on $(X,d)$ with $\lambda(X)=m_{i}+m_{i+1}<\infty$ and $\lambda(A)=\bar{\alpha}_{i}(A)+\bar{\alpha}_{i+1}(A)$. Since $X$ is nonempty, Inner and Outer Regularity of a Finite Borel Measure on a Metric Space, and Lipschitz Approximation of Indicators §lipschitz, applied to $\lambda$, $B$ and $\varepsilon$, gives a Lipschitz map $h:X\to\mathbb{R}$ with $0\le h\le1$ and a set $N\in\mathcal{B}(X)$ with $\lambda(N)\le\varepsilon$ such that $h=\mathbf{1}_{B}$ on $X\setminus N$. The map $h$ is continuous on $X$ by A Lipschitz Map is Uniformly Continuous, so (a) applies to it. Let $\bar{\alpha}$ be $\bar{\alpha}_{i}$ or $\bar{\alpha}_{i+1}$. The functions $h$ and $\mathbf{1}_{B}$ are measurable (claim 3 of Borel Measurability and Bounded Integration on a Metric Space and Step 0(b)) and bounded, hence integrable with respect to $\bar{\alpha}$ by claim 6(b) of that lemma, and $\int_{X}\mathbf{1}_{B}\,d\bar{\alpha}=\bar{\alpha}(B)$ by The Integral of an Indicator Function is the Measure of the Set together with claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $w=h-\mathbf{1}_{B}$ is integrable, in particular measurable by Integrable Function and the Lebesgue Integral, and $\int_{X}h\,d\bar{\alpha}-\bar{\alpha}(B)=\int_{X}w\,d\bar{\alpha}$. Since $|w|\le1$ and $w=0$ off $N$, Step 0(a) gives $\bigl|\int_{X}w\,d\bar{\alpha}\bigr|\le\bar{\alpha}(N)\le\lambda(N)\le\varepsilon$. Hence, using (a),

$$\bar{\alpha}_{i}(B)\le\int_{X}h\,d\bar{\alpha}_{i}+\varepsilon\le\int_{X}h\,d\bar{\alpha}_{i+1}+\varepsilon\le\bar{\alpha}_{i+1}(B)+2\varepsilon .$$

As $\varepsilon$ was arbitrary, claim 1 of Comparison of Real Numbers with Arbitrary Positive Slack gives $\bar{\alpha}_{i}(B)\le\bar{\alpha}_{i+1}(B)$. By induction, $\bar{\alpha}_{i}(B)\le\bar{\alpha}_{i'}(B)$ whenever $i\le i'$.

(c) Let $i\le i'$ and $B\in\mathcal{B}(X)$. By Basic Properties of a Measure §additivity, $m_{i}=\bar{\alpha}_{i}(B)+\bar{\alpha}_{i}(X\setminus B)$ and $m_{i'}=\bar{\alpha}_{i'}(B)+\bar{\alpha}_{i'}(X\setminus B)$, all terms real. Subtracting and using (b) for $X\setminus B$, and Step 3(f),

$$0\le\bar{\alpha}_{i'}(B)-\bar{\alpha}_{i}(B)\le m_{i'}-m_{i}\le1-\bigl(1-\tfrac{1}{i+1}\bigr)=\tfrac{1}{i+1}. \qquad\text{(5.1)}$$

Step 6 (the limit measure). For $B\in\mathcal{B}(X)$ the set $\{\bar{\alpha}_{i}(B):i\in\mathbb{N}\}$ is nonempty and bounded above by $1$ (by Basic Properties of a Measure §monotone, $\bar{\alpha}_{i}(B)\le m_{i}\le1$); let $\mu(B)$ be its least upper bound. For every $i$, (5.1) shows that $\bar{\alpha}_{i}(B)+\tfrac{1}{i+1}$ is an upper bound of $\bar{\alpha}_{i'}(B)$ for $i'\ge i$, and by Step 5(b) also for $i'<i$; hence

$$0\le\mu(B)-\bar{\alpha}_{i}(B)\le\tfrac{1}{i+1}\qquad(i\in\mathbb{N}), \qquad\text{(6.1)}$$

and $(\bar{\alpha}_{i}(B))_{i\in\mathbb{N}}$ converges to $\mu(B)$ by claim 3 of Order Properties of Limits of Real Sequences, the sequence $(\tfrac{1}{i+1})_{i\in\mathbb{N}}$ converging to $0$ by claim 3 of The Archimedean Property of the Real Numbers (for a positive real $\varepsilon$ it gives $n$ with $\tfrac{1}{n}<\varepsilon$, and then $\tfrac{1}{i+1}<\tfrac{1}{n}<\varepsilon$ for $i\ge n$).

We check Measure, Measure Space, and Probability Measure. First $\mu(\varnothing)=0$, as every $\bar{\alpha}_{i}(\varnothing)=0$. Let $(B_{k})_{k\in\mathbb{N}}$ be pairwise disjoint members of $\mathcal{B}(X)$ with union $B$. For $p\in\mathbb{N}$ and every $i$, Basic Properties of a Measure §additivity and Basic Properties of a Measure §monotone give $\sum_{k=1}^{p}\bar{\alpha}_{i}(B_{k})=\bar{\alpha}_{i}\bigl(\bigcup_{k=1}^{p}B_{k}\bigr)\le\bar{\alpha}_{i}(B)\le\mu(B)$; letting $i\to\infty$, using the convergence just proved and Arithmetic of Limits of Real Sequences §sums for the finite sum, claim 1 of Order Properties of Limits of Real Sequences gives $\sum_{k=1}^{p}\mu(B_{k})\le\mu(B)$. So the partial sums of $\sum_{k}\mu(B_{k})$ are bounded above, and by Measure, Measure Space, and Probability Measure the sum $S=\sum_{k}\mu(B_{k})$ is their least upper bound, with $S\le\mu(B)$. Conversely, for every $i$, the countable additivity of $\bar{\alpha}_{i}$ and Measure, Measure Space, and Probability Measure show that $\bar{\alpha}_{i}(B)$ is the least upper bound of the partial sums $\sum_{k=1}^{p}\bar{\alpha}_{i}(B_{k})$, each of which is at most $\sum_{k=1}^{p}\mu(B_{k})\le S$ termwise by (6.1); so $\bar{\alpha}_{i}(B)\le S$ for every $i$, whence $\mu(B)\le S$. Thus $\mu(B)=\sum_{k}\mu(B_{k})$, and $\mu$ is a Borel measure on $(X,d)$ by Borel Measure on a Metric Space. Finally $\mu(X)$ is the least upper bound of the numbers $m_{i}$, which satisfy $1-\tfrac{1}{i+1}\le m_{i}\le1$ by Step 3(f); so $\mu(X)\le1$, and for every positive real $\varepsilon$ some $i$ has $\tfrac{1}{i+1}<\varepsilon$ by claim 3 of The Archimedean Property of the Real Numbers, giving $1-\varepsilon\le\mu(X)$; claim 2 of Comparison of Real Numbers with Arbitrary Positive Slack yields $\mu(X)=1$.

Step 7 (the defect measures). Fix $i\in\mathbb{N}$ and put $\rho_{i}(B)=\mu(B)-\bar{\alpha}_{i}(B)$ for $B\in\mathcal{B}(X)$, a real number in the interval from $0$ to $\tfrac{1}{i+1}$ by (6.1). Then $\rho_{i}(\varnothing)=0$. For pairwise disjoint $(B_{k})_{k\in\mathbb{N}}$ in $\mathcal{B}(X)$ with union $B$, the partial sums of $\sum_{k}\mu(B_{k})$ and of $\sum_{k}\bar{\alpha}_{i}(B_{k})$ converge to $\mu(B)$ and to $\bar{\alpha}_{i}(B)$, by the countable additivity of these finite measures and Measure, Measure Space, and Probability Measure; so by claim 3 of Arithmetic of Limits of Real Sequences the partial sums of the series of nonnegative terms $\sum_{k}\rho_{i}(B_{k})$ converge to $\rho_{i}(B)$, and by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion the least upper bound of these partial sums is $\rho_{i}(B)$. By Measure, Measure Space, and Probability Measure, $\rho_{i}$ is a measure on $(X,\mathcal{B}(X))$, with $\rho_{i}(X)\le\tfrac{1}{i+1}$. Since $\mu(A)=\bar{\alpha}_{i}(A)+\rho_{i}(A)$ for every $A\in\mathcal{B}(X)$, $\mu$ is the measure $\gamma$ of Nonnegative Combinations of Two Finite Measures, and the Average of Two Couplings §combination for $\alpha=\bar{\alpha}_{i}$, $\beta=\rho_{i}$, $s=t=1$. Hence for every measurable $f:X\to\mathbb{R}$ with $|f|\le M$, integrable with respect to $\bar{\alpha}_{i}$ and $\rho_{i}$ by claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, that claim gives $\int_{X}f\,d\mu=\int_{X}f\,d\bar{\alpha}_{i}+\int_{X}f\,d\rho_{i}$, and claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space gives

$$\Bigl|\int_{X}f\,d\mu-\int_{X}f\,d\bar{\alpha}_{i}\Bigr|=\Bigl|\int_{X}f\,d\rho_{i}\Bigr|\le M\rho_{i}(X)\le\tfrac{M}{i+1}. \qquad\text{(7.1)}$$

Step 8 (weak convergence). Let $f:X\to\mathbb{R}$ be bounded, with bound $M$, and continuous on $X$; it is measurable by claim 3 of Borel Measurability and Bounded Integration on a Metric Space and integrable with respect to every $\mu_{n}$ and $\mu$ by claim 6(b) of that lemma. Let $\eta$ be a positive real number. The choices are made in this order: $\eta$, then $i$, then $N$. By claim 3 of The Archimedean Property of the Real Numbers choose $i\in\mathbb{N}$ with $\tfrac{1}{i}<\tfrac{\eta}{4M+1}$, so that $\tfrac{2M}{i+1}<\tfrac{\eta}{2}$. For every $n$, $f=f\mathbf{1}_{K_{i}}+f\mathbf{1}_{X\setminus K_{i}}$, where both summands are measurable by Step 0(b) and bounded by $M$; by claim 2 of Linearity and Monotonicity of the Lebesgue Integral and Step 0(a), applied with $Z=X$, $\nu=\mu_{n}$, $A=X\setminus K_{i}$ and $u=f\mathbf{1}_{X\setminus K_{i}}$, and by Step 1,

$$\Bigl|\int_{X}f\,d\mu_{n}-\int_{X}f\mathbf{1}_{K_{i}}\,d\mu_{n}\Bigr|\le M\mu_{n}(X\setminus K_{i})\le\tfrac{M}{i+1}.$$

By (4.2) and Limit of a Sequence of Real Numbers there is $N\in\mathbb{N}$ with $\bigl|\int_{X}f\mathbf{1}_{K_{i}}\,d\mu_{n_{j}}-\int_{X}f\,d\bar{\alpha}_{i}\bigr|<\tfrac{\eta}{2}$ for all $j\ge N$. For such $j$, the last display with $n=n_{j}$, this bound and (7.1), combined by the triangle inequality for the absolute value, give

$$\Bigl|\int_{X}f\,d\mu_{n_{j}}-\int_{X}f\,d\mu\Bigr|<\tfrac{M}{i+1}+\tfrac{\eta}{2}+\tfrac{M}{i+1}<\eta .$$

As $\eta$ was arbitrary, $\bigl(\int_{X}f\,d\mu_{n_{j}}\bigr)_{j\in\mathbb{N}}$ converges to $\int_{X}f\,d\mu$; as $f$ was arbitrary, $(\mu_{n_{j}})_{j\in\mathbb{N}}$ converges weakly to $\mu$ in the sense of Weak Convergence of Finite Borel Measures on a Metric Space. Together with Step 3(d) and Step 6, the strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ and the Borel measure $\mu$ with $\mu(X)=1$ are as claimed.