Proof of Weak Sequential Compactness of Borel Measures of Total Mass One on a Compact Metric Space

Throughout, $(K,d)$, $\mathcal{T}_d$, $\mathbb{R}$, $d_{\mathbb{R}}$, $\mathbb{N}$ and $(\mu_n)_{n\in\mathbb{N}}$ are as in the statement. Let $\mathcal{B}(K)$ be the Borel $\sigma$-algebra of $(K,d)$, and let $\mathcal{C}$ be the set of all maps $f:K\to\mathbb{R}$ that are continuous on $K$ as maps from $(K,d)$ to $(\mathbb{R},d_{\mathbb{R}})$. Write $\mathbf{1}$ and $\mathbf{0}$ for the constant maps on $K$ with values $1$ and $0$. Sums and scalar multiples of maps from $K$ to $\mathbb{R}$ are formed pointwise, as in Continuity of Sums and Products of Real-Valued Functions on a Metric Space; by claims 1 and 5 of that theorem, $\mathcal{C}$ contains every constant map on $K$ and contains $f+h$ and $cf$ whenever $f,h\in\mathcal{C}$ and $c\in\mathbb{R}$.

Conventions on sequences of real numbers. The results on sequences of real numbers used below, namely Every Cauchy Sequence of Real Numbers Converges, Arithmetic of Limits of Real Sequences, Order Properties of Limits of Real Sequences and Uniqueness of Limits and Boundedness of Convergent Real Sequences, together with convergence of a sequence of real numbers and the Cauchy condition for a sequence of real numbers, are phrased for sequences written $(a_n)_{n=1}^{\infty}$; these are read as sequences indexed by $\mathbb{N}$ in the sense of a sequence in a set, as in Convergence and the Cauchy Condition for Real Sequences Agree with Those in the Real Line as a Metric Space. Convergence of a sequence of real numbers always means convergence in the sense of Limit of a Sequence of Real Numbers, and every index sequence occurring below is a strictly increasing sequence in $\mathbb{N}$ in the sense of that definition.

Step 0 ($K$ is nonempty). By the definition of a measure, $\mu_1(\emptyset)=0$. If $K$ were empty we would get $1=\mu_1(K)=\mu_1(\emptyset)=0$, contradicting $0<1$, which holds by claim 6 of Elementary Order Arithmetic in an Ordered Field. Hence $K$ is nonempty.

Step 1 (continuous maps are bounded and integrable). Let $f\in\mathcal{C}$. Since $K$ is nonempty and compact, Extreme Value Theorem on a Compact Subset of a Metric Space gives $x_{\min},x_{\max}\in K$ with $f(x_{\min})\le f(x)\le f(x_{\max})$ for every $x\in K$. Put $M_f=\max\{|f(x_{\min})|,|f(x_{\max})|\}$, the maximum of two elements, so $0\le M_f$ by claim 1 of Properties of the Absolute Value in an Ordered Field, and by claims 3 and 6 of that lemma $|f(x)|\le M_f$ for every $x\in K$; thus $f$ is bounded. By claims 2 and 3 of Borel Measurability and Bounded Integration on a Metric Space, $f$ is measurable with respect to $\mathcal{B}(K)$ and the Borel $\sigma$-algebra of the real line. Hence, for every Borel measure $\nu$ on $(K,d)$ with $\nu(K)=1$, part (b) of claim 6 of Borel Measurability and Bounded Integration on a Metric Space shows that $f$ is integrable with respect to $\nu$ and that

In particular this holds with $\nu=\mu_n$ for every $n\in\mathbb{N}$.

Step 2 (two observations on sequences of real numbers).

(2a) (Subsequences of a convergent sequence.) Let $(a_m)_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}$ converging to $L\in\mathbb{R}$, and let $(\kappa_k)_{k\in\mathbb{N}}$ be a strictly increasing sequence in $\mathbb{N}$. Then $(a_{\kappa_k})_{k\in\mathbb{N}}$ converges to $L$. Indeed, let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and choose $N\in\mathbb{N}$ with $|a_m-L|<\varepsilon$ for every $m\in\mathbb{N}$ with $N\le m$. By Strictly Increasing Sequences of Natural Numbers Dominate Their Index we have $k\le\kappa_k$ for every $k\in\mathbb{N}$, so if $N\le k$ then $N\le\kappa_k$ by transitivity, claim 1 of Properties of the Order on the Natural Numbers, and therefore $|a_{\kappa_k}-L|<\varepsilon$.

(2b) (Extraction of a convergent subsequence.) Let $(a_m)_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}$ and let $M\in\mathbb{R}$ with $0\le M$ and $|a_m|\le M$ for every $m\in\mathbb{N}$. Then there are a strictly increasing sequence $(\kappa_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ and $L\in\mathbb{R}$ such that $(a_{\kappa_k})_{k\in\mathbb{N}}$ converges to $L$. Indeed, $-M\le M$: if $M=0$ then $-M=0=M$, while if $0<M$ then claim 1 of Elementary Order Arithmetic in an Ordered Field applied to $0<M$ with the element $-M$ gives $0+(-M)<M+(-M)$, which by claims 3 and 4 of Additive Cancellation and Elementary Additive Identities in a Field reads $-M<0$; since $-M<0$ gives $-M\le 0$, claim 2 of Elementary Order Arithmetic in an Ordered Field applied to $-M\le 0$ and $0<M$ gives $-M<M$. By claim 6 of Properties of the Absolute Value in an Ordered Field we have $-M\le a_m\le M$ for every $m\in\mathbb{N}$, so every term of $(a_m)_{m\in\mathbb{N}}$ lies in the closed interval $[-M,M]$. By A Closed Interval is Sequentially Compact in the Real Line that interval is sequentially compact in $(\mathbb{R},d_{\mathbb{R}})$, so there are a strictly increasing sequence $(\kappa_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ and an element $L$ of $[-M,M]$ such that $(a_{\kappa_k})_{k\in\mathbb{N}}$ converges to $L$ in $(\mathbb{R},d_{\mathbb{R}})$. By claim 1 of Convergence and the Cauchy Condition for Real Sequences Agree with Those in the Real Line as a Metric Space, $(a_{\kappa_k})_{k\in\mathbb{N}}$ then converges to $L$ in the sense of Limit of a Sequence of Real Numbers.

Step 3 (a countable family of test maps). By claim 2 of A Countable Uniformly Dense Family of Lipschitz Functions on a Compact Metric Space, applied to the nonempty compact metric space $(K,d)$, there is a countable set $\mathcal{G}$ whose elements are bounded maps from $K$ to $\mathbb{R}$ that are Lipschitz as maps from $(K,d)$ to $(\mathbb{R},d_{\mathbb{R}})$, and which has the property that for every $f\in\mathcal{C}$ and every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $g\in\mathcal{G}$ with $|f(x)-g(x)|\le\varepsilon$ for every $x\in K$. By A Lipschitz Map is Uniformly Continuous every element of $\mathcal{G}$ belongs to $\mathcal{C}$. Since $\mathbf{0}\in\mathcal{C}$ and $0<1$, that property applied with $f=\mathbf{0}$ and $\varepsilon=1$ produces an element of $\mathcal{G}$, so $\mathcal{G}$ is nonempty, and therefore Countable Set provides a sequence $(g_i)_{i\in\mathbb{N}}$ in $\mathcal{G}$ whose set of terms is $\mathcal{G}$. For $i\in\mathbb{N}$ write $M_i=M_{g_i}$ for the bound supplied by Step 1, so that $0\le M_i$ and $|g_i(x)|\le M_i$ for every $x\in K$.

Step 4 (nested selectors). Let $S$ be the set of all pairs $(r,\tau)$ in which $r\in\mathbb{N}$ and $\tau=(\tau_k)_{k\in\mathbb{N}}$ is a strictly increasing sequence in $\mathbb{N}$ such that the sequence of real numbers $\bigl(\int_K g_r\,d\mu_{\tau_k}\bigr)_{k\in\mathbb{N}}$ converges.

The set $S$ is nonempty. By Step 1, $0\le M_1$ and $|\int_K g_1\,d\mu_n|\le M_1$ for every $n\in\mathbb{N}$, so observation (2b) of Step 2 provides a strictly increasing sequence $\tau^{\ast}=(\tau^{\ast}_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that $\bigl(\int_K g_1\,d\mu_{\tau^{\ast}_k}\bigr)_{k\in\mathbb{N}}$ converges, so $s=(1,\tau^{\ast})\in S$.

Let $R$ be the set of all pairs $\bigl((r,\tau),(r',\tau')\bigr)$ of elements of $S$ such that $r'=r+1$ and $\tau'$ is a subsequence of $\tau$; thus $R$ is a binary relation on $S$. For every $(r,\tau)\in S$ there is $(r',\tau')\in S$ with $\bigl((r,\tau),(r',\tau')\bigr)\in R$. Indeed, by Step 1 the terms of the sequence $\bigl(\int_K g_{r+1}\,d\mu_{\tau_k}\bigr)_{k\in\mathbb{N}}$ are bounded in absolute value by $M_{r+1}$, and $0\le M_{r+1}$, so observation (2b) of Step 2 yields a strictly increasing sequence $(\lambda_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that, setting $\tau'_k=\tau_{\lambda_k}$ for $k\in\mathbb{N}$, the sequence $\bigl(\int_K g_{r+1}\,d\mu_{\tau'_k}\bigr)_{k\in\mathbb{N}}$ converges; by claims 2 and 3 of A Subsequence of a Subsequence is a Subsequence, $\tau'$ is a strictly increasing sequence in $\mathbb{N}$ and is a subsequence of $\tau$, so $(r+1,\tau')\in S$ and the pair lies in $R$.

By Axiom of Dependent Choice, applied to $S$, $R$ and the element $s$, there is a sequence $\bigl((r_m,\tau^{(m)})\bigr)_{m\in\mathbb{N}}$ in $S$ with $(r_1,\tau^{(1)})=s$ and $\bigl((r_m,\tau^{(m)}),(r_{m+1},\tau^{(m+1)})\bigr)\in R$ for every $m\in\mathbb{N}$. Thus $r_1=1$ and $r_{m+1}=r_m+1$ for every $m\in\mathbb{N}$, so $r_m=m$ for every $m\in\mathbb{N}$ by induction on $m$, using Principle of Induction for the Natural Numbers. Consequently, for every $m\in\mathbb{N}$, the sequence $\tau^{(m)}$ is strictly increasing in $\mathbb{N}$, the sequence $\tau^{(m+1)}$ is a subsequence of $\tau^{(m)}$, and $\bigl(\int_K g_m\,d\mu_{\tau^{(m)}_k}\bigr)_{k\in\mathbb{N}}$ converges; write $L_m$ for its limit, which is well defined by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences.

Step 5 (the diagonal sequence). For $j\in\mathbb{N}$ put $n_j=\tau^{(j)}_j$.

First we record that for all $i,j\in\mathbb{N}$ with $i\le j$ there is a strictly increasing sequence $(\lambda_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ with $\tau^{(j)}_k=\tau^{(i)}_{\lambda_k}$ for every $k\in\mathbb{N}$. For $t\in\mathbb{N}$ let $P(t)$ be the assertion that for every $i\in\mathbb{N}$ there is a strictly increasing sequence $(\lambda_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ with $\tau^{(i+t)}_k=\tau^{(i)}_{\lambda_k}$ for every $k\in\mathbb{N}$. The assertion $P(1)$ holds because $\tau^{(i+1)}$ is a subsequence of $\tau^{(i)}$ for every $i$. Assume $P(t)$ and let $i\in\mathbb{N}$. Choose $(\lambda_k)_{k\in\mathbb{N}}$ strictly increasing with $\tau^{(i+t)}_k=\tau^{(i)}_{\lambda_k}$ for every $k$, and, using $P(1)$ at $i+t$ together with the associativity of addition on $\mathbb{N}$ from Arithmetic of Addition on the Natural Numbers, choose $(\kappa_k)_{k\in\mathbb{N}}$ strictly increasing with $\tau^{(i+(t+1))}_k=\tau^{(i+t)}_{\kappa_k}$ for every $k$. Then $\tau^{(i+(t+1))}_k=\tau^{(i)}_{\lambda_{\kappa_k}}$ for every $k$, and $(\lambda_{\kappa_k})_{k\in\mathbb{N}}$ is strictly increasing by claim 2 of A Subsequence of a Subsequence is a Subsequence, so $P(t+1)$ holds. By Principle of Induction for the Natural Numbers, $P(t)$ holds for every $t\in\mathbb{N}$. Now let $i,j\in\mathbb{N}$ with $i\le j$. If $i=j$, take $\lambda_k=k$, which is strictly increasing by claim 6 of Properties of the Order on the Natural Numbers. Otherwise $i<j$, so by claim 7 of Properties of the Order on the Natural Numbers there is $t\in\mathbb{N}$ with $j=i+t$, and $P(t)$ applies.

Next, $(n_j)_{j\in\mathbb{N}}$ is strictly increasing. Fix $j\in\mathbb{N}$ and choose a strictly increasing sequence $(\kappa_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ with $\tau^{(j+1)}_k=\tau^{(j)}_{\kappa_k}$ for every $k\in\mathbb{N}$. By Strictly Increasing Sequences of Natural Numbers Dominate Their Index we have $j+1\le\kappa_{j+1}$, and $j<j+1$ by claim 6 of Properties of the Order on the Natural Numbers, so $j<\kappa_{j+1}$ by claim 1 of that lemma. Applying claim 1 of A Subsequence of a Subsequence is a Subsequence to the strictly increasing sequence $\tau^{(j)}$ therefore gives

Step 6 (convergence along the diagonal). We first show that for every $i\in\mathbb{N}$ the sequence $\bigl(\int_K g_i\,d\mu_{n_j}\bigr)_{j\in\mathbb{N}}$ converges to $L_i$. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and choose $N\in\mathbb{N}$ such that $|\int_K g_i\,d\mu_{\tau^{(i)}_k}-L_i|<\varepsilon$ for every $k\in\mathbb{N}$ with $N\le k$. Put $J=\max\{i,N\}$, the maximum of two elements of $\mathbb{N}$, which is available because $\le$ is a total order on $\mathbb{N}$ by claims 1, 2 and 3 of Properties of the Order on the Natural Numbers. Let $j\in\mathbb{N}$ with $J\le j$. Then $i\le j$, so by Step 5 there is a strictly increasing sequence $(\lambda_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ with $\tau^{(j)}_k=\tau^{(i)}_{\lambda_k}$ for every $k$, whence $n_j=\tau^{(j)}_j=\tau^{(i)}_{\lambda_j}$. Since $j\le\lambda_j$ by Strictly Increasing Sequences of Natural Numbers Dominate Their Index and $N\le J\le j$, transitivity (claim 1 of Properties of the Order on the Natural Numbers) gives $N\le\lambda_j$, and therefore $|\int_K g_i\,d\mu_{n_j}-L_i|<\varepsilon$.

Now let $f\in\mathcal{C}$; we show that $\bigl(\int_K f\,d\mu_{n_j}\bigr)_{j\in\mathbb{N}}$ is a Cauchy sequence of real numbers. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 8 of Elementary Order Arithmetic in an Ordered Field there is $\varepsilon_1\in\mathbb{R}$ with $0<\varepsilon_1$ and $\varepsilon_1+\varepsilon_1=\varepsilon$, and then, by the same claim, $\eta\in\mathbb{R}$ with $0<\eta$ and $\eta+\eta=\varepsilon_1$; thus $\eta+\eta+\eta+\eta=\varepsilon$. By Step 3 there is $g\in\mathcal{G}$ with $|f(x)-g(x)|\le\eta$ for every $x\in K$, and there is $i\in\mathbb{N}$ with $g=g_i$. Put $h=f+(-1)g$, so $h\in\mathcal{C}$ and, pointwise, $h(x)=f(x)-g(x)$, whence $|h(x)|\le\eta$ for every $x\in K$. As in Step 1, $h$ is measurable with respect to $\mathcal{B}(K)$ and the Borel $\sigma$-algebra of the real line, so part (b) of claim 6 of Borel Measurability and Bounded Integration on a Metric Space gives $|\int_K h\,d\mu_{n_j}|\le\eta\,\mu_{n_j}(K)=\eta$ for every $j\in\mathbb{N}$, while claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied with the coefficients $1$ and $-1$, gives $\int_K h\,d\mu_{n_j}=\int_K f\,d\mu_{n_j}-\int_K g\,d\mu_{n_j}$. Write $A_j=\int_K f\,d\mu_{n_j}$ and $B_j=\int_K g\,d\mu_{n_j}$, so that $|A_j-B_j|\le\eta$ for every $j\in\mathbb{N}$. By the first part of this step, $(B_j)_{j\in\mathbb{N}}$ converges to $L_i$, so there is $J\in\mathbb{N}$ with $|B_j-L_i|<\eta$ for every $j\in\mathbb{N}$ with $J\le j$. Let $j,l\in\mathbb{N}$ with $J\le j$ and $J\le l$. Using claim 5 of Properties of the Absolute Value in an Ordered Field three times, together with claim 2 of that lemma to rewrite $|L_i-B_l|$ as $|B_l-L_i|$, and then claims 1 and 3 of Elementary Order Arithmetic in an Ordered Field to combine the resulting inequalities,

Hence $\bigl(\int_K f\,d\mu_{n_j}\bigr)_{j\in\mathbb{N}}$ is a Cauchy sequence, so by Every Cauchy Sequence of Real Numbers Converges it converges to a real number, which is unique by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences. Denote that real number by $\Lambda(f)$.

Step 7 ($\Lambda$ is additive, homogeneous and positive). By Step 6, $\Lambda$ is a map from $\mathcal{C}$ to $\mathbb{R}$, and for every $f\in\mathcal{C}$ the sequence $\bigl(\int_K f\,d\mu_{n_j}\bigr)_{j\in\mathbb{N}}$ converges to $\Lambda(f)$.

Let $f,h\in\mathcal{C}$ and $c\in\mathbb{R}$. By Step 1 both $f$ and $h$ are integrable with respect to every $\mu_{n_j}$, so claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied with the coefficients $1$ and $1$ and with the coefficients $c$ and $0$ respectively, gives

for every $j\in\mathbb{N}$. By claims 1 and 3 of Arithmetic of Limits of Real Sequences the right-hand sides converge to $\Lambda(f)+\Lambda(h)$ and to $c\,\Lambda(f)$, while by Step 6 the left-hand sides converge to $\Lambda(f+h)$ and to $\Lambda(cf)$; uniqueness of limits, claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences, therefore gives $\Lambda(f+h)=\Lambda(f)+\Lambda(h)$ and $\Lambda(cf)=c\,\Lambda(f)$.

Suppose next that $0\le f(x)$ for every $x\in K$. By part (a) of claim 6 of Borel Measurability and Bounded Integration on a Metric Space, $\mathbf{0}$ is integrable with respect to $\mu_{n_j}$ with $\int_K \mathbf{0}\,d\mu_{n_j}=0\cdot\mu_{n_j}(K)=0$, so the monotonicity assertion in claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $0\le\int_K f\,d\mu_{n_j}$ for every $j\in\mathbb{N}$. The constant sequence with value $0$ converges to $0$ directly from Limit of a Sequence of Real Numbers, so claim 1 of Order Properties of Limits of Real Sequences gives $0\le\Lambda(f)$.

Finally, part (a) of claim 6 of Borel Measurability and Bounded Integration on a Metric Space gives $\int_K \mathbf{1}\,d\mu_{n_j}=1\cdot\mu_{n_j}(K)=1$ for every $j\in\mathbb{N}$; the constant sequence with value $1$ converges to $1$, so uniqueness of limits gives $\Lambda(\mathbf{1})=1$.

Step 8 (conclusion). The pair $(K,d)$ is a metric space with $K$ nonempty and compact in $(K,\mathcal{T}_d)$, the set $\mathcal{C}$ and the map $\mathbf{1}$ are exactly as in Riesz-Markov Representation Theorem on a Compact Metric Space, and by Step 7 the map $\Lambda:\mathcal{C}\to\mathbb{R}$ satisfies the additivity, homogeneity and positivity hypotheses of that theorem. Hence claim 1 of Riesz-Markov Representation Theorem on a Compact Metric Space provides a Borel measure $\mu$ on $(K,d)$ with $\mu(K)=\Lambda(\mathbf{1})=1$ such that every $f\in\mathcal{C}$ is integrable with respect to $\mu$ and $\Lambda(f)=\int_K f\,d\mu$.

Let $f:K\to\mathbb{R}$ be bounded and continuous on $K$ as a map from $(K,d)$ to $(\mathbb{R},d_{\mathbb{R}})$. Then $f\in\mathcal{C}$, so by Step 6 the sequence $\bigl(\int_K f\,d\mu_{n_j}\bigr)_{j\in\mathbb{N}}$ converges to $\Lambda(f)=\int_K f\,d\mu$. Since $\mu_{n_j}(K)=1$ for every $j$ and $\mu(K)=1$, every measure occurring here is finite, and what has just been shown is exactly the condition in Weak Convergence of Finite Borel Measures on a Metric Space for $(\mu_{n_j})_{j\in\mathbb{N}}$ to converge weakly to $\mu$. By Step 5 the sequence $(n_j)_{j\in\mathbb{N}}$ is strictly increasing in $\mathbb{N}$, so $(\mu_{n_j})_{j\in\mathbb{N}}$ is a subsequence of $(\mu_n)_{n\in\mathbb{N}}$. This proves the theorem.