Measures on compact metric spaces

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The limit argument of this course replaces sequences of finite point configurations by limiting measures and limiting sets. This lesson collects the tools, with proofs, on a compact metric space \(K\): regularity of finite Borel measures; weak convergence and its inequalities for open and closed sets; compactness of bounded sequences of measures; Hausdorff limits of closed sets; and conditional measures of a measure on a product \(Z\times[0,1]\) given the first coordinate. Two results are taken from other courses: the Riesz representation theorem, Theorem 2.2 of the programme's lesson Haar measure on locally compact groups, and the monotone class theorem, 136B of [Fremlin], the text of the core course Measure and Integration. Later lessons also use the Radon–Nikodym theorem [Fremlin, 232E] and the Lebesgue decomposition [Fremlin, 232I] from that course.

All measures are nonnegative, finite and defined on the Borel sets. The closed \(\rho\)-neighbourhood of a set \(F\) is \(F^{[\rho]}=\{x:d(x,F)\le\rho\}\), and the open one is \(\{x:d(x,F)<\rho\}\).

1. Regularity

Lemma 1.1. Let \(\mu\) be a finite Borel measure on a metric space \(X\). For every Borel set \(A\) and every \(\epsilon>0\) there are a closed \(F\) and an open \(U\) with \(F\subseteq A\subseteq U\) and \(\mu(U\setminus F)<\epsilon\). If \(X\) is compact, \(F\) is compact.

Proof. Let \(\mathcal A\) be the family of Borel sets with this property. It contains every closed set \(F\): the open sets \(\{x:d(x,F)<1/m\}\) decrease to \(F\), so their measures decrease to \(\mu(F)\). It is closed under complements, exchanging the roles of \(F\) and \(U\). For a sequence \(A_i\in\mathcal A\), choose \(F_i\subseteq A_i\subseteq U_i\) with \(\mu(U_i\setminus F_i)<\epsilon2^{-i-1}\); then \(U=\bigcup U_i\) is open, and for \(m\) large the closed set \(F=F_1\cup\dots\cup F_m\) satisfies \(\mu(\bigcup A_i\setminus F)<\epsilon/2\), so \(\mu(U\setminus F)<\epsilon\). Thus \(\mathcal A\) is a \(\sigma\)-algebra containing the closed sets, hence all Borel sets. In a compact space closed sets are compact. \(\square\)

2. Weak convergence

Measures \(\mu_n\) on a compact metric space \(K\) converge weakly to \(\mu\) if \(\int f\,d\mu_n\to\int f\,d\mu\) for every continuous \(f:K\to\mathbb R\). The limit is unique, by Lemma 1.1 and Urysohn functions: for closed \(F\), the functions \(\max(0,1-m\,d(\cdot,F))\) decrease to the indicator of \(F\).

Lemma 2.1. Let \(\mu_n\to\mu\) weakly on \(K\).

(a) \(\mu_n(K)\to\mu(K)\); \(\liminf\mu_n(U)\ge\mu(U)\) for open \(U\); \(\limsup\mu_n(F)\le\mu(F)\) for closed \(F\).

(b) If \(A\) is Borel and \(\mu(\partial A)=0\), then \(\mu_n(A)\to\mu(A)\), and the restrictions \(\mu_n|_A\) converge weakly to \(\mu|_A\).

(c) If \(\pi:K\to K'\) is a continuous map into a compact metric space, the image measures \(\pi_*\mu_n\) converge weakly to \(\pi_*\mu\).

Proof. (a) Take \(f=1\). For open \(U\) and closed \(F=K\setminus U\), the continuous functions \(f_m=\min(1,m\,d(\cdot,F))\) increase to the indicator of \(U\); so \(\liminf_n\mu_n(U)\) is at least \(\lim_n\int f_m\,d\mu_n=\int f_m\,d\mu\) for every \(m\), and the right side tends to \(\mu(U)\). The closed case follows by complements and the convergence of total masses.

(b) The interior \(A^\circ\) and closure \(\bar A\) satisfy \(\mu(A^\circ)=\mu(\bar A)=\mu(A)\), and (a) squeezes \(\mu_n(A)\) between \(\mu_n(A^\circ)\) and \(\mu_n(\bar A)\). For a continuous \(f\ge0\) and \(t\ge0\), the boundary of \(A\cap\{f>t\}\) lies in \(\partial A\cup\{f=t\}\), which is \(\mu\)-null for all but countably many \(t\); so \(\mu_n(A\cap\{f>t\})\to\mu(A\cap\{f>t\})\) for almost every \(t\), and \(\int_Af\,d\mu_n=\int_0^{\max f}\mu_n(A\cap\{f>t\})\,dt\) converges by dominated convergence. Signed \(f\) is a difference of two nonnegative ones.

(c) \(\int f\,d\pi_*\mu_n=\int f\circ\pi\,d\mu_n\), and \(f\circ\pi\) is continuous. \(\square\)

3. Compactness

Lemma 3.1. For a compact metric space \(K\), the space \(C(K)\) of continuous functions with the supremum norm has a countable dense subset.

Proof. For \(m\ge1\) choose a finite set \(x_1,\dots,x_r\) such that every point is within \(1/m\) of one of them, put \(\psi_i(x)=\max(0,2/m-d(x,x_i))\) and \(\Psi=\sum_i\psi_i\ge1/m\). For \(f\in C(K)\) with modulus of continuity \(\omega\), and rational numbers \(q_i\) with \(|q_i-f(x_i)|\le1/m\), \[ \Bigl|f(x)-\sum_iq_i\,\frac{\psi_i(x)}{\Psi(x)}\Bigr|\le\omega(2/m)+\frac1m , \] because only the \(x_i\) with \(d(x,x_i)<2/m\) contribute and the weights \(\psi_i/\Psi\) add up to one. The functions \(\sum_iq_i\psi_i/\Psi\), over all \(m\) and rational \(q_i\), form a countable dense set. \(\square\)

Theorem 3.2 (compactness). Let \(K\) be a compact metric space. Every sequence of measures \(\mu_n\) on \(K\) with \(\sup_n\mu_n(K)<\infty\) has a weakly convergent subsequence. For countably many such sequences, one subsequence of indices can be chosen along which all of them converge.

Proof. Let \(f_1,f_2,\dots\) be dense in \(C(K)\) (Lemma 3.1). The numbers \(\int f_j\,d\mu_n\) are bounded by \(\|f_j\|\sup_n\mu_n(K)\), so a diagonal subsequence makes all of them converge. For any \(f\in C(K)\) and \(\epsilon>0\), some \(f_j\) has \(\|f-f_j\|<\epsilon\), so \(\int f\,d\mu_n\) is a Cauchy sequence; let \(\Lambda(f)\) be its limit. Then \(\Lambda\) is a positive linear functional on \(C(K)=C_c(K)\), and the Riesz representation theorem Haar measure on locally compact groups (Theorem 2.2) gives a measure \(\mu\) with \(\int f\,d\mu=\Lambda(f)\); it is finite, with \(\mu(K)=\Lambda(1)\). For countably many sequences, apply the diagonal argument to all the countably many sequences of numbers \(\int f_j\,d\mu^{(i)}_n\) at once. \(\square\)

4. Hausdorff limits

For nonempty closed subsets \(A,B\) of a compact metric space \(K\) put \[ \begin{aligned} d_H(A,B)=\max\Bigl({}&\sup_{a\in A}d(a,B),\\ &\sup_{b\in B}d(b,A)\Bigr). \end{aligned} \] This is a metric on the nonempty closed subsets of \(K\), the Hausdorff distance.

Theorem 4.1 (Blaschke selection). Every sequence of nonempty closed subsets of \(K\) has a subsequence that converges in Hausdorff distance to a nonempty closed set.

Proof. A Cauchy subsequence. For \(m\ge1\) fix a finite \(1/m\)-net \(F_m\) of \(K\), and for a closed set \(A\) let \(T_m(A)=\{x\in F_m:d(x,A)\le1/m\}\). Every point of \(A\) is within \(1/m\) of a point of \(F_m\), which then lies in \(T_m(A)\); and every point of \(T_m(A)\) is within \(1/m\) of \(A\). So \(d_H(A,T_m(A))\le1/m\). There are finitely many possible sets \(T_m(A)\); a diagonal subsequence \((A_n)\) has \(T_m(A_n)\) constant for \(n\ge n_m\), for every \(m\). Then \(d_H(A_n,A_{n'})\le2/m\) for \(n,n'\ge n_m\).

The limit. Passing to a further subsequence, \(d_H(A_n,A_{n+1})<2^{-n}\) for all \(n\ge1\). Call a sequence \((x_m)_{m\ge1}\) admissible if \(x_m\in A_m\) and \(d(x_m,x_{m+1})<2^{-m}\) for all \(m\); it is Cauchy and converges in the compact space \(K\), and its limit \(y\) satisfies \(d(x_n,y)<2^{1-n}\). Every \(x\in A_n\) is the \(n\)th term of an admissible sequence: choose \(x_{m+1}\in A_{m+1}\) within \(2^{-m}\) of \(x_m\) for \(m\ge n\), and \(x_{m-1}\in A_{m-1}\) within \(2^{-(m-1)}\) of \(x_m\) for \(m\le n\), both possible since \(d_H(A_m,A_{m+1})<2^{-m}\). Let \(A\) be the closure of the set of limits of admissible sequences; it is nonempty. Each \(x\in A_n\) is within \(2^{1-n}\) of such a limit, and each limit \(y\), with its sequence \((x_m)\), is within \(2^{1-n}\) of \(x_n\in A_n\); passing to the closure, \(d_H(A_n,A)\le2^{1-n}\). \(\square\)

Lemma 4.2. Let \(\Gamma_n\to\Gamma\) in Hausdorff distance.

(a) Every point of \(\Gamma\) is the limit of points \(x_n\in\Gamma_n\), and every limit of a subsequence of points \(x_n\in\Gamma_n\) lies in \(\Gamma\).

(b) If a compact set \(L\) is disjoint from \(\Gamma\), then \(L\) is disjoint from \(\Gamma_n\) for all large \(n\).

(c) If measures \(\mu_n\) are concentrated on \(\Gamma_n\), that is \(\mu_n(K\setminus\Gamma_n)=0\), and converge weakly to \(\mu\), then \(\mu(K\setminus\Gamma)=0\).

Proof. (a) For \(y\in\Gamma\) choose \(x_n\in\Gamma_n\) nearest to \(y\); \(d(x_n,y)\le d_H(\Gamma_n,\Gamma)\). If \(x_n\in\Gamma_n\) and \(x_{n_i}\to x\), then \(d(x,\Gamma)\le d(x,x_{n_i})+d_H(\Gamma_{n_i},\Gamma)\to0\), and \(\Gamma\) is closed. (b) \(L\) has positive distance \(\rho\) from \(\Gamma\), and \(\Gamma_n\) lies in the open \(\rho\)-neighbourhood of \(\Gamma\) once \(d_H(\Gamma_n,\Gamma)<\rho\). (c) The set \(K\setminus\Gamma\) is the union of the compact sets \(L_m=\{x:d(x,\Gamma)\ge1/m\}\). By (b) and Lemma 2.1(a), applied to the open set \(\{d(\cdot,\Gamma)>1/(2m)\}\supseteq L_m\), which misses \(\Gamma_n\) for large \(n\), its \(\mu\)-measure is at most \(\liminf\mu_n=0\). \(\square\)

5. Conditional measures

Proposition 5.1 (conditional measures). Let \(Z\) be a compact metric space, \(\nu\) a finite measure on \(Z\times[0,1]\), and \(\lambda=\pi_*\nu\) its image on \(Z\) under \(\pi(z,t)=z\). There are probability measures \(h_z\) on \([0,1]\), \(z\in Z\), such that for every Borel \(J\subseteq[0,1]\) the function \(z\mapsto h_z(J)\) is Borel and \[ \nu(F\times J)=\int_Fh_z(J)\,d\lambda(z) \] for every Borel \(F\subseteq Z\).

Proof. Distribution functions at dyadic points. Let \(D\) be the set of dyadic rationals in \([0,1]\). For \(t\in D\), the measure \(F\mapsto\nu(F\times[0,t])\) is at most \(\lambda\), so by the Radon–Nikodym theorem [Fremlin, 232E] it has a Borel density \(G_t\) with respect to \(\lambda\), and one may take \(0\le G_t\le1\) and \(G_1=1\). For \(t<t'\) in \(D\), \(\int_F(G_{t'}-G_t)\,d\lambda=\nu(F\times(t,t'])\ge0\) for all \(F\), so \(G_t\le G_{t'}\) \(\lambda\)-almost everywhere. For \(t\in D\), \(t<1\), and \(t_m\in D\) decreasing to \(t\), the functions \(G_{t_m}\) decrease almost everywhere to a limit \(G'\ge G_t\) with \(\int_FG'\,d\lambda\) equal to \(\lim\nu(F\times[0,t_m])=\nu(F\times[0,t])\), by monotone convergence; so \(G'=G_t\) almost everywhere. These are countably many conditions, so outside a Borel \(\lambda\)-null set \(N_0\) the function \(t\mapsto G_t(z)\) on \(D\) is nondecreasing, takes the value \(1\) at \(t=1\), and satisfies \(G_t(z)=\lim G_{t_m}(z)\) along every dyadic sequence \(t_m\downarrow t\).

The measures. For \(z\notin N_0\) and \(x\in[0,1)\) let \(\Phi_z(x)=\inf\{G_t(z):t\in D,\,t>x\}\), and \(\Phi_z(1)=1\). This function is nondecreasing, right-continuous, and equal to \(G_t(z)\) at dyadic \(t\). For a continuous \(f\) on \([0,1]\) the sums \[ \begin{aligned} \Lambda_z^{(m)}(f)={}&f(0)\Phi_z(0)\\ &+\sum_{i=1}^{2^m}f(i2^{-m})\bigl(\Phi_z(i2^{-m})\\ &\qquad-\Phi_z((i-1)2^{-m})\bigr) \end{aligned} \] form a Cauchy sequence, by uniform continuity of \(f\) and \(\Phi_z(1)=1\); their limit \(\Lambda_z(f)\) is a positive linear functional with \(\Lambda_z(1)=1\). By the Riesz representation theorem there is a probability measure \(h_z\) with \(\int f\,dh_z=\Lambda_z(f)\). For dyadic \(t<1\), take continuous \(f_j\) equal to \(1\) on \([0,t]\), \(0\) on \([t+2^{-j},1]\), and affine in between: then \(\Lambda_z(f_j)\) lies between \(\Phi_z(t)\) and \(\Phi_z(t+2^{-j})\), and letting \(j\to\infty\) gives \(h_z([0,t])=\Phi_z(t)=G_t(z)\) by right-continuity. For \(z\in N_0\) let \(h_z\) be the unit mass at \(0\).

All Borel sets. Let \(\mathcal J\) be the family of Borel \(J\subseteq[0,1]\) for which \(z\mapsto h_z(J)\) is Borel and the displayed formula holds for every Borel \(F\). It contains \([0,1]\), since \(\nu(F\times[0,1])=\lambda(F)\), and the intervals \([0,t]\), \(t\in D\), since \(h_z([0,t])=G_t(z)\) off the null set \(N_0\). It is closed under differences \(J'\setminus J\) with \(J\subseteq J'\), and under increasing countable unions by monotone convergence; so it is a Dynkin class. The intervals \([0,t]\), \(t\in D\), are closed under intersection and generate the Borel \(\sigma\)-algebra of \([0,1]\). By the monotone class theorem [Fremlin, 136B], \(\mathcal J\) contains all Borel sets. \(\square\)

The measure \(h_z\) is the conditional distribution of the second coordinate given the first. Two families with the property agree for \(\lambda\)-almost every \(z\) on every dyadic interval, hence as measures; this uniqueness is not used.

6. Exercises

6.1 (easy). Show that the unit masses \(\delta_{1/n}\) on \([0,1]\) converge weakly to \(\delta_0\), and that \(\delta_{1/n}((0,1])=1\) does not converge to \(\delta_0((0,1])\). Which part of Lemma 2.1 explains this?

6.2 (easy). Show that the sets \(\{0,1/n,2/n,\dots,1\}\) converge in Hausdorff distance to \([0,1]\), and that the measures \(\frac1{n+1}\sum_{i=0}^n\delta_{i/n}\) converge weakly to Lebesgue measure on \([0,1]\).

6.3 (medium). In Proposition 5.1 let \(\nu\) be the image of Lebesgue measure on \([0,1]\) under \(t\mapsto(t,t)\), with \(Z=[0,1]\). Find \(h_z\).

7. Solutions

6.1. For continuous \(f\), \(f(1/n)\to f(0)\). The set \((0,1]\) has boundary \(\{0\}\), which carries the full mass of \(\delta_0\), so Lemma 2.1(b) does not apply; (a) gives only \(\liminf\delta_{1/n}((0,1])\ge\delta_0((0,1])=0\).

6.2. Every point of \([0,1]\) is within \(1/(2n)\) of the finite set, which lies in \([0,1]\). For continuous \(f\), \(\frac1{n+1}\sum f(i/n)\) is a Riemann sum up to a term \(O(\|f\|/n)\), and converges to \(\int_0^1f\).

6.3. \(\lambda\) is Lebesgue measure on \(Z=[0,1]\), and \(\nu(F\times J)=|F\cap J|=\int_F\mathbf1_J(z)\,dz\). So \(h_z=\delta_z\) for almost every \(z\): the second coordinate is determined by the first.

References