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

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is used. Throughout, $\bar{B}$, $U$, $d_{\bar{B}}$, $d_{U}$, $h$ and $g$ are as in The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions, read at the dimension $m$, and $\mathcal{T}$ is the collection of subsets of $\mathbb{R}^{m}$ open in $(\mathbb{R}^{m},d_{E})$. Write $\mathcal{B}(\bar{B})$ and $\mathcal{B}(U)$ for the Borel $\sigma$-algebras of the metric spaces $(\bar{B},d_{\bar{B}})$ and $(U,d_{U})$.

Step 0 (a bound used three times). Let $(Z,d_{Z})$ be a metric space, let $\nu$ be a Borel measure on $(Z,d_{Z})$ with $\nu(Z)<\infty$, let $A$ belong to the Borel $\sigma$-algebra $\mathcal{B}(Z)$ of $(Z,d_{Z})$, let $M$ be a nonnegative real number, and let $u:Z\to\mathbb{R}$ be measurable with respect to $\mathcal{B}(Z)$ and the Borel $\sigma$-algebra of the real line, with $|u(z)|\le M$ for every $z\in Z$ and $u(z)=0$ for every $z\in Z\setminus A$. Then $u$ is integrable with respect to $\nu$ and

$$\Bigl|\int_{Z}u\,d\nu\Bigr|\le M\,\nu(A).$$

Indeed, $u$ is integrable with respect to $\nu$ 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 (vi), applied to $(Z,d_{Z})$, the set $A$, the measure $\nu$ and the function $u$, the restriction $u|_{A}$ is measurable with respect to $\mathcal{B}(A)$, is 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 (v), $\nu|_{A}$ is a Borel measure on $(A,d_{A})$ with $\nu|_{A}(A)=\nu(A)$, which is finite because $\nu(A)\le\nu(Z)<\infty$ by claim 2 of Basic Properties of a Measure. Since $|u|_{A}(z)|\le M$ for every $z\in A$, claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, applied on $(A,d_{A})$, gives $\bigl|\int_{A}u|_{A}\,d(\nu|_{A})\bigr|\le M\,\nu|_{A}(A)=M\,\nu(A)$.

Step 1 (transport to the closed unit ball). By claims 2 and 3 of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions, $h(x)\in U\subseteq\bar{B}$ for every $x\in\mathbb{R}^{m}$, so the map $h_{\bar{B}}:\mathbb{R}^{m}\to\bar{B}$ with $h_{\bar{B}}(x)=h(x)$ is well defined. By claim 3 of that lemma $h$ is continuous on $\mathbb{R}^{m}$ as a map into $(\mathbb{R}^{m},d_{E})$, so $h_{\bar{B}}$ is continuous on $\mathbb{R}^{m}$ as a map into $(\bar{B},d_{\bar{B}})$ by The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure (iii). By claim 3 of Borel Measurability and Bounded Integration on a Metric Space, $h_{\bar{B}}$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{m})$ and $\mathcal{B}(\bar{B})$.

For each $n\in\mathbb{N}$ let $\tilde{\mu}_{n}$ be the image measure of $\mu_{n}$ under $h_{\bar{B}}$, formed by claim 1 of that lemma for the measure space $(\mathbb{R}^{m},\mathcal{B}(\mathbb{R}^{m}),\mu_{n})$ and the measurable space $(\bar{B},\mathcal{B}(\bar{B}))$. By that claim $\tilde{\mu}_{n}$ is a measure on $(\bar{B},\mathcal{B}(\bar{B}))$, that is, a Borel measure on $(\bar{B},d_{\bar{B}})$ by Borel Measure on a Metric Space, and $\tilde{\mu}_{n}(\bar{B})=\mu_{n}(\mathbb{R}^{m})=1$.

Step 2 (extraction). The metric space $(\bar{B},d_{\bar{B}})$ is compact by claim 2 of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions. By Weak Sequential Compactness of Borel Measures of Total Mass One on a Compact Metric Space, applied to it and to the sequence $(\tilde{\mu}_{n})_{n\in\mathbb{N}}$, there are a strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}$ and a Borel measure $\tilde{\mu}$ on $(\bar{B},d_{\bar{B}})$ with $\tilde{\mu}(\bar{B})=1$ such that $(\tilde{\mu}_{n_{j}})_{j\in\mathbb{N}}$ converges weakly to $\tilde{\mu}$. These $(n_{j})_{j\in\mathbb{N}}$ and $\tilde{\mu}$ are fixed for the rest of the proof, before any test function is chosen.

Step 3 (cutoffs adapted to a tolerance). Let $\varepsilon\in\mathbb{R}$ satisfy $0<\varepsilon$. Since $(\mu_{n})_{n\in\mathbb{N}}$ is tight, Tight Family of Borel Measures on a Metric Space (ii) provides a set $K\subseteq\mathbb{R}^{m}$, compact in $(\mathbb{R}^{m},\mathcal{T})$, with $\mu_{n}(\mathbb{R}^{m}\setminus K)\le\varepsilon$ for every $n\in\mathbb{N}$. Distinguish two cases.

Case A: $K$ is empty. Then $\mathbb{R}^{m}\setminus K=\mathbb{R}^{m}$, so $1=\mu_{1}(\mathbb{R}^{m})\le\varepsilon$.

Case B: $K$ is nonempty. By claim 4 of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions there is a real $\rho$ with $0\le\rho<1$ and $\lVert h(x)\rVert\le\rho$ for every $x\in K$. Let $\sigma=2^{-1}(1+\rho)$ and let $\chi:\mathbb{R}^{m}\to\mathbb{R}$ be the map given by claim 5 of that lemma for this $\rho$, so that $\rho<\sigma<1$, $\chi$ is continuous on $\mathbb{R}^{m}$, $0\le\chi\le 1$, $\chi(y)=1$ whenever $\lVert y\rVert\le\rho$, and $\chi(y)=0$ whenever $\sigma\le\lVert y\rVert$. In particular

$$\chi(h(x))=1\qquad\text{for every }x\in K. \tag{i}$$

Let $\chi_{\bar{B}}$ be the restriction of $\chi$ to $\bar{B}$. It is continuous on $\bar{B}$ as a map from $(\bar{B},d_{\bar{B}})$ to $\mathbb{R}$, by claim 4 of Semicontinuity and Continuity Under Composition with a Continuous Map followed by The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure (ii), and $|\chi_{\bar{B}}|\le 1$; so it is measurable with respect to $\mathcal{B}(\bar{B})$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space and integrable with respect to each of $\tilde{\mu}_{n}$ and $\tilde{\mu}$ by claim 6(b) of that lemma. Since $h(x)\in\bar{B}$ for every $x$, the composition $\chi_{\bar{B}}\circ h_{\bar{B}}$ is the map $x\mapsto\chi(h(x))$, and claim 2 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables gives, for every $n$,

$$\int_{\mathbb{R}^{m}}\chi(h(x))\,\mu_{n}(dx)=\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}_{n}.$$

The map $x\mapsto 1-\chi(h(x))$ is continuous on $\mathbb{R}^{m}$, hence measurable with respect to $\mathcal{B}(\mathbb{R}^{m})$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space, takes values in the interval from $0$ to $1$, and vanishes on $K$ by (i). The set $\mathbb{R}^{m}\setminus K$ belongs to $\mathcal{B}(\mathbb{R}^{m})$ by Compact Subsets of a Metric Space are Closed and Borel (ii), so Step 0, applied with $Z=\mathbb{R}^{m}$, $\nu=\mu_{n}$, $A=\mathbb{R}^{m}\setminus K$ and $M=1$, gives $\int_{\mathbb{R}^{m}}(1-\chi(h(x)))\,\mu_{n}(dx)\le\mu_{n}(\mathbb{R}^{m}\setminus K)\le\varepsilon$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion and claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space,

$$\int_{\mathbb{R}^{m}}\bigl(1-\chi(h(x))\bigr)\,\mu_{n}(dx)=1-\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}_{n},$$

so $1-\varepsilon\le\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}_{n}$ for every $n\in\mathbb{N}$. By Step 2 the sequence $\bigl(\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}_{n_{j}}\bigr)_{j\in\mathbb{N}}$ converges to $\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}$, so comparison with the constant sequence of value $1-\varepsilon$, by claim 1 of Order Properties of Limits of Real Sequences, gives

$$1-\varepsilon\le\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}. \tag{ii}$$

Step 4 (the limit measure). We show that $\tilde{\mu}(U)=1$. By claim 2 of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions the set $U$ is open in $(\bar{B},d_{\bar{B}})$, hence $U\in\mathcal{B}(\bar{B})$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space.

Let $\varepsilon\in\mathbb{R}$ satisfy $0<\varepsilon$ and run Step 3 with it. In Case A we have $1\le\varepsilon$, so $1-\varepsilon\le 0\le\tilde{\mu}(U)$. In Case B, let $D=\{y\in\bar{B}:\lVert y\rVert<\sigma\}$. Then $D=\bar{B}\cap V$, where $V$ is the open ball of $(\mathbb{R}^{m},d_{E})$ with centre $0$ and radius $\sigma$, open by Open Ball in a Metric Space is Open; so $D$ is open in $(\bar{B},d_{\bar{B}})$ by claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology and $D\in\mathcal{B}(\bar{B})$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space. Since $\sigma<1$ we have $D\subseteq U$. The map $\chi_{\bar{B}}$ vanishes at every $y\in\bar{B}\setminus D$, because such a $y$ satisfies $\sigma\le\lVert y\rVert$. Step 0, applied with $Z=\bar{B}$, $\nu=\tilde{\mu}$, $A=D$ and $M=1$, therefore gives $\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}\le\tilde{\mu}(D)$, and $\tilde{\mu}(D)\le\tilde{\mu}(U)$ by claim 2 of Basic Properties of a Measure. With (ii) this gives $1-\varepsilon\le\tilde{\mu}(U)$ again.

Thus $1-\varepsilon\le\tilde{\mu}(U)$ for every positive real $\varepsilon$, so $1\le\tilde{\mu}(U)$ by claim 2 of Comparison of Real Numbers with Arbitrary Positive Slack; and $\tilde{\mu}(U)\le\tilde{\mu}(\bar{B})=1$ by claim 2 of Basic Properties of a Measure. Hence $\tilde{\mu}(U)=1$.

By The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure (v), applied to $(\bar{B},d_{\bar{B}})$ and the set $U\in\mathcal{B}(\bar{B})$, the restriction $\tilde{\mu}|_{U}$ is a Borel measure on $(U,d_{U})$ with $\tilde{\mu}|_{U}(U)=\tilde{\mu}(U)=1$. By claim 3 of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions the map $g$ is continuous on $U$ as a map from $(U,d_{U})$ into $(\mathbb{R}^{m},d_{E})$, hence measurable with respect to $\mathcal{B}(U)$ and $\mathcal{B}(\mathbb{R}^{m})$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. Let $\mu$ be the image measure of $\tilde{\mu}|_{U}$ under $g$, formed by claim 1 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables; by that claim $\mu$ is a measure on $(\mathbb{R}^{m},\mathcal{B}(\mathbb{R}^{m}))$ with $\mu(\mathbb{R}^{m})=\tilde{\mu}|_{U}(U)=1$, so $\mu\in\mathcal{P}(\mathbb{R}^{m})$ by Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iii).

Step 5 (weak convergence). Let $f:\mathbb{R}^{m}\to\mathbb{R}$ be bounded and continuous on $\mathbb{R}^{m}$, and let $M$ be a nonnegative real number with $|f(x)|\le M$ for every $x\in\mathbb{R}^{m}$. Being continuous, $f$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{m})$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space, and integrable with respect to every probability measure on $\mathbb{R}^{m}$ by claim 6(b) of that lemma. Let $\eta\in\mathbb{R}$ satisfy $0<\eta$. The choices are made in this order: $\eta$ determines $\varepsilon$, then Step 3 determines $K$ and, in Case B, $\rho$, $\sigma$ and $\chi$, then $F$, and finally the index $N$.

Put $\varepsilon=\eta\bigl(2(2M+1)\bigr)^{-1}$, a positive real number, and run Step 3 with this $\varepsilon$.

Case A. Here $1\le\varepsilon$, so $\eta=2(2M+1)\varepsilon\ge 2(2M+1)$ and therefore $2M<\eta$. By claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space we have $\bigl|\int_{\mathbb{R}^{m}}f\,d\mu_{n_{j}}\bigr|\le M$ for every $j$ and $\bigl|\int_{\mathbb{R}^{m}}f\,d\mu\bigr|\le M$, so by claim 5 of Properties of the Absolute Value in an Ordered Field the difference of the two integrals has absolute value at most $2M$, hence less than $\eta$, for every $j\in\mathbb{N}$.

Case B. Let $F:\bar{B}\to\mathbb{R}$ be the map given by claim 6 of The Radial Compactification of Euclidean Space: Norm Continuity, a Homeomorphism onto the Open Unit Ball, Radial Cutoffs, and Extension of Test Functions for the data $\rho$, $\sigma$, $\chi$ and $f$ of Step 3. Thus $F$ is continuous on $\bar{B}$ as a map from $(\bar{B},d_{\bar{B}})$, $|F|\le M$, $F(h(x))=\chi(h(x))f(x)$ for every $x\in\mathbb{R}^{m}$, and $F(y)=0$ whenever $y\in\bar{B}$ and $\sigma\le\lVert y\rVert$; in particular $F(y)=0$ for every $y\in\bar{B}\setminus U$, since such a $y$ satisfies $\lVert y\rVert=1$ and $\sigma<1$. By claims 3 and 6(b) of Borel Measurability and Bounded Integration on a Metric Space, $F$ is measurable with respect to $\mathcal{B}(\bar{B})$ and integrable with respect to $\tilde{\mu}_{n}$ for every $n$ and with respect to $\tilde{\mu}$.

(a) The approximating side. Since $F\circ h_{\bar{B}}$ is the map $x\mapsto\chi(h(x))f(x)$, claim 2 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables gives $\int_{\bar{B}}F\,d\tilde{\mu}_{n}=\int_{\mathbb{R}^{m}}\chi(h(x))f(x)\,\mu_{n}(dx)$ for every $n$, and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives

$$\int_{\mathbb{R}^{m}}f\,d\mu_{n}-\int_{\bar{B}}F\,d\tilde{\mu}_{n}=\int_{\mathbb{R}^{m}}\bigl(1-\chi(h(x))\bigr)f(x)\,\mu_{n}(dx).$$

The integrand is continuous, hence measurable; it has absolute value at most $M$, because $0\le 1-\chi(h(x))\le 1$ and $|f|\le M$; and it vanishes on $K$ by (i). Step 0, applied with $Z=\mathbb{R}^{m}$, $\nu=\mu_{n}$ and $A=\mathbb{R}^{m}\setminus K$, gives

$$\Bigl|\int_{\mathbb{R}^{m}}f\,d\mu_{n}-\int_{\bar{B}}F\,d\tilde{\mu}_{n}\Bigr|\le M\,\mu_{n}(\mathbb{R}^{m}\setminus K)\le M\varepsilon\qquad(n\in\mathbb{N}).$$

(b) The limit side. By claim 2 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables, applied to the image measure $\mu$ of $\tilde{\mu}|_{U}$ under $g$ and to the bounded measurable map $f$, we have $\int_{\mathbb{R}^{m}}f\,d\mu=\int_{U}f\circ g\,d(\tilde{\mu}|_{U})$. Since $F$ vanishes on $\bar{B}\setminus U$, The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure (vi), applied to $(\bar{B},d_{\bar{B}})$, the set $U$, the measure $\tilde{\mu}$ and the function $F$, gives $\int_{\bar{B}}F\,d\tilde{\mu}=\int_{U}F|_{U}\,d(\tilde{\mu}|_{U})$, and $F|_{U}$ is the map $y\mapsto\chi(y)f(g(y))$ on $U$. Writing $\chi_{U}$ for the restriction of $\chi$ to $U$, claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives

$$\int_{\mathbb{R}^{m}}f\,d\mu-\int_{\bar{B}}F\,d\tilde{\mu}=\int_{U}\bigl(1-\chi_{U}\bigr)\,(f\circ g)\,d(\tilde{\mu}|_{U}),$$

all the integrands being continuous on $U$, hence measurable with respect to $\mathcal{B}(U)$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space, and bounded, hence integrable by claim 6(b) of that lemma. Pointwise on $U$,

$$-M\bigl(1-\chi_{U}\bigr)\le\bigl(1-\chi_{U}\bigr)(f\circ g)\le M\bigl(1-\chi_{U}\bigr),$$

since $0\le 1-\chi_{U}\le 1$ and $|f|\le M$; integrating and using the monotonicity and the linearity of claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, together with claim 6 of Properties of the Absolute Value in an Ordered Field,

$$\Bigl|\int_{\mathbb{R}^{m}}f\,d\mu-\int_{\bar{B}}F\,d\tilde{\mu}\Bigr|\le M\int_{U}\bigl(1-\chi_{U}\bigr)\,d(\tilde{\mu}|_{U})=M\Bigl(1-\int_{U}\chi_{U}\,d(\tilde{\mu}|_{U})\Bigr),$$

the last step by claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space and $\tilde{\mu}|_{U}(U)=1$. Finally $\chi_{\bar{B}}$ vanishes on $\bar{B}\setminus U$, so The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure (vi) gives $\int_{U}\chi_{U}\,d(\tilde{\mu}|_{U})=\int_{\bar{B}}\chi_{\bar{B}}\,d\tilde{\mu}$, which is at least $1-\varepsilon$ by (ii). Hence

$$\Bigl|\int_{\mathbb{R}^{m}}f\,d\mu-\int_{\bar{B}}F\,d\tilde{\mu}\Bigr|\le M\varepsilon .$$

(c) Conclusion. The map $F$ is bounded and continuous on $\bar{B}$, so by Step 2 the sequence $\bigl(\int_{\bar{B}}F\,d\tilde{\mu}_{n_{j}}\bigr)_{j\in\mathbb{N}}$ converges to $\int_{\bar{B}}F\,d\tilde{\mu}$; by Limit of a Sequence of Real Numbers there is $N\in\mathbb{N}$ such that

$$\Bigl|\int_{\bar{B}}F\,d\tilde{\mu}_{n_{j}}-\int_{\bar{B}}F\,d\tilde{\mu}\Bigr|<\varepsilon\qquad\text{for every }j\in\mathbb{N}\text{ with }N\le j .$$

For such $j$, claim 5 of Properties of the Absolute Value in an Ordered Field, applied twice to the decomposition of $\int f\,d\mu_{n_{j}}-\int f\,d\mu$ into the three differences bounded in (a), (c) and (b), gives

$$\Bigl|\int_{\mathbb{R}^{m}}f\,d\mu_{n_{j}}-\int_{\mathbb{R}^{m}}f\,d\mu\Bigr|\le M\varepsilon+\varepsilon+M\varepsilon=(2M+1)\varepsilon=2^{-1}\eta<\eta .$$

In both cases we have produced, for the given $\eta$, an index $N$ beyond which the difference of the two integrals has absolute value less than $\eta$. As $\eta$ was an arbitrary positive real number, $\bigl(\int_{\mathbb{R}^{m}}f\,d\mu_{n_{j}}\bigr)_{j\in\mathbb{N}}$ converges to $\int_{\mathbb{R}^{m}}f\,d\mu$ in the sense of Limit of a Sequence of Real Numbers. As $f$ was an arbitrary bounded continuous real-valued map on $\mathbb{R}^{m}$, the subsequence $(\mu_{n_{j}})_{j\in\mathbb{N}}$ converges weakly to $\mu$ on $(\mathbb{R}^{m},d_{E})$ in the sense of Weak Convergence of Finite Borel Measures on a Metric Space.