Proof of A Compact Subset of a Metric Space as a Metric Space: Compactness, Local Compactness, Separability, Continuous Functions and Finite Borel Measures

Throughout, maps given by explicit formulas, such as the enumerations of Step 3 and the map $(m,i)\mapsto x_{m}(i)$, are maps by Maps and Relations Given by Formulas (i) (with The Graph Reading of a Formula Tree Is Predicative and Agrees with Its Reading, So the Reading Is Predicative (iii), Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (i) and Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (iii)).

Step 1 (the topology of $K$; proves (i)). By Subsets of a Metric Space: the Restricted Metric, Sequences, Limits in Closed Subsets and Completeness of Closed Subsets (i), $(K,d_{K})$ is a metric space, and by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (v) its open sets are the sets $V\cap K$ with $V$ open in $(S,d)$. So $\mathcal{T}_{d_{K}}$ is the subspace topology of $\mathcal{T}_{d}$ on $K$, and compactness of the space $(K,\mathcal{T}_{d_{K}})$ is, by Compact Topological Space and Compact Subset (ii), the hypothesis that $K$ is compact in $(S,\mathcal{T}_{d})$. In particular $K$ is a compact subset of $(K,\mathcal{T}_{d_{K}})$, and by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (i), applied to the metric space $(K,d_{K})$, every cover of $K$ by open subsets of $(K,d_{K})$ has a finite subcover.

Step 2 (proves (ii)). Let $x\in K$; we show that the radius $1$ works. The closed ball $\bar{B}_{d_{K}}(x,1)$ of $(K,d_{K})$ is closed in $(K,\mathcal{T}_{d_{K}})$ by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (iv), applied to $(K,d_{K})$, and it lies in $K$, which is compact in $(K,\mathcal{T}_{d_{K}})$ by Step 1. Hence $\bar{B}_{d_{K}}(x,1)$ is compact by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (i).

Step 3 (proves (iii)). If $K=\emptyset$, the empty set is countable (Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (iii)) and dense in $K$. Let now $K\neq\emptyset$ and fix $x_{0}\in K$. Since $\mathbb{Q}$ is countable (Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (ix)), there is a surjection $q:\mathbb{N}\to\mathbb{Q}$; write $\rho_{m}=\kappa(q(m))$, with $\kappa:\mathbb{Q}\to\mathbb{R}$ the canonical embedding.

Let $m\in\mathbb{N}$ with $\rho_{m}>0$. The open balls $B_{d_{K}}(x,\rho_{m})$, $x\in K$, are open in $(K,d_{K})$ by Open Ball in a Metric Space is Open (i), and they cover $K$, since $x\in B_{d_{K}}(x,\rho_{m})$. By Step 1 a finite set $\mathcal{V}_{m}$ of these balls covers $K$. It is nonempty, as $K\neq\emptyset$, and countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (iii), so there is a surjection $e_{m}:\mathbb{N}\to\mathcal{V}_{m}$. Each ball $e_{m}(i)$ has a centre in $K$, and choosing one for every $i\in\mathbb{N}$ by Countable Choice Follows from Dependent Choice gives a sequence $y:\mathbb{N}\to K$ with

$$e_{m}(i)=B_{d_{K}}\bigl(y(i),\rho_{m}\bigr)\qquad(i\in\mathbb{N}),$$

so the balls $B_{d_{K}}(y(i),\rho_{m})$, $i\in\mathbb{N}$, cover $K$.

For $m\in\mathbb{N}$ let $X_{m}$ be the set of sequences $x:\mathbb{N}\to K$ whose balls $B_{d_{K}}(x(i),\rho_{m})$, $i\in\mathbb{N}$, cover $K$ if $\rho_{m}>0$, and the set of all sequences $\mathbb{N}\to K$ if $\rho_{m}\le0$. Each $X_{m}$ is nonempty: it contains the sequence $y$ just constructed if $\rho_{m}>0$, and the constant sequence $x_{0}$ otherwise. So countable choice (Countable Choice Follows from Dependent Choice) chooses $x_{m}\in X_{m}$ for every $m\in\mathbb{N}$, and $(x_{m})_{m\in\mathbb{N}}$ is a family of sequences in $K$.

Let $D$ be the image of the map $\mathbb{N}\times\mathbb{N}\to K$, $(m,i)\mapsto x_{m}(i)$. As $\mathbb{N}\times\mathbb{N}$ is countable (Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (ii)), $D$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (vi). It is dense: let $x\in K$ and $r>0$. By the density of the rationals (Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema (iii)) there is a rational number strictly between $0$ and $r$; it equals $\kappa(q(m))$ for some $m$, so $0<\rho_{m}<r$. Since $x_{m}\in X_{m}$ and $\rho_{m}>0$, the balls $B_{d_{K}}(x_{m}(i),\rho_{m})$ cover $K$, so there is $i$ with $x\in B_{d_{K}}(x_{m}(i),\rho_{m})$, and then

$$d_{K}\bigl(x,x_{m}(i)\bigr)<\rho_{m}<r ,$$

so $x_{m}(i)\in D\cap B_{d_{K}}(x,r)$.

Step 4 (proves (iv)). For continuous $g:K\to\mathbb{R}$, its support is a closure, hence closed in $(K,\mathcal{T}_{d_{K}})$ by Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (ii), and lies in $K$, which is compact in $(K,\mathcal{T}_{d_{K}})$ by Step 1; so it is compact by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (i). For continuous $f:S\to\mathbb{R}$, $f|_{K}$ is continuous on the subset $K$ by Restricting a Continuous Map to a Smaller Subset Preserves Continuity (ii), hence continuous on the metric space $(K,d_{K})$ by Continuity on a Subset Is Continuity on the Metric Subspace (ii).

Step 5 (proves (v)). $K$ is closed in $(S,\mathcal{T}_{d})$ by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (ii), so $K\in\mathcal{B}(S)$ by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii), and the displayed equality is Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (vii). The Borel $\sigma$-algebra $\mathcal{B}(K)$, generated by the open sets of $(K,d_{K})$, is a $\sigma$-algebra on $K$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii)). For $B\in\mathcal{B}(S)$ we have $\iota^{-1}(B)=B\cap K\in\mathcal{B}(K)$ by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (vi), so $\iota$ is measurable.

Step 6 (proves (vi)). $\mu_{K}$ is a finite measure. By monotonicity (Basic Properties of a Measure (iii)), $\mu(K)\le\mu(S)<\infty$, so $\mu(K)$ is finite (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (iii)). By Step 5, $\mathcal{B}(K)=\{B\in\mathcal{B}(S):B\subseteq K\}$, so $\mu_{K}$ is a finite measure on $(K,\mathcal{B}(K))$ by Restricting a Measure to a Measurable Set (ii), and $\mu_{K}(K)=\mu(K)$.

Two measures on $S$ agree. The image measure $\mu_{K}\circ\iota^{-1}$ is a finite measure on $(S,\mathcal{B}(S))$ by Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (i), and $\mathbf{1}_{K}\mu$ is the measure with density $\mathbf{1}_{K}$. For $B\in\mathcal{B}(S)$, by Image Measure (i), Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Integrals over Measurable Subsets (i) and Measure with a Density (i),

$$(\mu_{K}\circ\iota^{-1})(B)=\mu(B\cap K)=\int_{S}\mathbf{1}_{B\cap K}\,d\mu=\int_{S}\mathbf{1}_{B}\mathbf{1}_{K}\,d\mu=\int_{B}\mathbf{1}_{K}\,d\mu=(\mathbf{1}_{K}\mu)(B).$$

So $\mathbf{1}_{K}\mu=\mu_{K}\circ\iota^{-1}$; in particular $\mathbf{1}_{K}\mu$ is finite.

The integral formula. A bounded measurable $f:S\to\mathbb{R}$ is integrable for both finite measures (Bounded Measurable Functions on a Finite Measure Space Are Integrable (i)). Change of variables with $T=\iota$, for which $f\circ\iota=f|_{K}$ (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iii)), and integration against the density $h=\mathbf{1}_{K}$ (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (vii)) give

$$\int_{K}f|_{K}\,d\mu_{K}=\int_{S}f\,d(\mu_{K}\circ\iota^{-1})=\int_{S}f\,d(\mathbf{1}_{K}\mu)=\int_{S}f\,\mathbf{1}_{K}\,d\mu .$$

Step 7 (proves (vii)). By Step 5, $\iota$ is measurable, so $\nu\circ\iota^{-1}$ is a finite measure on $(S,\mathcal{B}(S))$ by Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (i), and $(\nu\circ\iota^{-1})(B)=\nu(\iota^{-1}(B))=\nu(B\cap K)$ by Image Measure (i). A bounded measurable $f:S\to\mathbb{R}$ is integrable for the finite measure $\nu\circ\iota^{-1}$ (Bounded Measurable Functions on a Finite Measure Space Are Integrable (i)), and $f\circ\iota=f|_{K}$, so change of variables with $T=\iota$ and $g=f$ (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iii)) gives

$$\int_{S}f\,d(\nu\circ\iota^{-1})=\int_{K}f|_{K}\,d\nu .$$