TheoremBase

Enumerate a countable dense set; for each m, continuity from below picks finitely many closed balls of radius 1/m whose union misses mass less than eps/2^m. The intersection over m of these finite unions is closed and totally bounded, hence compact in the complete space, and countable subadditivity bounds the mass of its complement by eps.

Proof

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.

Let B(X)\mathcal{B}(X) be the Borel σ\sigma-algebra of (X,d)(X,d), on which μ\mu is a measure by Borel Measure on a Metric Space, and let ε\varepsilon be a positive real number. By Tight Family of Borel Measures on a Metric Space §tight we must find a set K⊆XK\subseteq X, compact in (X,Td)(X,\mathcal{T}_{d}), with μ(X∖K)≤ε\mu(X\setminus K)\le\varepsilon.

Step 1 (the empty space). Suppose X=∅X=\varnothing and put K=∅K=\varnothing. The set KK is closed in (X,Td)(X,\mathcal{T}_{d}) by claim 1 of Complements, Unions and Intersections of Closed Sets in a Topological Space, and it is totally bounded in (X,d)(X,d) in the sense of Totally Bounded Subset of a Metric Space, the empty finite set F=∅F=\varnothing serving for every radius. Hence KK is compact in (X,Td)(X,\mathcal{T}_{d}) by A Closed Totally Bounded Subset of a Complete Metric Space is Compact §compact, applied to the complete metric space (X,d)(X,d) and the set KK, and μ(X∖K)=μ(∅)=0≤ε\mu(X\setminus K)=\mu(\varnothing)=0\le\varepsilon by Measure, Measure Space, and Probability Measure. From now on we assume X≠∅X\neq\varnothing.

Step 2 (an enumerated dense set, and coverings by closed balls). By Separable Metric Space there is a countable set D⊆XD\subseteq X that is dense in XX for Td\mathcal{T}_{d}, that is, cl⁡X(D)=X\operatorname{cl}_{X}(D)=X by Dense Subset of a Topological Space. Let x∈Xx\in X and rr a positive real number. Since x∈cl⁡X(D)x\in\operatorname{cl}_{X}(D), the implication from condition 1 to condition 3 of Characterization of the Closure in a Metric Space by Open Balls gives a∈Da\in D with d(x,a)<rd(x,a)<r. Taking any x∈Xx\in X shows that DD is nonempty, so by Countable Set there is a sequence (qi)i∈N(q_{i})_{i\in\mathbb{N}} in DD such that every point of DD equals qiq_{i} for some i∈Ni\in\mathbb{N}. Consequently, for every x∈Xx\in X and every m∈Nm\in\mathbb{N} there is i∈Ni\in\mathbb{N} with d(qi,x)=d(x,qi)<1md(q_{i},x)=d(x,q_{i})<\tfrac{1}{m}, using the symmetry condition 3 of Metric Space; that is, xx lies in the closed ball Bˉd(qi,1m)={y∈X:d(qi,y)≤1m}\bar{B}_{d}(q_{i},\tfrac{1}{m})=\{y\in X:d(q_{i},y)\le\tfrac{1}{m}\} of Closed Ball in a Metric Space.

Step 3 (finite unions of closed balls). For m,N∈Nm,N\in\mathbb{N} let Cm,N=⋃i=1NBˉd(qi,1m)C_{m,N}=\bigcup_{i=1}^{N}\bar{B}_{d}(q_{i},\tfrac{1}{m}). Each of these closed balls is closed in (X,Td)(X,\mathcal{T}_{d}) by claim 3 of Elementary Properties of the Closed Ball in a Metric Space, so Cm,NC_{m,N} is closed in (X,Td)(X,\mathcal{T}_{d}) by claim 2 of Complements, Unions and Intersections of Closed Sets in a Topological Space, and Cm,N∈B(X)C_{m,N}\in\mathcal{B}(X) by claim 1 of Borel Measurability and Bounded Integration on a Metric Space. For fixed mm we have Cm,N⊆Cm,N+1C_{m,N}\subseteq C_{m,N+1} for every NN, and ⋃N∈NCm,N=X\bigcup_{N\in\mathbb{N}}C_{m,N}=X by Step 2.

Step 4 (choice of the radii counts). Fix m∈Nm\in\mathbb{N}. By the monotonicity Basic Properties of a Measure §monotone, μ(Cm,N)≤μ(X)\mu(C_{m,N})\le\mu(X) for every NN, and μ(X)\mu(X) is real by hypothesis; so every μ(Cm,N)\mu(C_{m,N}) is real and the set of these values is bounded above by μ(X)\mu(X). By the continuity from below Basic Properties of a Measure §continuity-below, applied to the increasing sequence (Cm,N)N∈N(C_{m,N})_{N\in\mathbb{N}} with union XX, the sequence (μ(Cm,N))N∈N(\mu(C_{m,N}))_{N\in\mathbb{N}} converges to μ(X)\mu(X). The number ε(12)m\varepsilon\bigl(\tfrac{1}{2}\bigr)^{m} is positive, so by Limit of a Sequence of Real Numbers there is Nm∈NN_{m}\in\mathbb{N} with μ(X)−μ(Cm,Nm)<ε(12)m\mu(X)-\mu(C_{m,N_{m}})<\varepsilon\bigl(\tfrac{1}{2}\bigr)^{m}. Put Cm=Cm,NmC_{m}=C_{m,N_{m}}. By Basic Properties of a Measure §differences,

μ(X∖Cm)=μ(X)−μ(Cm)<ε(12)m.\mu(X\setminus C_{m})=\mu(X)-\mu(C_{m})<\varepsilon\bigl(\tfrac{1}{2}\bigr)^{m}.

The numbers NmN_{m} are chosen for all m∈Nm\in\mathbb{N} after ε\varepsilon has been fixed.

Step 5 (the compact set). Let K=⋂m∈NCmK=\bigcap_{m\in\mathbb{N}}C_{m}. It is closed in (X,Td)(X,\mathcal{T}_{d}) by claim 3 of Complements, Unions and Intersections of Closed Sets in a Topological Space, applied to the family (Cm)m∈N(C_{m})_{m\in\mathbb{N}} of closed sets. It is totally bounded in (X,d)(X,d): given a positive real η\eta, claim 3 of The Archimedean Property of the Real Numbers gives m∈Nm\in\mathbb{N} with 1m<η\tfrac{1}{m}<\eta; the set F={q1,…,qNm}F=\{q_{1},\dots,q_{N_{m}}\} is finite, and every y∈Ky\in K lies in CmC_{m}, hence satisfies d(qi,y)≤1m<ηd(q_{i},y)\le\tfrac{1}{m}<\eta for some i≤Nmi\le N_{m}, so that K⊆⋃a∈FBd(a,η)K\subseteq\bigcup_{a\in F}B_{d}(a,\eta) with the open balls of Open Ball in a Metric Space, as Totally Bounded Subset of a Metric Space requires. Since (X,d)(X,d) is complete, A Closed Totally Bounded Subset of a Complete Metric Space is Compact §compact, applied to (X,d)(X,d) and the closed totally bounded set KK, shows that KK is compact in (X,Td)(X,\mathcal{T}_{d}).

Step 6 (the measure of the complement). We have X∖K=⋃m∈N(X∖Cm)X\setminus K=\bigcup_{m\in\mathbb{N}}(X\setminus C_{m}) with every X∖Cm∈B(X)X\setminus C_{m}\in\mathcal{B}(X), so the countable subadditivity Basic Properties of a Measure §subadditivity gives

μ(X∖K)≤∑m∈Nμ(X∖Cm),\mu(X\setminus K)\le\sum_{m\in\mathbb{N}}\mu(X\setminus C_{m}),

the sum being that of Measure, Measure Space, and Probability Measure. Its terms am=μ(X∖Cm)a_{m}=\mu(X\setminus C_{m}) are real, with 0≤am<ε(12)m0\le a_{m}<\varepsilon\bigl(\tfrac{1}{2}\bigr)^{m} by Step 4. For every n∈Nn\in\mathbb{N}, the geometric sum formula of Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric, read with r=12r=\tfrac{1}{2}, gives ∑m=1n(12)m=1−(12)n≤1\sum_{m=1}^{n}\bigl(\tfrac{1}{2}\bigr)^{m}=1-\bigl(\tfrac{1}{2}\bigr)^{n}\le1, hence

∑m=1nam≤ε∑m=1n(12)m≤ε.\sum_{m=1}^{n}a_{m}\le\varepsilon\sum_{m=1}^{n}\bigl(\tfrac{1}{2}\bigr)^{m}\le\varepsilon .

Thus the partial sums are bounded above by ε\varepsilon, so by Measure, Measure Space, and Probability Measure the sum ∑m∈Nam\sum_{m\in\mathbb{N}}a_{m} is their least upper bound, which is at most ε\varepsilon. Therefore μ(X∖K)≤ε\mu(X\setminus K)\le\varepsilon, with KK compact in (X,Td)(X,\mathcal{T}_{d}). As ε\varepsilon was an arbitrary positive real number, the set {μ}\{\mu\} is tight in (X,d)(X,d) by Tight Family of Borel Measures on a Metric Space §tight.

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