Proof of Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight

Throughout, open balls of positive radius are open (Open Ball in a Metric Space is Open (i)), unions of open sets are open, open sets are Borel (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (i)), and $\mu(S)$ is a real number.

Fix a real number $\varepsilon>0$. If $S=\emptyset$, take $K=\emptyset$, which is compact 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) (the empty subfamily of any open cover covers it); then $\mu(S\setminus K)=\mu(\emptyset)=0<\varepsilon$. Otherwise let $D\subseteq S$ be countable and dense; it is nonempty, as it meets every open ball, so by Countable Set (i) there is a sequence $(x_{i})_{i\in\mathbb{N}}$ with $D=\{x_{i}:i\in\mathbb{N}\}$. For $N\in\mathbb{N}$ let $A_{N}=\{x_{1},\dots,x_{N}\}$, a nonempty finite set (the image of $[N]$ under $i\mapsto x_{i}$, Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi)), and for real $r>0$ let

$$U_{N}(r)=\bigcup_{i=1}^{N}B_{d}(x_{i},r),$$

an open set.

Step 1 (a chain of open sets of small measure). Put $t=\varepsilon/2$. Let $b$ be the set of pairs $(r,W)$ with $r>0$ real, $W\subseteq S$ open and $\mu(W)<t$, and let $R$ relate $(r,W)\in b$ to $(r',W')\in b$ when $r'=r/2$, $W\subseteq W'$ and $S\setminus W'\subseteq U_{N}(r)$ for some $N\in\mathbb{N}$. We show that every $(r,W)\in b$ is related by $R$ to some element of $b$.

Let $r'=r/2$. The open sets $U_{N}(r')\cup W$ increase with $N$, and their union is $S$: every $y\in S$ has, by density, some $x_{i}$ with $d(x_{i},y)<r'$, so $y\in U_{i}(r')$. By continuity from below (Basic Properties of a Measure (viii)), $\mu(S)$ is the supremum of the numbers $\mu(U_{N}(r')\cup W)$. Since $\mu(W)<t$, the number $\mu(S)+\mu(W)-t$ is smaller than $\mu(S)$, hence not an upper bound of them, and there is $N$ with

$$\mu\bigl(U_{N}(r')\cup W\bigr)>\mu(S)+\mu(W)-t. \tag{1}$$

Let

$$V=\{y\in S:\operatorname{dist}(y,A_{N})\le r'\},$$

with the distance to the nonempty set $A_{N}$. It is closed by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (vi), so $W'=W\cup(S\setminus V)$ is open. If $d(x_{i},y)<r'$ with $i\le N$, then $\operatorname{dist}(y,A_{N})\le d(y,x_{i})<r'$; hence $U_{N}(r')\subseteq V$, and $W'$ contains no point of $U_{N}(r')\setminus W$. As $U_{N}(r')\setminus W=(U_{N}(r')\cup W)\setminus W$, Basic Properties of a Measure (iii), Basic Properties of a Measure (v), Basic Properties of a Measure (iv) and (1) give

$$\mu(W')\le\mu\bigl(S\setminus(U_{N}(r')\setminus W)\bigr)=\mu(S)-\mu\bigl(U_{N}(r')\cup W\bigr)+\mu(W)<t .$$

Finally, $S\setminus W'\subseteq V\subseteq U_{N}(r)$: if $\operatorname{dist}(y,A_{N})\le r'<r$, then $r$ is not a lower bound of the distances $d(y,x_{i})$, $i\le N$, whose infimum is $\operatorname{dist}(y,A_{N})$, so $d(x_{i},y)<r$ for some $i\le N$. Thus $(r',W')\in b$, $W\subseteq W'$, and $(r,W)$ is related to $(r',W')$ by $R$.

Since $\mu(\emptyset)=0<t$, the pair $(1,\emptyset)$ lies in $b$. By the axiom of dependent choice there is a sequence $((r_{m},W_{m}))_{m\in\mathbb{N}}$ in $b$ with $(r_{1},W_{1})=(1,\emptyset)$ and $(r_{m},W_{m})$ related to $(r_{m+1},W_{m+1})$ by $R$ for every $m$. So each $W_{m}$ is open with $\mu(W_{m})<t$, $W_{m}\subseteq W_{m+1}$, $r_{m+1}=r_{m}/2$, and $S\setminus W_{m+1}\subseteq U_{N}(r_{m})$ for some $N$. The radii become arbitrarily small: the set $\{r_{m}:m\in\mathbb{N}\}$ is nonempty and bounded below by $0$, so it has an infimum $\iota\ge0$ (Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema (i)); if $\iota>0$, then $2\iota$ is not a lower bound, so $r_{m}<2\iota$ for some $m$, and $r_{m+1}=r_{m}/2<\iota$, a contradiction. Hence $\iota=0$, and for every real $\delta>0$ there is $m$ with $r_{m}<\delta$.

Step 2 (the compact set). Let

$$K=S\setminus\bigcup_{m\in\mathbb{N}}W_{m}.$$

It is closed in $(S,\mathcal{T}_{d})$, its complement $\bigcup_{m}W_{m}$ being open. It is totally bounded: given $\delta>0$, choose $m$ with $r_{m}<\delta$; then $K\subseteq S\setminus W_{m+1}\subseteq U_{N}(r_{m})$ for some $N$, so every $y\in K$ lies in $B_{d}(x_{i},\delta)$ for some $x_{i}$ in the finite set $A_{N}$. Since $(S,d)$ is complete, $K$ is compact in $(S,\mathcal{T}_{d})$ by A Closed Totally Bounded Subset of a Complete Metric Space is Compact (ii) (which assumes the axiom of dependent choice).

Step 3 (the mass outside $K$; proves (i)). The sets $W_{m}$ increase and their union is $S\setminus K$, so by continuity from below (Basic Properties of a Measure (viii))

$$\mu(S\setminus K)=\sup_{m}\mu(W_{m})\le t=\frac{\varepsilon}{2}<\varepsilon .$$

As $\varepsilon$ was arbitrary, the family consisting of $\mu$ alone is tight by Tight Family of Borel Measures on a Metric Space (i).