Proof of A Tight Family of Finite Borel Measures Has an Increasing Sequence of Compact Sets Carrying All but 1/j of Every Mass

Call a pair $(j,K)$ with $j\in\mathbb{N}$ and $K\subseteq S$ a state if $K$ is compact in $(S,\mathcal{T}_{d})$ and $\mu_{i}(S\setminus K)<1/j$ for every $i\in I$; here $S\setminus K\in\mathcal{B}(S)$, since $K$ is closed (Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (ii)) and closed sets are Borel (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii)). With compactness taken in its open-cover form (Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (i)) and $1/j$ formed from the image of $j$ in $\mathbb{R}$, which is positive, predicatively in $j$ (The Natural Numbers Read in ℝ: a Predicative Condition and a Function from ℕ₀ to ℝ (ii), The Natural Numbers Read in ℝ: a Predicative Condition and a Function from ℕ₀ to ℝ (i), Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vi)), the states form a subset $b$ of $\mathbb{N}\times\mathcal{P}(S)$; let $R$ be the relation on $b$ (Maps and Relations Given by Formulas (iii)) with $((j,K),(j',K'))\in R$ if and only if $j'=j+1$ and $K\subseteq K'$.

Every state $(j,K)$ has an $R$-successor. Indeed, by Tight Family of Borel Measures on a Metric Space (i) with $\varepsilon=1/(j+1)$ there is a set $C\subseteq S$, compact in $(S,\mathcal{T}_{d})$, with $\mu_{i}(S\setminus C)<1/(j+1)$ for every $i\in I$. The set $K\cup C=\bigcup\{K,C\}$ 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 (ii), as $\{K,C\}=\{K\}\cup\{C\}$ is finite (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 (iii), 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 (v)), and

$$\mu_{i}\bigl(S\setminus(K\cup C)\bigr)\le\mu_{i}(S\setminus C)<\frac{1}{j+1}\qquad(i\in I),$$

so $(j+1,K\cup C)$ is a state and $((j,K),(j+1,K\cup C))\in R$. Tightness with $\varepsilon=1$ likewise gives a starting state $(1,K_{1})$.

By Axiom of Dependent Choice (i) there is a sequence $(s_{m})_{m\in\mathbb{N}}$ of states with $s_{1}=(1,K_{1})$ and $(s_{m},s_{m+1})\in R$ for every $m\in\mathbb{N}$. By induction on $m$, the first coordinate of $s_{m}$ is $m$; write $s_{m}=(m,K_{m})$. Then $(K_{j})_{j\in\mathbb{N}}$ is a sequence of subsets of $S$, each compact in $(S,\mathcal{T}_{d})$, with $K_{j}\subseteq K_{j+1}$ because $(s_{j},s_{j+1})\in R$, and with

$$\mu_{i}(S\setminus K_{j})<\frac{1}{j}\qquad\text{for all }i\in I\text{ and }j\in\mathbb{N}$$

because each $s_{j}$ is a state. This proves (i).