Every borel probability on a polish space is tight

Assuming countable choice, Borel probability measures on Polish spaces are inner regular: Assume countable choice. If $P$ is Polish and $\mu$ is a Borel probability measure on $P$, then for every Borel $A\subseteq P$ and $\epsilon >0$ there is a compact $K\subseteq A$ with $\mu (A\setminus K)<\epsilon$.

Tight family of probability measures: A family $\mathcal{A}$ of Borel probabilities on a metric space S is tight if, for every $\epsilon >0$, there is a compact $K\subseteq S$ such that $\mu (S\setminus K)<\epsilon$ for every $\mu \in \mathcal{A}$. One K must work for the whole family. Compactness is def-metric-compactness. The empty family is tight, witnessed by the empty compact set.

Continuity from below for measures: Let $(E_n{)}_{n\in \mathbb{N}}$ be an increasing sequence of measurable sets for a measure $\mu$, so $E_n\subseteq E_{n+1}$. Then

$\mu \left({\bigcup }_{n\in \mathbb{N}}{E}_n\right)={\sup }_{n\in \mathbb{N}}\mu (E_n)\mathrm{.}$

No finiteness hypothesis is required.

Finite and countable subadditivity of measures: Let $\mu$ be a measure and let $(E_k{)}_{k\in \mathbb{N}}$ be measurable. Then

$\mu \left({\bigcup }_{k\in \mathbb{N}}{E}_k\right)\le {\sum }_{k=0}^{\mathrm{\infty }}\mu (E_k)\mathrm{.}$

For every $m\in \mathbb{N}$ one also has

$\mu \left({\bigcup }_{k<m}{E}_k\right)\le {\sum }_{k<m}\mu (E_k),$

including $m=0$, where both sides are $0$.

A complete, totally bounded metric space is compact, proved from countable choice used exactly once: Assume the Axiom of Countable Choice (def-countable-choice). Let $(X,d)$ be a metric space (def-metric-space) that is complete (def-complete-metric-space) and totally bounded (def-totally-bounded). Then $(X,d)$ is compact (def-metric-compactness).

Where the axiom is spent, and why the weaker principle suffices. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used exactly once, at step 3.1, to fix one finite $1\mathrm{/}(n+1)$-net together with a listing of it for every $n\in \mathbb{N}$ at once. The family of sets being chosen from is written down before any selection is made and does not depend on the earlier selections, which is precisely the situation countable choice covers and dependent choice (def-dependent-choice) is not needed for. Everything after step 3.1 is canonical: at each stage the construction takes the least admissible index in the listing already fixed.

As always on this page, the claim is an upper bound on the cost of the proof given here, not an assertion that ${\mathrm{A}\mathrm{C}}_{\omega }$ is necessary for the theorem.