Quantile coupling on the real line

Probability laws correspond to distribution functions: Assume the Axiom of Countable Choice.

1. Let $X$ be a real random variable, let ${\mathbb{P}}_X$ be its law, and let $F_X(x)=\mathbb{P}(X\le x)$. Then $F_X$ is nondecreasing and right-continuous, satisfies ${\lim }_{x\to -\mathrm{\infty }}{F}_X(x)=0,{\lim }_{x\to +\mathrm{\infty }}{F}_X(x)=1,$ and obeys ${\mathbb{P}}_X((a,b])=F_X(b)-F_X(a)(a<b)\mathrm{.}$
2. Conversely, if $F:\mathbb{R}\to \mathbb{R}$ is nondecreasing and right-continuous with ${\lim }_{x\to -\mathrm{\infty }}F(x)=0,{\lim }_{x\to +\mathrm{\infty }}F(x)=1,$ then there is a unique Borel probability measure $\mu$ on $\mathbb{R}$ such that $\mu ((a,b])=F(b)-F(a)(a<b),$ equivalently $F(x)=\mu ((-\mathrm{\infty },x])(x\in \mathbb{R})\mathrm{.}$

A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included: Let $n\ge 1$, assume the Axiom of Countable Choice (def-countable-choice), and let $a_i\le b_i$ be reals for $i<n$. Write

$R^{\circ }:=\{x\in {\mathbb{R}}^n:a_i<x_i<b_i\text{ for every }i<n\},\bar{R}:=[a,b]=\{x\in {\mathbb{R}}^n:a_i\le x_i\le b_i\text{ for every }i<n\}$

(def-multidimensional-rectangle-and-volume). Then $R^{\circ }$ is open and $\bar{R}$ is closed, so both are Borel and Lebesgue measurable, and every set $R$ with $R^{\circ }\subseteq R\subseteq \bar{R}$ is Lebesgue measurable with

${\lambda }_n(R)={\prod }_{i<n}(b_i-a_i)\mathrm{.}$

In particular this covers the four one-dimensional face conventions in each coordinate — the open box, the closed box $[a,b]$, the half-open box $B(a,b)={\prod }_{i<n}(a_i,b_i]$ of def-half-open-box, and every mixture of them, in any combination of coordinates — and it gives measure $0$ to all of them whenever $a_i=b_i$ for some $i<n$. For a half-open box with infinite parameters the value is already ${\lambda }_n(B)=\mathrm{vol}(B)$ (thm-lebesgue-measure-is-a-complete-measure).

Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used: Let $I\subseteq \mathbb{R}$ be order-convex (def-interval) and let $f:I\to \mathbb{R}$ be monotone (def-monotone-function). Then the set

$D:=\{c\in I:f\text{ is discontinuous at }c\}$

(def-classification-of-discontinuities) is at most countable (def-countable).

More precisely, the proof exhibits an injection $J:D\to \mathbb{N}$ (def-injection-surjection-bijection) built from one fixed enumeration of the rationals: at a discontinuity $c$ interior to $I$ the value $J(c)$ is read off the least index of a rational lying in the gap $({\lim }_{x\to c^{-}}f(x),{\lim }_{x\to c^{+}}f(x))$, which is a nonempty open interval by thm-monotone-discontinuities-are-jumps. The map $J$ is therefore determined by $f$ and by the fixed enumeration, and no choice principle is used: least indices are canonical by thm-well-ordering-principle, and nothing anywhere in the proof is selected without being determined.

Portmanteau theorem: For Borel probabilities ${\mu }_n,\mu$ on a metric space S, the following are equivalent: (i) ${\mu }_n\Rightarrow \mu$; (ii) integrals converge for all bounded uniformly continuous real tests; (iii) ${\mathrm{lim\hspace{0.17em}sup}}_n{\mu }_n(F)\le \mu (F)$ for every closed F; (iv) ${\mathrm{lim\hspace{0.17em}inf}}_n{\mu }_n(G)\ge \mu (G)$ for every open G; (v) ${\mu }_n(A)\to \mu (A)$ for every Borel A with $\mu (\mathrm{\partial }A)=0$.

Every at most countable subset of ${\mathbb{R}}^n$ is Lebesgue null; in particular ${\lambda }_1(\mathbb{Q})=0$: Let $n\ge 1$ and assume the Axiom of Countable Choice (def-countable-choice). Every at most countable subset $E\subseteq {\mathbb{R}}^n$ (def-countable) is Lebesgue measurable with

${\lambda }_n(E)=0,$

so $E$ is a ${\lambda }_n$-null set (def-measure-null-set-and-almost-everywhere). In particular every singleton is null, and on the real line the set ${\mathbb{Q}}_{\mathbb{R}}$ of rational reals (lem-rat-embeds-dense) satisfies ${\lambda }_1({\mathbb{Q}}_{\mathbb{R}})=0$.