Proof of Random Vectors: Coordinates, Functions, Pairs and Almost-Sure Equality

Throughout, random variables, random vectors and random elements are the measurable functions of Probability Spaces, Events and Random Elements (iv), Probability Spaces, Events and Random Elements (v) and Probability Spaces, Events and Random Elements (iii); a random vector in $\mathbb{R}^{n}$ is a random element of the metric space $(\mathbb{R}^{n},d_{E})$, whose Borel $\sigma$-algebra is $\mathcal{B}(\mathbb{R}^{n})$ (Borel Sigma-Algebra of a Metric Space (iii)).

Coordinates. By Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (i), $X:\Omega\to\mathbb{R}^{n}$ is measurable if and only if each coordinate $X_{k}$ ($k\in[n]$) is measurable, that is, a random variable.

Functions of random elements. Here $X\in\mathrm{Meas}(\Omega,\mathcal{F};S,\mathcal{B}(S))$ and $g\in\mathrm{Meas}(S,\mathcal{B}(S);T,\mathcal{B}(T))$ (Measurable Map, and the Set of Measurable Maps (ii)). By Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iv), applied to the measure $P$, the map $X$ in place of $T$ and the map $g$ in place of $S$, the function $g(X)=g\circ X$ lies in $\mathrm{Meas}(\Omega,\mathcal{F};T,\mathcal{B}(T))$, that is, it is a random element of $T$, and

$$P_{X}\circ g^{-1}=(P\circ X^{-1})\circ g^{-1}=P\circ(g\circ X)^{-1}=P_{g(X)}$$

(The Law of a Random Element (i)). The particular case is $S=\mathbb{R}^{n}$ and $T=\mathbb{R}^{m}$ with their Euclidean distances.

Pairs. By Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (ii), the coordinates of $(X,Y)$ are

$$(X,Y)_{k}=X_{k}\quad(k\in[m]),\qquad(X,Y)_{m+j}=Y_{j}\quad(j\in[n]),$$

and every index in $[m+n]$ has one of these two forms (Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (i)). By (i) applied to $X$ and to $Y$, all these coordinates are random variables, so $(X,Y)$ is a random vector in $\mathbb{R}^{m+n}$ by (i).

Almost-sure equality. The function $X-Y$ is measurable and so is $\lVert X-Y\rVert$ (Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (i), Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (iii)). Since a vector has norm $0$ exactly when it is $0$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i)),

$$\{X\neq Y\}=\{\lVert X-Y\rVert>0\}\in\mathcal{F}$$

by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v).

Now let $X=Y$ almost surely: by Probability Spaces: Notation (vii), A Property Holding Almost Everywhere (i) and Null Set of a Measure (i), there is $N\in\mathcal{F}$ with $\{X\neq Y\}\subseteq N$ and $P(N)=0$. Let $B\in\mathcal{B}(\mathbb{R}^{n})$. If $X(\omega)\in B$ and $\omega\notin N$, then $Y(\omega)=X(\omega)\in B$; hence $\{X\in B\}\subseteq\{Y\in B\}\cup N$, and by Basic Properties of a Measure (iii) and Basic Properties of a Measure (vii),

$$P_{X}(B)=P(X\in B)\le P(Y\in B)+P(N)=P_{Y}(B).$$

Exchanging $X$ and $Y$ gives $P_{Y}(B)\le P_{X}(B)$. So $P_{X}(B)=P_{Y}(B)$ for every $B\in\mathcal{B}(\mathbb{R}^{n})$, that is, $P_{X}=P_{Y}$ (The Law of a Random Element (i)).