Proof of A Random Vector Is Gaussian with Mean m and Covariance Σ Exactly When Its Law Is N(m,Σ)

Throughout, Borel functions of random vectors are random vectors with the image laws (Random Vectors: Coordinates, Functions, Pairs and Almost-Sure Equality (ii)).

Step 1 ("only if"). If $X$ is Gaussian with mean $m$ and covariance matrix $\Sigma$, then $P_{X}=N(m,\Sigma)$ by The Gaussian Law with Given Mean and Covariance Matrix (i).

Step 2 (a model of the law). Let $P_{X}=N(m,\Sigma)$. By The Law of a Gaussian Random Vector Is Determined by Its Mean and Covariance Matrix (iii) there are a probability space $(\Omega',\mathcal{F}',P')$ and a Gaussian random vector $X'$ on it with mean $m$ and covariance matrix $\Sigma$, and $P'_{X'}=N(m,\Sigma)=P_{X}$ by The Gaussian Law with Given Mean and Covariance Matrix (i). Consequently, if $A\in\mathcal{B}(\mathbb{R}^{d})$ and $X'\in A$ almost surely, with exceptional event $N'$, then $X\in A$ almost surely. Indeed, $\mathbb{R}^{d}\setminus A$ is Borel, so, the law being $P_{X}(C)=P(X^{-1}(C))$ for Borel $C$ (The Law of a Random Element (i)),

$$P(X\notin A)=P_{X}(\mathbb{R}^{d}\setminus A)=P'_{X'}(\mathbb{R}^{d}\setminus A)=P'(X'\notin A)\le P'(N')=0,$$

using $\{X'\notin A\}\subseteq N'$ (Basic Properties of a Measure (iii)); as an element of $[0,\infty]$ that is at most $0$ equals $0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)), $P(X\notin A)=0$, and $X\in A$ almost surely, with exceptional event $\{X\notin A\}$.

Step 3 ("if", $\Sigma=0_{d}$). By The Law of a Gaussian Random Vector Is Determined by Its Mean and Covariance Matrix (i), $X'=m$ almost surely. The singleton $\{m\}$ is Borel (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (iii)), so by Step 2, $X=m$ almost surely. Hence $X$ is Gaussian (Gaussian Random Vectors and Jointly Gaussian Random Variables (i)), with $\mathbb{E}X=m$ and $\operatorname{Cov}(X)=0_{d}$ by Square-Integrability, Mean and Covariance Matrix of a Gaussian Random Vector (iii).

Step 4 ("if", $\Sigma\neq0_{d}$). Choose $r$ and $B$ by A Nonzero Positive Semidefinite Matrix Factors as BBᵀ with BᵀB Invertible (i), let $M=(B^{\top}B)^{-1}B^{\top}$, and let

$$g(x)=M\bigl(x+(-m)\bigr),\qquad h(x)=\bigl(x+(-m)\bigr)+\bigl((-1)B\bigr)g(x)\qquad(x\in\mathbb{R}^{d}),$$

with matrix-vector products. These maps are well defined (Maps Formed Pointwise from Vector and Matrix Operations (i)) and are built from the identity and constants by sums and matrix-vector products with constant matrices, so they are continuous (Continuity of Identity Maps, Compositions and Reciprocals (i), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)), hence Borel (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iv)). Since $((-1)B)y=(-1)(By)$ (Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v)), comparing coordinates shows that $h(x)=0$ exactly when $x=m+Bg(x)$. In particular

$$A=\{x\in\mathbb{R}^{d}:x=m+Bg(x)\}=h^{-1}(\{0\})$$

is a Borel set, $\{0\}$ being Borel by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (iii). By The Law of a Gaussian Random Vector Is Determined by Its Mean and Covariance Matrix (i), $g(X')$ is a standard normal vector in $\mathbb{R}^{r}$ and $X'=m+Bg(X')$ almost surely, that is, $X'\in A$ almost surely. Since $P_{X}=P'_{X'}$, the random vector $W=g(X)$, with $W(\omega)=g(X(\omega))$ for $\omega\in\Omega$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)), has the law of $g(X')$, which is $\gamma_{r}$ (Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (iii)), so $W$ is a standard normal vector (Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (iii)); and by Step 2, $X\in A$ almost surely, that is,

$$X=m+BW\quad\text{almost surely}.$$

So $(r,W,m,B)$ is a Gaussian representation of $X$, $X$ is Gaussian, and $\mathbb{E}X=m$, $\operatorname{Cov}(X)=BB^{\top}=\Sigma$ by Square-Integrability, Mean and Covariance Matrix of a Gaussian Random Vector (ii).