Proof of Pairwise Uncorrelated Jointly Gaussian Random Variables are Independent

All covariances below are defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector.

Reduction to a full factorization. We first show that it suffices to prove: for all Borel sets $C_1,\dots,C_p$,

with the finite product notation. Indeed, let Borel sets $B_1,\dots,B_p$ and a nonempty subset $S\subseteq\{1,\dots,p\}$ be given, as in Independence of Events and of Random Variables. Apply the full factorization with $C_i=B_i$ for $i\in S$ and $C_i=\mathbb{R}$ for $i\notin S$. Since $\{X_i\in\mathbb{R}\}=\Omega$, the intersection on the left reduces to $\bigcap_{i\in S}\{X_i\in B_i\}$, and since $P(\Omega)=1$, the factors with $i\notin S$ equal $1$; hence

As $S$ and the Borel sets were arbitrary, the events $\{X_1\in B_1\},\dots,\{X_p\in B_p\}$ are independent for every Borel choice, i.e., $X_1,\dots,X_p$ are independent random variables.

Induction. Fix Borel sets $C_1,\dots,C_p$. We prove by induction on $k\in\{1,\dots,p\}$ that

For $k=1$ both sides equal $P(X_1\in C_1)$ and there is nothing to prove.

Let $1\le k<p$ and assume the identity holds for $k$. The initial segment $(X_1,\dots,X_k,X_{k+1})$ is a Gaussian random vector, being a subfamily of the Gaussian random vector $(X_1,\dots,X_p)$, by Affine Transformations of Gaussian Random Vectors are Gaussian. By hypothesis its cross-covariances with the last component vanish:

By Uncorrelated Jointly Gaussian Blocks are Independent, applied with the blocks $(X_1,\dots,X_k)$ and $(X_{k+1})$ (that is, with $d=k$ and $q=1$), the $\sigma$-algebras $\sigma(X_1,\dots,X_k)$ and $\sigma(X_{k+1})$ are independent.

The event $E=\bigcap_{l=1}^{k}\{X_l\in C_l\}$ belongs to $\sigma(X_1,\dots,X_k)$: each $\{X_l\in C_l\}=X_l^{-1}(C_l)$ is one of the generating sets of $\sigma(X_1,\dots,X_k)$ exhibited in Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, and a $\sigma$-algebra contains every finite intersection of its members, a finite intersection being the complement of the finite union of the complements. Likewise $\{X_{k+1}\in C_{k+1}\}\in\sigma(X_{k+1})$. Independence of the two $\sigma$-algebras therefore gives

and inserting the inductive hypothesis for $P(E)$ yields the identity for $k+1$. Taking $k=p$ gives the full factorization, which by the reduction completes the proof. $\blacksquare$