Proof of Almost Sure Modifications of Gaussian Random Vectors are Gaussian

Step 1: finite unions of null events are null. Let $M_1,M_2\in\mathcal{F}$ with $P(M_1)=P(M_2)=0$. By the $\sigma$-algebra operations, $M_2\setminus M_1$, $M_2\cap M_1$, and $M_1\cup M_2$ are events. Since $M_2=(M_2\setminus M_1)\cup(M_2\cap M_1)$ is a disjoint union, additivity and nonnegativity of the measure $P$ give $P(M_2\setminus M_1)\le P(M_2)=0$. Since $M_1\cup M_2=M_1\cup(M_2\setminus M_1)$ is a disjoint union, additivity gives $P(M_1\cup M_2)=P(M_1)+P(M_2\setminus M_1)=0$. Iterating, any union of finitely many events of probability $0$ has probability $0$. Also, for any event $A$, the disjoint union $A\cup(\Omega\setminus A)=\Omega$ and $P(\Omega)=1$ give $P(\Omega\setminus A)=1-P(A)$; in particular an event has probability $1$ exactly when its complement has probability $0$, and an event containing the complement of an event of probability $0$ has probability $1$, by the monotonicity argument above.

Step 2: transfer of a representation. Let $\bigl(m,(\mu_i),(a_{ij}),(Z_j)\bigr)$ be any Gaussian representation of $(Y_1,\dots,Y_d)$, so that the events

satisfy $P(A_i)=1$ (these are events for the same reason as the sets $\{X_i=Y_i\}$: differences of random variables are random variables, as noted in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process, sums and scalar multiples of random variables being random variables in the same way, and $\{0\}$ is a Borel set). Put $B_i=\{X_i=Y_i\}$, with $P(B_i)=1$ by hypothesis, and let

By Step 1, $P(N)=0$. For every $\omega\in\Omega\setminus N$ and every $i$,

so the event $\bigl\{X_i=\mu_i+\sum_{j=1}^{m}a_{ij}Z_j\bigr\}$ contains $\Omega\setminus N$ and therefore has probability $1$ by Step 1. Since $Z_1,\dots,Z_m$ are unchanged, $\bigl(m,(\mu_i),(a_{ij}),(Z_j)\bigr)$ is a Gaussian representation of $(X_1,\dots,X_d)$, and $(X_1,\dots,X_d)$ is a Gaussian random vector.

Step 3: moments. By claim 2 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector applied to the shared representation,

By claim 3 of the same lemma these quantities do not depend on the choice of representation, so the mean vectors and covariance matrices of the two tuples coincide. $\blacksquare$