Proof of Wiener Integrals Against a Vector Brownian Motion are Jointly Gaussian

Part 1. Fix $j$, $s>0$, and continuous $f:[0,s]\to\mathbb{R}$. By the specification in the statement, the Wiener integral is the It^{o} integral of the constant family $(f(u))_{u\in(0,s]}$ with respect to the It^{o} integrator $(W^{j},1)$ of Brownian Motion is an Ito Integrator with Unit Intensity, taken with respect to the natural filtration $(\mathcal{F}^{W^{j}}_t)_{t\ge0}$ of $W^{j}$, and by Ito Integrable Process and the Ito Integral a version may be chosen $\mathcal{F}^{W^{j}}_{s}$-measurable, where $\mathcal{F}^{W^{j}}_{s}=\sigma(W^{j}_u:0\le u\le s)$. Since the generating events of $\mathcal{F}^{W^{j}}_{s}$ lie in $\sigma(W^{j}_t:t\ge0)$, minimality of the generated $\sigma$-algebra gives $\mathcal{F}^{W^{j}}_{s}\subseteq\sigma(W^{j}_t:t\ge0)$, proving claim 1.

Part 2. First take, for every $j$, $s$, $f$, the $\sigma(W^{j}_t:t\ge0)$-measurable versions of claim 1. For each fixed $j$, claim 3 of Wiener Integrals of Continuous Functions are Jointly Gaussian states that the family $\Gamma_j$ consisting of all $W^{j}_t$ ($t\ge0$) and all Wiener integrals against $W^{j}$ is jointly Gaussian; moreover every member of $\Gamma_j$ is measurable with respect to $\mathcal{A}_j=\sigma(W^{j}_t:t\ge0)$, so the $\sigma$-algebra generated by the members of $\Gamma_j$ is contained in $\mathcal{A}_j$ by minimality. The $\sigma$-algebras $\mathcal{A}_1,\dots,\mathcal{A}_m$ are independent by property (ii) of Vector Brownian Motion, and sub-$\sigma$-algebras of independent $\sigma$-algebras are independent directly from Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras: the defining product identity is demanded there for a smaller collection of admissible events. Hence Independent Jointly Gaussian Families are Jointly Gaussian applies to the groups $\Gamma_1,\dots,\Gamma_m$ (with the index sets declared in the statement, whose disjoint union over $j$ is the index set of claim 2): the combined family is jointly Gaussian.

Centering: each Wiener integral has expectation $0$ and each $W^{j}_t$ with $t>0$ satisfies $W^{j}_t=\int_0^t1\,dW^{j}_u$ almost surely, hence also $\mathbb{E}[W^{j}_t]=0$, by claims 1 and 2 of Wiener Integrals of Continuous Functions are Jointly Gaussian; and $W^j_0=0$ almost surely by property (i) of Standard Brownian Motion, so $\mathbb{E}[W^j_0]=0$.

For arbitrary versions: any version differs from the measurable version by almost-sure equality, and a family obtained from a jointly Gaussian family by replacing members with almost surely equal random variables is jointly Gaussian with the same expectations and covariances, by Almost Sure Modifications of Gaussian Random Vectors are Gaussian (applied to each finite subfamily). This proves claim 2.

Part 3. For $i=j$ and $s,t>0$ the covariance formulas are exactly those of claim 2 of Wiener Integrals of Continuous Functions are Jointly Gaussian, using $W^{j}_t=\int_0^t1\,dW^{j}_u$ almost surely (claim 1 there) for the mixed and pure Brownian cases, covariances being unchanged under almost-sure replacement; in particular $\operatorname{Cov}(W^i_{s'},W^j_t)=\int_0^{\min(s',t)}1\,du=\min(s',t)$ for $s',t>0$. For the cases with a time equal to $0$: $W^j_0=0$ almost surely, and the covariance of a square-integrable random variable with an almost surely vanishing one is $0$, matching $\int_0^0=0$ and $\min(0,t)=0$ under the stated convention. For $i\ne j$, the two random variables belong to distinct groups $\Gamma_i$ and $\Gamma_j$ of Part 2 (after almost-sure replacement by measurable versions, which changes no covariance), and members of distinct groups are uncorrelated by the covariance statement of Independent Jointly Gaussian Families are Jointly Gaussian. $\blacksquare$