I'm trying to figure out the connections between two contructions of Gaussian measure.


Let $(U, \langle\cdot,\cdot\rangle\_U)$ be a seprable Hilbert space, and $\mathcal{B}(U)$ be the Borel sigma-algebra.




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[Definition 2.1, page 10](http://www.math.chalmers.se/%7Estig/underv/doktorandkurs/SPDE0708/spdenotes.pdf) gave the following definition of Gaussian measure:


A measure $\mu$ on $(U, \mathcal{B}(u), \mu)$ is Gaussian if for all $v\in U$, there are $m\_v\in \mathbb{R}$, $\sigma\_v\in \mathbb{R}^+$ such that if $\sigma\_v>0$,


$$
\mu(u \in U\colon \langle v,u\rangle\_U \in A) = \frac{1}{\sqrt{2\pi\sigma\_v}}\int\_A\exp(-\frac{(s-m\_v)^2}{2\sigma\_v^2})ds,
$$


for all $A\in\mathcal{B}(\mathbb{R})$.




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However, I recall from when we construct the Brownian motion. See, page 23 of <https://www.springer.com/gp/book/9780387287201>


Let $U$ be the space of continuous funcitons $u(t)$ on $[0,1]$, and $D$ be a cylindrical set


$$
D = \{u\in U: (u(t\_1), u(t\_2), \ldots, u(t\_N))\in E\in \mathcal{B}(\mathbb{R}^N)\}
$$


Then the measure is defined by


$$
\mu(D) = \int\_E \prod\_{i=1}^N (\frac{1}{\sqrt{2\pi(t\_i-t\_{i-1})}} \exp(-\frac{(h\_i-h\_{i-1})}{2(t\_i-t\_{i-1})}))dh\_1dh\_2\cdots dh\_N
$$


on $0<t\_1<t\_2<\cdots<t\_N<1$.




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My questions is: Is it possible to define the infintite Gaussian measure on finite-dimensional distributions similar to we did for Brownian motion? I don't know how to find the connection bewtween the inner product $\langle v,u\rangle\_U$ and a finite-dimensional distribution.



Infinite-dimensional Gaussian measure vs finite-dimensional Wiener measure
If $U$ is a Banach space, then the natural thing to do is replacing the $\langle v,\cdot\rangle\_U$ by any continuous form $\phi\in U^\*$. Then the definition of a Gaussian measure is a measure $\mu$ such that the pushforward $\phi^\*\mu$ is Gaussian for all $\phi\in U^\*$ (note that this works in finite dimension too). In general, there are subtleties about the mean being in $U$ or $U^{\*\*}$, but for centred examples everything works well.


Your definition of Brownian motion implies that the every form $\phi:u\mapsto\alpha\_1u(t\_1)+\cdots+\alpha\_n\phi(t\_n)$ is Gaussian (it is in fact equivalent, using the above remark). Now one can show that the subspace of those forms, $\mathrm{Vect}(\mathrm{ev}\_t,t\in[0,1])$, is actually dense in $U^\*$ in the weak-$\*$ topology. Let us show that this proves that $\phi^\*\mu$ is Gaussian for all $\phi\in U^\*$.


Let $\phi\in U^\*$. Let $\phi\_n\in U^\*$ be a sequence of functions in $\mathrm{Vect}(\mathrm{ev}\_t,t\in[0,1])$ that converges to $\phi\in U^\*$ in the weak-$\*$ topology (here we use the fact that the topology is metrisable, which is true since $U$ is second countable). We show that $\phi^\*\mu$ is the weak limit of $\phi\_n^\*\mu$ (in the sense of probability measures on $\mathbb R$), and that the limit is Gaussian.


The first point is a matter of unfolding the definitions: for every bounded continuous $f:\mathbb R\to\mathbb R$, $f\circ\phi\_n(u)\to f\circ\phi(u)$ for all $u$, and since $f$ is bounded we apply Lebesgue's dominated convergence theorem to get
$$ \int f(x)\phi\_n^\*\mu(x) = \int f\circ\phi\_n(u)\mu(du) \to \int f\circ\phi(u)\mu(du) = \int f(x)\phi^\*\mu(dx). $$
It remains to show that a weak limit of Gaussian measures stays Gaussian. This can be seen by showing first that the means $m\_n$ and variances $\sigma^2\_n$ have to be bounded, then up to extraction we can assume they converge to $m$ and $\sigma$, and at this point one can see that the characteristic function of the limit is the limit of the characteristic functions $t\mapsto\exp(im\_nt-\sigma^2\_nt^2/2)$, which is of course the characteristic function of $\mathcal N(m,\sigma^2)$


