Question about Wiener measure In this post user SM2 provides a way of constructing the Wiener measure $\nu$ on the measurable space $(C([0,\infty),\mathcal{B}(C([0,\infty))))$ where $\mathcal{B}(C([0,\infty)))$ is the Borel $\sigma$-algebra of $C([0,\infty))$ induced by the topology of uniform convergence on compact sets. 
In 3), for a Wiener process $\{W_t\}_{t\in[0,\infty)}$ on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, SM2 defines the measurable map $g:\Omega\to C([0,\infty))$ which I'm guessing is just $$\omega\in\Omega\mapsto g(\omega)\in C([0,\infty))$$ with $g(\omega)$ given by $$t\in[0,\infty)\mapsto g(\omega)(t)=W_t(\omega).$$ At the end SM2 defines $\nu:\mathcal{B}(C([0,\infty)))\to[0,1]$ by setting $\nu(A)=\mathbb{P}(g^{-1}(A))$ for $A\in\mathcal{B}(C([0,\infty))))$ so that $\nu$ is the Wiener measure.
My question is, if we don't work with $g$ but instead define $$\nu^*:\mathcal{B}(C([0,\infty)))\to[0,1]$$ by specifying $\nu^*(\{f\in C([0,\infty)):f(t)\leq a\})=\mathbb{P}(\{\omega\in\Omega:W_t(\omega)\leq a\})$ for each $t\in[0,\infty)$ and $a\in\mathbb{R}$ then will $\nu^*$ and $\nu$ coincide?
 A: This does not uniquely define a measure on $\mathcal B(C([0,\infty)))$. The problem is that the marginals of a process at a single time are not enough to determine the full process. For a play example, instead of a stochastic process in contininuous time let us look at a process $X$ in discrete time $\{0,1,\ldots,N\}$. It may be tempting to say the law of $X$ is fully determined by looking at the law of $X_n$ for $n=0,1,\ldots,N$, but really we know this is not true - for a couple of extreme examples, suppose $X_m$ and $X_n$ are identically distributed for all $m,n$, then we could have $X_m=X_n$ for all $m,n$ or $X_m,X_n$ independent for $m\neq n$, and looking at each $X_n$ individually will in no way distinguish these extremely different cases.
For the finite case there is an easy work around (namely, looking at the joint distribution of $X=(X_0,\ldots,X_N)$ which is a random variable in $\mathbb R^{N+1}$), but what about the infinite case? Thankfully, the Kolmogorov consistency theorem tells us that if we know all of the finite dimensional distributions, that is, the law of $(X_{t_1},\ldots,X_{t_n})$ for any finite subset $\{t_1,\ldots,t_n\}\subset T$ (where $T=\mathbb R$ or $\mathbb N$, whatever our time scale is), and these are consistent with one another, then this uniquely determines a measure on the relevant space. For continuous processes such as the Wiener process, we typically need a slight modification of this argument to guarantee a continuous version, but it still works. This is why SM2's construction is the correct one.
Let's end with a concrete counterexample, that is, a measure on $C([0,\infty))$ with the same marginals as $\nu$ which nonetheless defines a different process. Let $Z:\Omega\longrightarrow\mathbb R$ be any measurable function which is standard normal with respect to $\mathbb P$ and define $\Gamma_t:=\sqrt{t}Z$. Then $\Gamma:\Omega\longrightarrow C([0,\infty))$ is measurable and
$$
\mathbb P(\Gamma_t\le a)=\frac1{\sqrt{2\pi t}}\int_{-\infty}^ae^{-x^2/2t}dx=\mathbb P(W_t\le a)
$$
for any $t\ge0$ and $a\in\mathbb R$. However, $\Gamma$ and $W$ have different laws. For example, $\mathbb P(\Gamma_1> 1,\Gamma_2\le 0)=0$ but $\mathbb P(W_1>1,W_2\le0)>0$. So $\nu^*$ in your question could be equal to either $\nu$ or the law of $\Gamma$.
