Let $(\Omega,\mathscr{F},P)$ be a probability space.
Let $X,Y:(\Omega,\mathscr{F})\rightarrow (( \mathbb{R}^n)^I, \otimes_{t\in I} \mathscr{B}_{\mathbb{R}^n})$ be measurable functions.
Let $\pi_F:(\mathbb{R}^n)^I \rightarrow (\mathbb{R}^n)^F$ be the projection map for each finite subset $F$ of $I$.
If $(\pi_F)_*(X_*P)= (\pi_F)_*(Y_*P)$ for all finite subsets $F$ of $I$, then is $X_*P=Y_*P$?
Here's a proof of this that I think must be 'wrong':
PROOF
First of all, note that there are two versions of Kolmogorov extension theorem (See the post).
Let $\pi_i:(\mathbb{R}^n)^I\rightarrow \mathbb{R}^n$ be the $i$-th projection map for each $i\in I$.
Define $\mu_i := (\pi_i)_*(X_*P)$ for each $i\in I$.
Then, by the theorem 1 in the above given link, there exists a unique probability measure $\mu$ on $(\mathbb{R}^n)^I$ such that $(\pi_i)_*\mu =\mu_i$. Since $X_*P, Y_*P$ also satisfy this, we have $X_*P=\mu=Y_*P$. Hence, we can conclude $X_*P=Y_*P$.
However, I think there's something wrong in my proof. As the answer in the above link points, theorem $1$ is related to independency and theorem 2 is related to dependency. Since $X_i$'s need not be independent and I used theorem 1, there must be something wrong.. Is my proof correct?
