If $(X_t)\sim (Y_t)$, does $\mathbb P(\bigcap_{n\in\mathbb N}(X_{t_n}\in A))=\mathbb P(\bigcap_{n\in\mathbb N}(Y_{t_n}\in A))$ hold?

Question

Let $(X_t)$ and $(Y_t)$ two stochastic process that has same law. This mean that they have same finite dimensional distribution, i.e. $$(X_{t_1},...,X_{t_m})\sim (Y_{t_1},...,Y_{t_m}).$$

So, indeed, $$\mathbb P\left(\bigcap_{n=1}^m\{X_{t_n}\in A\}\right)=\mathbb P\left(\bigcap_{n=1}^m\{Y_{t_n}\in A\}\right).$$

But what about the equality $$\mathbb P\left(\bigcap_{n\in\mathbb N^*}\{X_{t_n}\in A\}\right)=\mathbb P\left(\bigcap_{n\in\mathbb N^*}\{Y_{t_n}\in A\}\right) \ \ ?$$

If $\{X_{t_i}\in A\}$ are not increasing or decreasing, I guess it's not correct, but I can't find a counter example. Any idea ?

Answer 1

This is just continuity from above of probability measures. $$ \mu(\cap_{n=1}^{\infty} B_n)=\lim_{N\to \infty}\mu(\cap_{n=1}^N B_n) $$ for any finite measure $\mu$ and sequence of measurable sets $(B_n)_{n\in\mathbb{N}}$.

Plugging in $B_n=(X_{t_n}\in A_n)$ yields the thing you want.