Let $\Omega = (0,1), F = $ Borel sets, $P = $ Lebesgue measure. Show $X_n(\omega) = \sin(2\pi \omega), n = 1, 2, \ldots$ are uncorrelated but not independent.
To show they are uncorrelated, I found that $$E[X_n(\omega)] = \int_{0}^{1} \sin(2\pi n \omega) d\omega = 0,$$ and therefore $$E[X_n - E[X_n]]E[X_m - E[X_m]] = 0.$$
I'm a little bit stuck on showing that they're not independent. I know I have to find $B_i \in F$ such that $$P(\cap \{X_i \in B_i\}) = \prod P(X_i \in B_i).$$ (Can this be over countably many $B_i$, or does the definition only allow for finitely many?)
My original thought was essentially that of the linked answer, namely,
Consider $X_1$ and $X_2$: $$ P(X_1,X_2\in[0,1])=P([0,1/4])=\frac{1}{4} $$ but $$ P(X_1\in[0,1])=\frac{1}{2},\\ P(X_2\in[0,1])=\frac{1}{4}. $$
However, I believe this is wrong because shouldn't $P(X_2 \in [0,1]) = 1/2$, since $\sin(4\pi \omega)$ is nonnegative on $[0,1/4]$ and $[2/4, 3/4]$?
You're right about $\mathbb P(X_2 \in [0, 1])$ being $1/2$, not $1/4$. The idea of that answer is right, but it does indeed fail to be a counterexample, as you showed.
All that's happening here is that the interval $[0, 1]$ is an unlucky set to try. As an alternate approach, try the same tactic but with some interval of the form $[1-\epsilon, 1]$ where $\epsilon$ is fairly small.