Let $\Omega =(0,1)$, $\mathcal{B}=$Borel Set and $P=$ Lebesgue measure. Suppose $X_n(\omega) =\sin (2\pi n\omega),n=1,2, \dots$ are uncorrelated but not independent.
The solution to uncorrelated part is really easy, but stuck on the independent part. I started with the set $A(a)=\{ \omega : X_n(\omega)\in [0,a),X_m(\omega)\in [0,a), m\neq n \}$, As $a$ becomes arbitrarily small the set $A(a)$ contains all points such that $X_n(\omega)$ and $X_m(\omega)$ intersect when they are $0$.
Again let $B(a)=\{ \omega : X_n(\omega)\in [0,a)\}$ and $C(a)=\{ \omega : X_m(\omega)\in [0,a)\}$. So we if independent then $P(A(a))=P(B(a)) P(C(a))$.
It is easy to check that if $a\to 0$, $\# A(a) < \min\{ \# B(a), \# C(a) \}$, here $\# D=$number of elements in $D$. How to claim after that, that $P(A(a))<P(B(a)) P(C(a))??$
(May be everything I wrote is wrong, in that case please help with right solution please!)
Consider $X_1$ and $X_2$: $$ P(X_1,X_2\in[0,1])=P([0,1/4])=\frac{1}{4} $$ but $$ P(X_1\in[0,1])=\frac{1}{2},\\ P(X_2\in[0,1])=\frac{1}{4}. $$ Hence, $P(X_1\in[0,1])P(X_2\in[0,1])\neq P(X_1,X_2\in[0,1])$ and thus $X_1$ and $X_2$ are not independent.