Question: Let $U$ be uniform distributed on (0; 1]. For any $n\in \mathbb N$ define
$ X_n = \begin{cases} 1 & \sin(2^n \pi U)<0, \\ 0 & \text{otherwise}. \end{cases} $
Show that $X_{1};X_{2};X_{3}$ are independent and B(1; 0.5)-distributed.
My attempt/idea: I drew a graph for the functions $$ h_{n}(u) = \begin{cases} 1 & \sin(2^n \pi U)<0, \\ 0 & \text{otherwise}. \end{cases} $$ for n = 1, 2, 3.
But I'm still not able to get it. Help with proving that $_1,_2,_3$ are independent will really be appreciated. Thank you!
$$P(X_1=1)=P(\sin(2\pi U)<0)=P(U\in(\tfrac12,1))=\tfrac12$$ $$P(X_2=1)=P(\sin(4\pi U)<0)=P(U\in(\tfrac14,\tfrac12)\cup(\tfrac34,1))=\tfrac12$$ $$P(X_3=1)=P(\sin(8\pi U)<0)=P(U\in(\tfrac18,\tfrac14)\cup(\tfrac38,\tfrac12)\cup(\tfrac58,\tfrac34)\cup(\tfrac78,1))=\tfrac12$$ Thus,$$X_1,X_2,X_3\text{~Bernoulli}(\tfrac12)$$ Now, to show independence, we can brute force our approach by checking all cases (I admit that this is not necessarily the best approach, but at least it suffices to answer the question.)
In order for $X_1,X_2,X_3$ to be independent, we must show that all of the following are true: $X_1,X_2$ are independent, $X_1,X_3$ are independent, $X_2,X_3$ are independent, and $X_1,X_2,X_3$ are independent. Here I only check the first, as the process for showing the rest of the conditions hold is very similar.
As $X_1,X_2$ are independent if and only if $P(X_1=k,X_2=i)=P(X_1=k)P(X_2=i)$ $\forall{i,k}$: $$P(X_1=1,X_2=1) = P(U\in(\tfrac34,1))=\tfrac14=\tfrac12*\tfrac12=P(X_1=1)P(X_2=1)$$ $$P(X_1=1,X_2=0) = P(U\in(\tfrac12,\tfrac34))=\tfrac14=\tfrac12*\tfrac12=P(X_1=1)P(X_2=0)$$ $$P(X_1=0,X_2=1) = P(U\in(\tfrac14,\tfrac12))=\tfrac14=\tfrac12*\tfrac12=P(X_1=0)P(X_2=1)$$ $$P(X_1=0,X_2=0) = P(U\in(0,\tfrac14))=\tfrac14=\tfrac12*\tfrac12=P(X_1=0)P(X_2=0)$$
And $X_1,X_2$ can only take the values $0$ and $1$ so we have verified all cases and thus shown their independence.
Note: to do this process for showing $$P(X_1=i,X_2=j,X_3=k)=P(X_1=i)P(X_2=j)P(X_3=k)\text{ }\forall{i,j,k},$$ note that each variable can take on $2$ values, so we need to verify $2^3=8$ total cases.