Given the following exercise:
Let $(\Omega, \mathcal F, P) = (]0,1[, \mathcal B(]0,1[), \lambda)$ with the lebesgue measure $\lambda$. Define $$Y_n(\omega) := \mathbb 1_{\{ \lfloor 2^n \omega \rfloor \text{is even}\}}$$ Show that $(Y_n)_{n\in \mathbb N}$ is an independent sequence of Bernoulli($\frac{1}{2}$)-distributed random variables.
I first showed that the $Y_n$ are indeed Ber($1/2$)-distributed:
It is easy to see that $\lfloor 2^n \omega \rfloor$ is even for $\omega \in [\frac{k}{2^n}, \frac{k+1}{2^n}[$, where $k \in \{0, 2, 4, \dots, 2^n - 2\}$ and odd for $\omega \in [\frac{k}{2^n}, \frac{k+1}{2^n}[$, where $k \in \{1, 3, 5, \dots, 2^n - 1\}$. Since this is a partition of intervals of the same size of the interval $[0,1[$ it follows that $$P(Y_n = 1) = P(Y_n =0) = \frac{1}{2}.$$ Now I am not sure how to show independence. It seems really obvious to me but I feel like actually formally showing it is technical and messy. Any simple way to show it?
Let $m<n$.
Step I: For any two Bernoulli random variables $X,Y$, they are independent if and only if $P[X=1,Y=1]=P[X=1]\cdot P[Y=1]$. Hence, in our case it is enough to show that $P[Y_m=1, Y_n=1]=1/4$. Equivalently, we may show $P[Y_n=1|\ Y_m=1]=1/2$.
Step II: On the event $Y_m=1$ we have $\omega\in [\frac{k}{2^m},\frac{k+1}{2^m}]$ where $k$ is even. For these $\omega$'s, when do we have $Y_n=1$? divide each interval of the form $[\frac{k}{2^m},\frac{k+1}{2^m}]$ into $2^{n-m}$ equal sub intervals. Each of them will be of size $\frac{1}{2^n}$, and exactly in half of them we will have $Y_n=1$.
Well, that's it. Sorry for not being very rigorous, the details can be quite long. But the idea is here.