Let $X$ be a uniformly distributed random variable in $[0,1]$ and define $X_n:=\lfloor 2^nX\rfloor 2^{-n}$ for all $n$. I want to show that $\bigcap_{n\geq 0} \sigma(X_n,X_{n+1}...)=\sigma(X)$.
My idea was the following:
Proof: $\subseteq$ Let me remark that for all $n$, $X_n$ is $\sigma(X)$ measurable since $X_n=f(X)$ where $f(x)=\lfloor 2^nx\rfloor 2^{-n}$ is measurable. Therefore $\sigma(X_n,X_{n+1},...)\subset \sigma(X)$ for all $n$ and hence $\bigcap_{n\geq 0} \sigma(X_n,X_{n+1}...)\subseteq\sigma(X)$.
$\supseteq$ Here let me remark that $$\frac{2^nX}{2^n}\leq X_n\leq \frac{2^nX+1}{2^n}$$ where $\frac{2^nX}{2^n}\rightarrow X$ and $\frac{2^nX+1}{2^n}\rightarrow X$, so in particular $X_n\rightarrow X$. But since the limit exists we also know that $\limsup_{n\rightarrow \infty}X_n=X$ but we know that $\limsup_n$ is $\sigma(X_n,X_{n+1},...)$ measurable. But then this implies that $X$ is $\sigma(X_n,X_{n+1},...)$ measurable for all $n$ and hence $\bigcap_{n\geq 0} \sigma(X_n,X_{n+1}...)\supseteq\sigma(X)$.
Does this work?