Let $\mathcal{F}_n \subseteq \mathcal{F}$ for n=1,2,3,..... be independent sequence of $\sigma$-algebra. Define- $$\mathcal{T}_n = \sigma( \bigcup\limits_{m=n}^{\infty}\mathcal{F}_n) $$
Is there a way to say that $\mathcal{F}_x$ is independent of $\mathcal{T}_n$ for x < n where x = 1,2,3.... and n = 2,3.....
Remark- Similar thing comes somewhere in the middle of the proof of Kolmogorov's 0-1 law.
You have to prove that for all $A\in \mathcal{F}_x$ and $B \in \mathcal{T}_n$, $P(A\cap B)= P(A) P(B) $.
I think the quickest way to prove it is to prove that :
Proof : 0) By definition, $\mathcal{F}_m \subset \mathcal{V}_A$, $m\geq n$
1) It is straightforward that $\emptyset \in \mathcal{V}_A$
2) If $B\in \mathcal{V}_A$,
\begin{align*} P(A)= P(A\cap \Omega) &= P(A\cap (B\cup B^c) \\ &= P(A\cap B\cup A\cap B^c) \\ &= P(A\cap B)+P( A\cap B^c) \\ &= P(A) P(B) +P( A\cap B^c) \end{align*} so $$ P( A\cap B^c) =P(A)-P(A) P(B)=P(A)(1-P(B))=P(A)P(B^c)$$ and $$ B^c \in \mathcal{V}_A$$
3) Let $B_1, \dots, B_k, \dots \in \mathcal{V}_A$ and $$ B := \bigcup_{k=1}^{\infty}B_k$$ From 2) it is equivalent to prove that $B^c=\bigcap_{k=1}^{\infty}B_k^c\in \mathcal{V}_A$ \begin{align*} P(B^c\cap A) &= P\Big(A\cap \bigcap_{k=1}^{\infty}B_k^c\Big) \\ & = \lim_{m\rightarrow +\infty} P\Big(A\cap \bigcap_{k=1}^{m}B_k^c\Big) \end{align*} Here we have to regroup the $B_k^c$ belonging from the same $\mathcal{F}_{i_k}$ and use the independence of the sequence $(\mathcal{F}_{m})$ : we will have then $P\Big(A\cap \bigcap_{k=1}^{m}B_k^c\Big) = P(A) P\Big(\bigcap_{k=1}^{m}B_k^c\Big)$, so $$P(B^c\cap A)=P(A)P(B^c) \quad \Rightarrow \quad B^c\in \mathcal{V}_A \quad \Rightarrow \quad B=\bigcup_{k=1}^{\infty}B_k\in \mathcal{V}_A \qquad \square$$
So $\mathcal{V}_A$ is a $\sigma$-algebra containing $\mathcal{F}_m$, $m\geq n$, which gives (by definition of $\mathcal{T}_n$) $\mathcal{T}_n\subset \mathcal{V}_A$, so : $$\forall B \in \mathcal{T}_n, \qquad P(A\cap B)=P(A)P(B) $$
Since $A$ can be chosen as you wish in $\mathcal{F}_x$, the proof is complete.