Are random variables independent of their tail sigma-algebra? Let $X_1, X_2, ...$ be independent random variables.
Define $$\mathscr{T}_n = \sigma(X_{n+1}, X_{n+2}, \ldots)$$ and $$\mathscr{T} = \bigcap_{n} \mathscr{T}_n,$$ the tail σ-algebra of $(X_1, X_2, \ldots)$.
Are $\sigma(X_1), \sigma(X_2), ...$ independent of $\mathscr{T}$?
If so, why?
If not, why, and what about $$\sigma(X_1), \sigma(X_2), ..., \sigma(X_k) \ \;\forall k \in \mathbb{N}\quad?$$
All I got so far is that if $X_1, X_2, \ldots$ were events instead of random variables, $X_1, X_2, \ldots, X_k \ \forall k \in \mathbb{N}$ would be independent of some events in $\mathscr{T}$ such as $\limsup X_n$.
 A: [This answers the original edition of the question, which did not assume $X_1,X_2,X_3,\ldots$ are independent.]
No.  Suppose $R\sim\mathrm{Uniform}(0,1)$ and $(Y_1,Y_2,Y_3,\ldots)\mid R\sim\mathrm{i.i.d. Bernoulli}(R)$.
Then the strong law of large numbers implies that
$$
\Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \mid R\right) = 1.
$$
So
\begin{align}
& \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \right) \\[10pt]
= {} & \operatorname{E} \left( \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \mid R\right) \right) = \operatorname{E}(1) = 1.
\end{align}
Let $X_n= (Y_1+\cdots+Y_n)/n$.  The event that $\lim_{n\to\infty} X_n = r$ is in the tail sigma-algebra of $X_1,X_2,X_3,\ldots$. But $X_1$ is not independent of that event, since $\Pr(X_1=1\mid R=r)=r$.
If you want a tail event whose probability is positive, observe that $\lim\limits_{n\to\infty} X_n \overset{\text{a.s.}}= R \sim\mathrm{Uniform}(0,1)$, so $\Pr(X_1=1) = \operatorname{E}(\Pr(X_1=1\mid \lim\limits_{n\to\infty} X_n)) = \operatorname{E}(\lim\limits_{n\to\infty} X_n) = 1/2$, and find $\Pr(X_1=1\mid \lim\limits_{n\to\infty} X_n>1/2)$.
A: This is merely a slight extension of Nate's comment, but it should answer your question: 
On $(\Omega,\mathcal{A},\mathbf{P})$ we have a sequence $(X_i)_{i\in \mathbb{N}}$ of independent RVs $X_i:(\Omega,\mathcal{A})\to(\Omega_i,\mathcal{A}_i)$ and the corresponding sequence $\sigma(X_i)_{i\in \mathbb{N}}\subset\mathcal{A}$ of independent $\sigma$-algebras, where we have $\sigma(X_i):=\{X^{-1}(A):A\in\mathcal{A}_i\}$.
Now we apply Kolmogorov's 0-1 law which states, that all events in the tail $\sigma$-algebra (also called terminal $\sigma$-algebra) $\mathcal{T}$ are trivial (i.e. almost sure happen or or almost sure don't)
$$
A\in\mathcal{T}:\mathbf{P}(A)\in\{0,1\}
$$
Now, this means that $\mathcal{T}$ is independent to the other $\sigma(X_i)_{i\in \mathbb{N}}$, since for $A\in\mathcal{T},B\in\sigma(X_i)$ it holds
$$
\mathbf{P}(A\cap B)=\mathbf{P}(A)\mathbf{P}(B) \tag 1
$$
This must be true, because either we have $\mathbf{P}(A)=0$, which means - since $(A\cap B)$ is a subset of $A$ - that we have $\mathbf{P}(A\cap B)=0=\mathbf{P}(A)\mathbf{P}(B)$, or we have $\mathbf{P}(A)=1$. In the latter case we have because of $\mathbf{P}(A\cup B)=1$ again the stated equality of $(1)$
$$
\mathbf{P}(A\cap B)=\mathbf{P}(A)+\mathbf{P}(B)-\mathbf{P}(A\cup B)=1+\mathbf{P}(B)-1=\mathbf{P}(B)=\mathbf{P}(A)\mathbf{P}(B)
$$
This proves that $\mathcal{T}$ is independent to the other $\sigma(X_i)_{i\in \mathbb{N}}$.
