Independence of sums of infinitely many random variables Let $\{X_i \}^\infty_1$ be independent random variables (that is, any finitely numbers of random variables are independent). Furthermore, if $S_n=\sum_1^n X_i$ converges almost surely to a random variable $S$, then are $S_n$ and $S-S_n$ independent?
It is easy to show that $S_n$ and $S_m-S_n$ are independent for every $m>n$, but it seems that there is a big gap between the finite version and infinite version. Since letting $m\rightarrow \infty$ for the both sides of 
$$P(S_m-S_n>a, S_n>b)=P(S_m-S_n>a)P(S_n>b)$$
may not yield 
$$P(S-S_n>a, S_n>b)=P(S-S_n>a)P(S_n>b)$$
I notice that someone asks a similar question here. But till now there are no answers and I don't know if the result for the independence of two sigma-algebras will answer my question.
 A: The event $S - S_n > a$, i.e. $\sum_{j=n+1}^\infty X_j > a$, is in the $\sigma$-algebra generated by $X_j$ for $j \ge n+1$, e.g. up to a set of probability $0$ it can be written
as 
$$ \bigcap_{K = 1}^\infty \bigcup_{M = 1}^\infty \bigcap_{m = M}^\infty
\left\{\sum_{j=n+1}^m X_j > a + 1/K \right\}$$
while the event $S_n > b$ is in the $\sigma$-algebra generated by $X_j$ for $j \le n$, so they are independent.
A: Yes, $S_n$ and $S-S_n$ are independent. To prove this, let $\mathcal{H}$ denote the $\sigma$-algebra generated by $\{ X_i \}_{1 \leq i \leq n}$, and let $\mathcal{F}$ denote the $\sigma$-algebra generated by $\{X_j\}_{j >n}$.
Then $\mathcal{H}$ and $\mathcal{F}$ are independent $\sigma$-algebras (because the random vectors $(X_1, ...,X_n)$ and $(X_{n+1},...X_{n+m})$ are independent, for any $m \in \mathbb{N}$).
Since $S_n$ is $\mathcal{H}$-measurable and $S-S_n$ is $\mathcal{F}$-measurable, the independence of $\mathcal{H}$ and $\mathcal{F}$ implies the independence of $S_n$ and $S-S_n$.
