Is $\{w:\lim_{n \rightarrow \infty}S_n/n \in A\}$ in tail $\sigma$-field? Suppose $\{X_n\}$ are independent random variables on $(\Omega, \mathcal{F},P)$. Set $S_n:=\sum_{i=1}^{n}X_i$.
Suppose $\frac{1}{n}S_n \rightarrow Y \;a.s.$ for some real-valued random variable $Y$. How to show that for any set $A \subset R$ 
$$E:=\{w:Y \in A\} \in \mathcal{T}?$$
Here $\mathcal{T}$ represents the tail $\sigma$-field. 

The trouble I meet:
If for any $t \in N^{+}$,
$$\{\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \in A\} = \{\lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i}{n} \in A\},$$
we actually finished the proof.
My puzzle about above equation:
It's clear that the left hand side implies the right hand side. But I think the right hand side does not necessarily implies the left hand side. What if $X_j(w) = \infty$ for some $j < t$?
I didn't use the fact that $Y$ is a real-valued random variable. I think it may be the key. But How to use it?
 A: Case 1: For all $n\in \mathbb{N}^+$, $X_n$ are real-valued random variables. 
In this case, for all $n\in \mathbb{N}^+$ and for all $w\in \Omega$, $X_n(w)\neq \pm\infty$. 
So, we have that, for any $A\subseteq \mathbb{R}$ and for any $t\in \mathbb{N}^+$    
$$\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}=\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i(w)}{n} \in A\right\}$$
and we can easily conclude that 
$$\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}\in \mathcal{T} \:\:\:\:\:\:\:\: (*)$$
where $\mathcal{T}$ is the tail $\sigma$-field. 
Remark for case 1: On the other hand, since $\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \rightarrow Y \;a.s.$, we have that
$$P\left(\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}\:\Delta\:\{w:Y(w) \in A\}\right)=0$$
But, this information, even combined with $(*)$, is NOT enough to prove that that 
$\{w:Y(w) \in A\} \in \mathcal{T}$. 
Case 2: The random variables $X_n$ may have infinity values ($\pm\infty$).
In this case, the result is false, as shown in the following counterexemple. 
Let, for each $n\in \mathbb{N}^+$, let $\Omega_n=\{0,1\}$, $\mathcal{F}_n=2^{\Omega_n}$ and $P_n$ the probability defined on $\mathcal{F}_n$ by $P_n({0})=1$, $P_n({1})=0$. 
Let $\Omega=\prod_{n=1}^{+\infty}\Omega_n$, $\mathcal{F}=\prod_{n=1}^{+\infty} \mathcal{F}_n$ and $P=\prod_{n=1}^{+\infty} P_n$. For each $n\in \mathbb{N}^+$, $\pi_n: \Omega \to \Omega_n$ is the projection on the $n$th factor.   
Let us,  for each $n\in \mathbb{N}^+$, define the random variable $X_n$ by: 
$$ 
X_n(w)= \left\{\begin{aligned} &0 &\textrm{ if } \pi_n(w)=0; \\ &+\infty  &\textrm{ if }  \pi_n(w)=1  \end{aligned} \right.
$$
It is easy to see that $\{X_n\}$ are independent random variables on $(\Omega, \mathcal{F},P)$. Let $Y$ be a random variable defined on $(\Omega, \mathcal{F},P)$ by 
$$ 
Y(w)= \left\{\begin{aligned} &0 &\textrm{ if, for all } n\in \mathbb{N}^+, \pi_n(w)=0; \\ &1  &\textrm{ if there is }  n\in \mathbb{N}^+ \textrm{ s.t. }  \pi_n(w)=1 \end{aligned} \right.
$$
Now, let $$E= \left\{w : \textrm{ for all } n\in \mathbb{N}^+, \pi_n(w)=0 \right\}$$
It is easy to see that 
$$\left\{w :\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} = Y(w)\right\}=E$$
and 
$$P(E)=1$$
So $\frac{\sum_{i=1}^{n}X_i}{n}$ converges $Y$ a.s.
Now take $A=\{0\}$, we have
$$\left\{w :\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}=\left\{w : Y(w) \in A \right\}=E$$
And, it is easy to see that $E\notin \mathcal{T}$.
A: If the random variables take values in $(\mathbb R, \mathscr B(\mathbb R))$ rather than $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$, then none of the random variables or $Y$ can ever be infinite. I ran into a similar problem.

If the random variables indeed take values in $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$...
Say for example $t=2$ and $X_1(\omega) = \infty$, then...I still don't see what's the problem. Actually, we know that if indeed
$$Y = \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \ \text{a.s.}$$
then $\forall t \in \mathbb N$,
$$\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} = \lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i}{n} \ \text{a.s.}$$
This is because the limit of a sequence of numbers or functions doesn't depend on any finite subset of terms. Hence, the equality in question follows from the aforementioned.
So yeah, assuming the limit indeed exists, I don't think it should matter w/c of the random variables are infinite. Are you thinking something like:
$$Y = \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} = \infty \ \text{a.s.}$$
$$\lim_{n \rightarrow \infty}\frac{\sum_{i=2}^{n}X_i}{n} < \infty \ \text{a.s.}$$
?
That would contradict the above fact. If for example $Y = X_1 = \infty \ \text{a.s.}$, then it indeed follows that
$$\lim_{n \rightarrow \infty}\frac{\sum_{i=2}^{n}X_i}{n} = \infty \ \text{a.s.}$$

Another thing to note is almost surely. Perhaps the equality in question actually holds only almost surely.
