Application of Kolmogorov 0-1 Law for $S_x := \{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}\leq x\}$ I have come across the following exercise during a course on probability, and I'm nearly certain it has to be proved using the Kolmogorov 0-1 Law, but the theorem is only stated in the lecture notes, and no examples on how to apply it were given.

Let $A_1, A_2, \dots$ be any independent sequence of events and let $S_x := \{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}\leq x\}$. Prove that for each $x\in\mathbb{R}$ we have $\mathbb{P}(S_x)=\{0,1\}$.

I've found that the answer is trivial for any $x\in[0,1)^C$, but that's obviously not the point of the exercise.
So far it seems like it is not possible to construct a sequence $B_i$ of independent events such that $S_x$ is in the tail-field, since there is seemingly no way to prove independence of the $B_i$ without more knowledge about $\mathbb{P}:\Omega\rightarrow [0,1]$.
A friend suggested maybe it is possible that you can make a probability triple with a tail-field as its sigma-algebra. Then if we can show that $f_n(\omega)=\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}$ is a measurable function, then  $\lim_{n\rightarrow\infty}f_n^{-1}([-\infty,x])$ would have to be inside the tail-field, hence also have probability $0$ or $1$. 
It seems like if the last approach were to work, we must first show that $1_{A_{i}}$ is measurable in this new probability triple. Meaning we have to show that both $A_i$ and $A_i^C$ have to be in the tail-field. But since I have no prior experience with tail-fields it is not obvious at all how one could construct such a thing (if it's even possible). Any help would be appreciated.
One last note: $\{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}\leq x\} \iff \{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}\leq x\}$.

I found a similar question here, but there seems to be a big jump  in the logic of the first answer.
 A: First we notice that for any $m\geq 1$ we have the following,
\begin{align}
\{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}\leq x\} 
&\iff \{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{m-1}1_{A_{i}} 
+ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}\leq x\} \\
&\iff \{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}\leq x\}.
\end{align}
Now let $\mathcal{T}:=\bigcap_{n=1}^{\infty}\sigma(A_i:i\geq n)$ be the tail field. From the Kolmogorov 0-1 Law we know that if we can show $S_x$ to be in $\mathcal{T}$, that we must have $P(S_x)\in\{0,1\}$. To achieve this we will show that
\begin{equation}
\limsup_{n\rightarrow\infty} \frac{1}{n}\sum_{i=m}^{n}1_{A_{i}} \quad \text{and} \quad \liminf_{n\rightarrow\infty} \frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}
\end{equation}
are both $\mathcal{T}$-measurable. Let $m\geq 1$ and notice that $1_{A_m}$ is $\sigma(A_i:i\geq m)$-measurable. This in turn means that the simple functions $\frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}$ are also $\sigma(A_i:i\geq m)$-measurable. Finally the $\limsup$ and $\liminf$ of measurable functions is again  measurable. Combining this with what we proved at the start we conclude that if $\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}$ converges pointwise to some function, we must have the following for any $m\geq 1$
\begin{equation}
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}}
= \limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}.
\end{equation}
Since the r.h.s. is measurable for all $m\geq 1$, then the l.h.s. must be $\mathcal{T}$-measurable. This proves the claim.
