Let $\{A_n\}$ be a sequence of independent events. Show that, for every $x$, the event $$A=\left\{\omega:\frac1{n}\sum_{k=1}^n I_{A_k}(\omega) \rightarrow x\right\}$$ has probability either $0$ or $1$.
I think if I show this event as a tail event then the event has probability $0$ or $1$ by Kolmogorov $0$-$1$ law. Any Help?
Best Answer
The event under consideration is independent from any finite number of $A_n$. The sum converges if its tail converges, so to measure it, you need to know the tail behavior of the $A_n$'s. The event $A$ is determined by the sequence $\{A_1,A_2, \dots\}$ but not by a finite set of the form $\{A_1,A_2, \ldots A_m\}$. Hence it belongs to the tail $σ-$ algebra.
This is the basic idea, which you can formulate as follows. Let $$σ(A_m, A_{m+1}, \ldots)$$ be the $σ-$ algebra generated by all events $A_k$ with $k\ge m$. The tail $σ-$ algebra $\mathcal T$ is defined as $$\mathcal T:=\bigcap_{n=1}^{+\infty}σ(A_m, A_{m+1}, \ldots)$$ Now, the event $A:=\left\{\omega:\frac1n\sum_{k=1}^n I_{A_k}(\omega) \rightarrow x\right\}$, for any $m>0$ depends only on the events $A_k$ with $k\ge m$ (see an explanation below) and hence it belongs to $σ(A_m, A_{m+1},\ldots)$ for every $m\ge 1$ (we can remove the first $m$ terms and the tail behavior remains unchanged). Hence $A$ belongs to the intersection of these $σ-$ algebras which is by the definition the tail $σ-$ algebra $\mathcal T$.
By the Kolmogorov $0-1$ law, every event in $T$ has probability either $0$ or $1$ (although it is usually hard to say which).
To see why $A$ depends only on the events $A_k$ for $k\ge m$, fix $m<n$ and write: $$\frac1n \sum_{k=1}^nI_{A_k}(ω)=\frac1n \sum_{k=1}^mI_{A_k}(ω)+\frac1n \sum_{k=m+1}^nI_{A_k}(ω)$$ Letting $n \to +\infty$ this gives \begin{align}\lim_{n\to+\infty}\frac1n \sum_{k=1}^nI_{A_k}(ω)&=\lim_{n\to+\infty}\frac1n \sum_{k=1}^{m}I_{A_k}(ω)+\lim_{n\to+\infty}\frac1n \sum_{k=m+1}^nI_{A_k}(ω)\\[0.2cm]&=0+\lim_{n\to+\infty}\frac1n \sum_{k=m+1}^nI_{A_k}(ω)=\lim_{n\to +\infty}\frac1n\sum_{k=1}^{n}I_{A_{m+k}(ω)}\end{align} This shows that the limit does not depend on the first $m$ events and since $m$ was arbitrary (but fixed) this shows the assertion.