# 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.

• Did you mean $\sum_{i = m}^n A_i \leq x$? If so, then the event is a tail even as it belongs to $\sigma(A_k; k \geq m)$ for any $m,$ hence the result. Feb 23, 2019 at 1:02
• Kolmogorov's 0-1 law tells that, if $X_n$'s are independent random variables and $\mathcal{T}=\bigcap_{n=1}^{\infty} \sigma(X_k:k\geq n)$ is the corresponding tail $\sigma$-algebra, then every $\mathcal{T}$-measurable event is $\mathbb{P}$-trivial, i.e., $\mathbb{P}(A)\in\{0,1\}$ for any $A\in\mathcal{T}$. Now with $X_n=\mathbf{1}_{A_n}$, it is easy to check that $S_x \in \mathcal{T}$, i.e., $S_x \in \sigma(A_k:k\geq n)$ for any $n$, and so, $\mathbb{P}(S_x) \in\{0,1\}$. Feb 23, 2019 at 1:03
• @WillM. No, it's exactly like above. Feb 23, 2019 at 1:07
• What you wrote doesn't make sense really, what is $i$ then? Also, you are overthinking an easy peasy question. Simply note that whatever the value of $m$ maybe, $m/n \to 0$ as $n \to \infty,$ hence, the first $m$ variables are irrelevant, this is why the event is a tail event. Feb 23, 2019 at 1:08
• Then the obvious idea is to show that $S_x \in \tau$, i.e., $S_x \in \sigma(A_k : k \geq n)$ for each $n$. But this is obvious since, for each $n$, both $$\limsup_{N\to\infty} \frac{1}{N}\sum_{k=n}^{N}\mathbf{1}_{A_k}, \qquad \liminf_{N\to\infty} \frac{1}{N}\sum_{k=n}^{N}\mathbf{1}_{A_k}$$ are limits of $\sigma(A_k : k \geq n)$-measurable functions, hence themselves being $\sigma(A_k : k \geq n)$-measurable as well. Feb 23, 2019 at 1:38

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

$$$$\limsup_{n\rightarrow\infty} \frac{1}{n}\sum_{i=m}^{n}1_{A_{i}} \quad \text{and} \quad \liminf_{n\rightarrow\infty} \frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}$$$$

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 $$m\geq 1$$

$$$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}1_{A_{i}} = \limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=m}^{n}1_{A_{i}}.$$$$

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.