# Kolmogorov 0-1 law with almost sure convergence

Recently I've tried to solve next problem: consider the sequence of $\{X_n\}_{n=1}^{\infty}$ of independent random variables and $X_n \rightarrow X$ in probability. Prove that the limit $X$ is degenerate ($\exists c \in R: P(X = c) = 1$).

The solution is: as we have convergence in probability we can find a subsequence $n_k$ in $X_n$ such that $X_n \rightarrow X$ almost sure. After that, we need use Kolmogorov 0-1 law, but I don't really understand how.

So, the question is: Consider sequence $X_n$ of independent random variables defined on probability space. We know that $X_n$ converge almost sure to $X$.

I would like to use Kolmogorov 0-1 law to show that $X$ is degenerate.

For that, I need to show that $X$ is measurable with respect to tail sigma algebra. Why that's true?

• What happens to the convergence behavior of a converging sequence if you change it at finitely many places? – Michael Greinecker Mar 16 '18 at 19:55
• The tail $\sigma$-algebra consists of all events independent of each finite subset of the $X_n$. So let $E$ be an event in the tail $\sigma$-algebra and $F$ an event in the $\sigma$-algebra generated by some finite subset of the $X_n$, and show that $$\mathbb P(E\cap F) = \mathbb P(E)\mathbb P(F).$$ – Math1000 Mar 16 '18 at 19:58
• @MichaelGreinecker nothing, I guess – Nikita Durasov Mar 16 '18 at 20:17
• This is the point. – zhoraster Mar 16 '18 at 20:21
• @zhoraster don't know how it helps, really – Nikita Durasov Mar 16 '18 at 21:15

Let $(\Omega,\mathcal{A},\mathbb{P})$ your probability space and $X_n,X:\Omega \to \mathbb{R}$ your random variables such that $X_n \to X$ a.e . Denote $\mathcal{C}_{\infty} = \cap_{m=1}^{\infty} \sigma(X_m,X_{m+1},...)$ the tail sigma algebra of $X_n$.

Now let $A=\{\omega \in \Omega : X_n(\omega) \to X(\omega)\}$ which $\mathbb{P}(A)=1$ and

$C=\{\omega \in \Omega : \lim_{n \to \infty}X_n(\omega) \text{ exists \}}$. Now you can show that $A \subseteq C$ and $C\in \mathcal{C}_{\infty}\text{(exercise)}$. So we have that $\mathbb{P}(C)=1.$

Note: In general it is not true that $A=C$. If $A\neq C$ am not sure if it is possible to formally show that $A \in \mathcal{C}_{\infty}$(am not even sure if this is true) but if $A=C$ you can procceed like this:

Let $b \in \mathbb{R}$ and define $B = C\cap X^{-1}((-\infty,b)).$

Fix $m \in \mathbb{N}$ and i'll show that $B \in \sigma(X_m,X_{m+1},...)$ and this will imply that $B \in \mathcal{C}_{\infty}\text{ (why? :P)}.\\$

For this let $\epsilon>0$ and define $B_{n}^{\epsilon} = C\cap X_{n}^{-1}((-\infty,b-\epsilon))$ for $n \geq m.$ Now observe that

$B_{n}^{\epsilon} \in \mathcal{\sigma}(X_m,X_{m+1},...)$ for all $n\geq m$ and $B=\bigcup_{k=1}^{\infty}\bigcup_{l=m}^{\infty}\bigcap_{n=l}^{\infty}B_{n}^{\frac{1}{k}}\ \text{( exercise)}$.

And this gives us that $B \in \mathcal{\sigma}(X_m,X_{m+1},...).\\$

So we have that $B=C\cap X^{-1}((-\infty,b)) \in \mathcal{C}_{\infty}$ so the set

$C\cap X^{-1}((-\infty,b])\in \mathcal{C}_{\infty} \ \text{( exercise)}\\$.

So for every $b \in \mathbb{R}$ you have that $\mathbb{P}(C\cap X^{-1}((-\infty,b]))\in \{0,1\}$.

Now let $C_b=C\cap X^{-1}((-\infty,b]).$ Since $C_b \nearrow C$ as $b\nearrow \infty$ we have that $\mathbb{P}(C_b)\nearrow \mathbb{P}(C)=1$.

So there exist $b_0$ such that $\mathbb{P}(C_{b_0})=1$ which means that the set

$S=\{b\in \mathbb{R} : \mathbb{P}(C_b)=1\}$ is non empty and since $C_b \searrow \emptyset$ when $b\searrow -\infty$ we have that there exist $b_1$ such that $\mathbb{P}(C_{b_0})=0$ which means that $S$ is bounded above.

Let $s*=\inf(S)$ now for every $n \in \mathbb{N}$ we have $\mathbb{P}(C\cap X^{-1}((-\infty,s^*+\frac{1}{n}])=1$ so

$\mathbb{P}(C\cap X^{-1}((-\infty,s^*]))=\mathbb{P}(\bigcap_{n=1}^{\infty}C\cap X^{-1}((-\infty,s^*+\frac{1}{n}])=1\text{ (exercise).}$ So $s^*\in S.\\$

On the other hand since $s^*=\inf(S)$ we have that $\mathbb{P}(C\cap X^{-1}((-\infty,s^*-\frac{1}{n}]))=0$ for all $n \in \mathbb{N}$

so $\mathbb{P}(C\cap X^{-1}((-\infty,s^*)))=\mathbb{P}(\bigcup_{n=1}^{\infty}C \cap X^{-1}((-\infty,s^*-\frac{1}{n}]))=0\text{ (exercise).}$

Summarizing the above we have that $\mathbb{P}((-\infty,s^*))=0$ and $\mathbb{P}((-\infty,s^*])=1$ which means that $\mathbb{P}(X=s^*)=1.\\$

And eventually we have $\mathbb{P}(X=s^*)=1$ , $\mathbb{P}(X=s)=0$ for all $s<s^*$ and $\mathbb{P}(X=s)=1$ for all $s>s^*.$