Events

\begin{proof} (Of ) To prove this we need the following lemma:

>[!lemma|3.5.2] >For $B,(B_{n})_{n\in \mathbb{N}}$ Independent, any $S\in\sigma(B_{1},B_{2},\dots)$ is independent of $B$.

\begin{proof} (Of ) To show independence we must show $$\mathbb{P}(S\cap B)=\mathbb{P}(S)\mathbb{P}(B)$$If $\mathbb{P}(B)=0$ then the proof is trivial. Hence, let $\mathbb{P}(B)>0$. Then we define the semialgebra $$\mathcal{J}=\{ D_{i_{1}}\cap\dots \cap D_{i_{n}}:D_{i_{\ell}}=B_{i_{\ell}}\text{ or }B_{i_{\ell}}^{c},\forall n\ge 1 \}\cup \{ \emptyset,\Omega \}$$ Then, for $A\in \mathcal{J}$ we have by independence, $$\mathbb{P}(A)= \frac{\mathbb{P}(B\cap A)}{\mathbb{P}(B)}=\mathbb{P}(A|B)$$Now, we define a new probability measure $\mathbb{Q}:\sigma(B_{1},B_{2},\dots)\to[0,1]$ by $$\mathbb{Q}(S)= \frac{\mathbb{P}(S\cap B)}{\mathbb{P}(B)},\quad\text{for }S\in\sigma(B_{1},B_{2},\dots)$$ Then, $\mathbb{Q}$ is a prob measure since $\mathbb{Q}(\emptyset)=0,\mathbb{Q}(\Omega)=1$ and $\sigma$-additivity holds since $\mathbb{P}$ is a prob measure. But, we see that $\mathbb{Q}$ and $\mathbb{P}$ agree on $\mathcal{J}$ and since $\mathcal{J}\subseteq\sigma(B_{1},B_{2},\dots)$ then by Extension Theorem, they agree on $\sigma(\mathcal{J})=\sigma(B_{1},B_{2},\dots)$. Thus, $$\mathbb{P}(S)=\mathbb{Q}(S)= \frac{\mathbb{P}(S\cap B)}{\mathbb{P}(B)},\quad\forall S\in\sigma(B_{1},B_{2},\dots)$$ \end{proof}

Applying this lemma twice we get the following corollary:

>[!corollary|3.5.3] >Let $(A_{n})_{n\in \mathbb{N}},(B_{n})_{n\in \mathbb{N}}$ be sequences of mutually Independent. If $S\in\sigma(A_{1},A_{2},\dots)$ then $S,B_{1},B_{2},\dots$ are independent.

\begin{proof} (Of ) For $i_{1},\dots,i_{n}$, let $B=B_{i_{1}}\cap\dots \cap B_{i_{n}}$. Note $B,A_{1},A_{2},\dots$ are independent. Thus $$S\in\sigma(A_{1},A_{2},\dots)\implies \mathbb{P}(B\cap S)=\mathbb{P}(B)\mathbb{P}(S)\implies S,B_{1},B_{2},\dots \text{ are independent.}$$Where the first implication is by . \end{proof}

Now we can finish proving the theorem. If for any $n$, $A\in\sigma(A_{n},A_{n+1},\dots)$ then gives us that $$A,A_{1},A_{2},\dots,A_{n-1}\text{ are independent}$$since this holds $\forall n$ we have $$A,A_{1},A_{2},\dots \text{ are independent}.$$ Then, the gives us that $$\text{For }S\in\sigma(A_{1},A_{2},\dots),\text{ we have }S\perp\!\!\!\perp A.$$But, $A\in \tau \subseteq\sigma(A_{1},A_{2},\dots)$, so $A\in\sigma(A_{1},A_{2},\dots)$ and by the logic above we have: $$A\perp\!\!\!\perp A\implies \mathbb{P}(A)=\mathbb{P}(A\cap A)=\mathbb{P}(A)^{2}\implies \mathbb{P}(A)=0\text{ or }\mathbb{P}(A)=1$$ \end{proof} # Random Variables >[!thm] Kolmogorov 0-1 Law (rvs) >Let $(X_{n})_{n\in\mathbb{N}}$ be Independent. Let $\mathcal{F}_{n}=\sigma(X_{0},\dots,X_{n})$, $\forall n\in\mathbb{N}$ and let $\mathcal{G}_{n}=\sigma(X_{n+1},X_{n+2},\dots)$, $\forall n\in\mathbb{N}$ ($\mathcal{G}_{n}\supset\mathcal{G}_{n+1}\supset\dots$). Let $$\mathcal{G}=\bigcap_{n\in\mathbb{N}}\mathcal{G}_{n}$$where $\mathcal{G}$ is called the σ-algebra of tail events. We have that $$P(A)\in\{ 0,1 \}, \ \forall A\in\mathcal{G}$$