5
$\begingroup$

I have thought at this proof of the Kolmogorov 0-1 Law varying a little the sketch found in Probability essentials (Jean Jacod, Philip Protter). My questions are

  1. Is it a valid proof?
  2. Is it a bad proof? (And by bad I mainly mean that written properly will result too heavy)

Given $ \mathcal{B}_n $ the $\sigma$-algebra pulled back from $X_n$ $$\mathcal{C}_n := \sigma\left(\bigcup_{m\geq n} \mathcal{B}_m\right) $$ $$\mathcal{C}_\infty := \sigma\left(\bigcap_{n=1}^{\infty} \mathcal{C}_n\right) $$ $$\mathcal{D}_n := \sigma\left(\bigcup_{i=1}^{n} \mathcal{B}_i\right) $$ $$\mathcal{D}_\infty := \sigma\left(\bigcup_{i=1}^{\infty} \mathcal{C}_i \right) = \mathcal{C}_1 = \sigma\left(\bigcup_{i=1}^{\infty} \mathcal{D}_i \right) $$

I would like to apply what the book calls the Monotone Class Theorem (but Dinkin $\pi$ - $\lambda$ theorem might be a better name according to: https://math.stackexchange.com/a/1841800/88132) to $\bigcup_{i=1}^{\infty} \mathcal{D}_i $.

That set contains $\Omega$ (every $\mathcal{D}_i$ does) and it is closed under differences and finite intersections (I could show why... but come on! Every finite property of that set relies on the finite properties of a single $\mathcal{D}_i$ since $\mathcal{D}_1 \subseteq\mathcal{D}_2 \subseteq\mathcal{D}_3 \subseteq \dots$ should I be more verbose?).

The theorem tells me that $\mathcal{C}_{1} = \sigma \left(\bigcup_{i=1}^{\infty} \mathcal{D}_i\right) $ is the smallest class closed by difference and increasing limit.

This means that every element $A$ of $\mathcal{C}_{1}$ can be written as $$ A = \bigcup_{n=1}^{\infty} A_n \ \text{ where } \ A_n \in \mathcal{D}_n \ \text{ and } \ A_1 \subseteq A_2 \subseteq A_3 \dots $$ or $$ A = \bigcap_{n=1}^{\infty} A_n \ \text{ where } \ A_n \in \mathcal{D}_n \ \text{ and } \ A_1 \supseteq A_2 \supseteq A_3 \dots $$ This is the key point, for me it is in some way obvious that this is true and clearly follows the definition of increasing limit. Should I write something else? What?

Given this and the fact that the book shows that for every $A \in \mathcal{D}_i$ and $B \in \mathcal{C}_{i+1}$ $$ P(A \cap B) = P(A)P(B) $$ The same relation must hold for every $A \in \mathcal{D}_i$ and $B \in \mathcal{C}_{\infty}$ and therefore $$ P(A_n \cap A) = P(A_n)P(A) $$ Taking limits we get $$ P(A) = P(A)^2 $$ From which $$ P(A) = 0 \text{ or } 1.$$

$\endgroup$
0
$\begingroup$

The proof doesn't work since the key point is simply wrong. A generic element might be an infinite union of infinite intersections.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.