# Kolmogoroff 0-1 does this proof work?

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 $$P(A) = 0 \text{ or } 1.$$