# Proove a fact about $\sigma$-algebras in preparation for Kolmogorov's zero–one law

Let $(J_n)_{n\in\mathbb{N}}$ be finite sets which monotonically increase to $I\cong\mathbb{N}$ and $(\mathcal{A}_n)_{n\in\mathbb{N}}$ be a family of $\sigma$-algebras. I want to show that it holds $$A:=\bigcap_{J\subset I\;:\;|J|<\infty}\sigma\left(\bigcup_{i\in I\setminus J}\mathcal{A}_i\right)=\bigcap_{n\in\mathbb{N}}\sigma\left(\bigcup_{i\in I\setminus J_n}\mathcal{A}_i\right)=:B$$

Proof: Clearly "$\subseteq$". For the other direction: Let $J\subset I$ be a finite set. Since $(J_n)_n$ is monotonically increasing, there is a $N\in\mathbb{N}$ such that $J\subseteq J_N$ and we've got

$$\begin{split} \bigcap_{n\in\mathbb{N}}\sigma\left(\bigcup_{i\in J\setminus J_n}\mathcal{A}_i\right)&=\bigcap_{n=1}^N\sigma\left(\bigcup_{i\in J\setminus J_n}\mathcal{A}_i\right)\stackrel{\text{Monotonicity of }\sigma}{\subseteq}\bigcap_{n=1}^N\sigma\left(\bigcup_{i\in I\setminus J_n}\mathcal{A}_i\right)\\ &\subseteq \sigma\left(\bigcup_{i\in I\setminus J_N}\mathcal{A}_i\right)\stackrel{\text{Monotonicity of }\sigma}{\subseteq}\sigma\left(\bigcup_{i\in I\setminus J}\mathcal{A}_i\right) \end{split}$$

If we take the intersection of both sides over all finite $J\subset I$ we get $$\tilde{B}:=\bigcap_{J\subset I\;:\;|J|<\infty}\bigcap_{n\in\mathbb{N}}\sigma\left(\bigcup_{i\in J\setminus J_n}\mathcal{A}_i\right)\subseteq\bigcap_{J\subset I\;:\;|J|<\infty}\sigma\left(\bigcup_{i\in I\setminus J}\mathcal{A}_i\right)=A$$

While the right side perfectly matches $A$, I'm unsure whether or not the left side $\tilde{B}$ equals $B$. I would be really happy if someone could help.

For any $J\subset I$ with $|J|<+\infty$, there is a $J_N \in (J_n)_{n\in \mathbb{N}}$ such that $J \subset J_N$.
So $$\bigcup_{i\in I\setminus J_N}\mathcal{A}_i \subset \bigcup_{i\in I\setminus J}\mathcal{A}_i$$
then $$\sigma\left(\bigcup_{i\in I\setminus J_N}\mathcal{A}_i\right) \subset \sigma\left(\bigcup_{i\in I\setminus J}\mathcal{A}_i\right)$$
so $$\bigcap_{n\in\mathbb{N}}\sigma\left(\bigcup_{i\in I\setminus J_n}\mathcal{A}_i\right) \subset \sigma\left(\bigcup_{i\in I\setminus J_N}\mathcal{A}_i\right) \subset \sigma\left(\bigcup_{i\in I\setminus J}\mathcal{A}_i\right)$$
Since this is true for any $J\subset I$ with $|J|<+\infty$, we can take intersection for the rhs:
$$\bigcap_{n\in\mathbb{N}}\sigma\left(\bigcup_{i\in I\setminus J_n}\mathcal{A}_i\right)\subset \bigcap_{J\subset I\;:\;|J|<\infty}\sigma\left(\bigcup_{i\in I\setminus J}\mathcal{A}_i\right)$$