Below is the text of the theorem:

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$\mathcal{F}_{i,j}$ are sigma algebras indexed by $i$ and $j$.

I'm having some difficulties in understanding this proof.

Do the $\mathcal{A}_i$ contain $\Omega$ or is $\Omega \in \mathcal{A}_i$? Also, why would $\bigcup_j \mathcal{F_{i,j}} \subset\mathcal{A}_i$?

Any help would be appreciated.


$\Omega \in \mathcal{A}_i$ follows from the fact that $\Omega \in \mathcal{F}_{i,j}$ for all $i,j$.

Regarding your second question: Let $A \in \bigcup_{j=1}^{m(i)} \mathcal{F}_{i,j}$ for some fixed $i$. Then there exists $j_0 \in \{1,\ldots,m(i)\}$ such that $A \in \mathcal{F}_{i,j_0}$ and therefore we can write

$$A = \bigcap_{j=1}^{m(i)} A_{i,j}$$


$$A_{ij} := \begin{cases} \Omega, & j \neq j_0, \\ A, & j = j_0 \end{cases} \in \mathcal{F}_{i,j}.$$

By the definition of $\mathcal{A}_i$, we get $A \in \mathcal{A}_i$. This proves

$$ \bigcup_{j=1}^{m(i)} \mathcal{F}_{i,j} \subseteq \mathcal{A}_i.$$

  • $\begingroup$ Thanks for your answer. Just to be sure, is it common practice to write in textbooks 'contains' either with respect to $\in$ or $\subset$? It's up to the reader to understand when is which? $\endgroup$ – An old man in the sea. Feb 4 '16 at 21:01
  • $\begingroup$ @Anoldmaninthesea. well...yes... but here $\mathcal{A}_i$ is a collection of sets, so $\Omega \subset \mathcal{A}_i$ doesn't even make sense. $\endgroup$ – saz Feb 4 '16 at 21:04
  • $\begingroup$ Just one more question, which I hope you don't mind. Could you please explain how we can know $\sigma(\mathcal{A}_i)=\mathcal{G}_i$? $\endgroup$ – An old man in the sea. Feb 4 '16 at 21:25
  • $\begingroup$ I think I get it now. $\cap_j A_{ij}$ is a finite intersection in $\cup_j \mathcal{F}_{ij}$, so must be in $\sigma (\cup_j \mathcal{F}_{ij})$ (which is the smallest sigma algebra). Similarly $\sigma (\mathcal{A}_{i})$ should be the smallest sigma algebra containing the same sets. Hence they must be the same. Am I correct? $\endgroup$ – An old man in the sea. Feb 4 '16 at 21:54
  • $\begingroup$ @Anoldmaninthesea.Yes, exactly. $\endgroup$ – saz Feb 5 '16 at 5:53

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