My question is in the interpreting the question that is this post: Union of sigma-algebras

Here is the question:

Let $\mathscr{E}_1$ and $\mathscr{E}_2$ be $\sigma$-algebras on the same set $E$.

Their union is not a $\sigma$-algebra, except in some special cases. The $\sigma$-algebra generated by $\mathscr{E}_1\cup\mathscr{E}_2$ is denoted by $\mathscr{E}_1\lor\mathscr{E}_2$. More generally, if $\mathscr{E}_i$ is a $\sigma$-algebra on $E$ for each $i$ in some (countable or uncountable) index set $I$, then $\mathscr{E}_I=\bigvee\limits_{i\in I}\mathscr{E}_i$ denotes the $\sigma$-algebra generated by $\bigcup_{i\in I}\mathscr{E}_i$ (a similar notation for intersection is superfluous, since $\bigcap_{i\in I}\mathscr{E}_i$ is always a $\sigma$-algebra).

Let $\mathscr{C}$ be the collection of all sets $A$ having the form $A=\bigcap\limits_{i\in J}A_i$ for some finite subset $J$ of $I$ and $\underline{\text{sets }A_i \text{ in }\mathscr{E}_i\text{, }i\in J}$. Show that $\mathscr{C}$ contains all $\mathscr{E}_i$ and therefore $\bigcup_{I}\mathscr{E}_i$. Thus, $\mathscr{C}$ generates the $\sigma$-algebra $\mathscr{E}_I$. Show that $\mathscr{C}$ is a p-system.

I have problems understanding the underline part of the question. Does $A_i$ mean the $i^{th}$ set of $\mathscr{E}_i$? What if the $A_1$ of $\mathscr{E}_1$, $A_2$ of $\mathscr{E}_2$, $A_3$ of $\mathscr{E}_3$, .... are all $\phi$? Does the author actually mean all the sets $A_k$ of $\mathscr{E}_i$, but mistakenly put the indices of $A$ and $\mathscr{E}$ to be the same index $i$?

I have also read the solutions to Union of sigma-algebras , one of the answers:

First assertion follows by considering the definition of $C$ with $J=\{i\}$.

This answer does not consider where $A_k\in\mathscr{E}_i$, where $i\neq k$ ...

Please help me understand the question.


To construct an element of $\mathscr C$ you must start with choosing a finite set $J\subseteq I$.

Then for every $i\in J$ you choose an element of $\mathscr E_i$ and label it as $A_i$.

Then finally you take the inclusion $A=\bigcap_{i\in J}A_i$.

(Be careful here: if $J=\varnothing$ then by convention $A=E$)

Then $A$ is an element of $\mathscr C$, and elements of $\mathscr C$ are exactly the sets that can be constructed this way.

  • $\begingroup$ So $A_i$ can be any element of $\mathscr{E}_i$? $A_i$ does not mean the $i^{th}$ element of $\mathscr{E}_i$? $\endgroup$ – green onion Jan 30 at 3:29
  • $\begingroup$ Indeed. There no order in $\matscr E_i $ defined so it makes no sense to speak of something as the $i $-th element of it. $\endgroup$ – drhab Jan 30 at 7:57
  • $\begingroup$ Does it mean for the same finite subset $J\subseteq I$ we might have different $A$ since we can have different $A_i$? $\endgroup$ – green onion Jan 31 at 0:45
  • $\begingroup$ Yes, every fixed finite set $J\subseteq I$ gives you a bunch of sets that are elements of $\mathscr C$. $\endgroup$ – drhab Jan 31 at 7:06

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