# Unions of $\sigma$-algebras

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 $$i\neq k$$ ...

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.

• So $A_i$ can be any element of $\mathscr{E}_i$? $A_i$ does not mean the $i^{th}$ element of $\mathscr{E}_i$? – green onion Jan 30 at 3:29
• 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. – drhab Jan 30 at 7:57
• Does it mean for the same finite subset $J\subseteq I$ we might have different $A$ since we can have different $A_i$? – green onion Jan 31 at 0:45
• Yes, every fixed finite set $J\subseteq I$ gives you a bunch of sets that are elements of $\mathscr C$. – drhab Jan 31 at 7:06