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Let $\Omega\not=\emptyset$ be a set. Suppose $\Gamma$ is uncountable and $A_i \subset 2^\Omega$ for each $i\in \Gamma$. Then each $B\in \sigma(A_i, i\in \Gamma)$ corresponds to a countable subcollection $\lbrace A_{i_j}, j\in \mathbb{N}\rbrace \subset \lbrace A_i, i\in \Gamma \rbrace$ such that $B \in \sigma (\lbrace A_{i_j}, j\in \mathbb{N} \rbrace)$.

I am not sure how to do show this since there is not much I can say about $B$. I.e., just because $B$ belongs to $\sigma(A_i, i\in \Gamma)$ does not mean there is an effective way to describe $B$. I don't even think it is correct to say that $B$ is generated by the sets $A_i$ by using countable set operations (but I am not exactly sure).

May I have a hint on how to proceed?

The Substitute
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3 Answers3

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Let $\mathcal{B}$ be the union of all the $\sigma$-algebras generated by countable subcollections of $\Gamma$. If you can show that $\mathcal{B}$ is a $\sigma$-algebra, then this will show that $\mathcal{B}$ is in fact the $\sigma$-algebra generated by $\Gamma$, itself. Meanwhile, clearly every element of $\mathcal{B}$ satisfies the property in the question.

Noah Schweber
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  • I want to see what you say to this one: http://math.stackexchange.com/questions/1518050/why-cant-we-describe-the-elements-in-a-sigma-algebra – David C. Ullrich Nov 07 '15 at 22:58
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Show that the set of $B$ such that $B$ is in the $\sigma$-algebra generated by some countable collection of $A_i$ is a $\sigma$-algebra. This algebra certainly contains every $A_i$, so it contains the $\sigma$-algebra generated by all the $A_i$.

David C. Ullrich
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While there isn't exactly an effective way to describe $\sigma(A_i, i\in\Gamma)$, there is a construction of it in just-barely-uncountably-many stages which will let you get a handle on how its elements are all built up. By transfinite recursion (which, we'll see, really only goes up to the first uncountable ordinal $\omega_1$), define: $$ \begin{align} S_0 &= \{A_i\mid i\in \Gamma\}, \\ S_{\alpha +1} &= \text{set of all complements of $X\in S_{\alpha}$ and countable unions of $(X_n)_{n\ge 0}$ with all $X_n\in S_{\alpha}$,} \\ S_{\lambda} &= \bigcup_{\alpha<\lambda} S_{\alpha} \quad\text{for limit ordinals $\lambda$.}\\ \end{align} $$

It's easy to show that the construction stops adding new sets at stage $\omega_1$, so $S_{\omega_1}$ is a $\sigma$-algebra containing all the $A_i, i\in \Gamma$. Furthermore, it's contained in every $\sigma$-algebra containing all the $A_i$. Thus, $$ S_{\omega_1} = \sigma(A_i, i\in\Gamma). $$

Now, by (transfinite) induction on $\alpha < \omega_1$, it's also easy to see that any $B\in S_{\omega_1}$ involves only countably many of the $A_i$.

The essential fact used in establishing that last claim, as well as the claim that $S_{\omega_1}$ is a $\sigma$-algebra, is: the union of countably many countable sets is countable.

BrianO
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  • http://math.stackexchange.com/questions/54172/the-sigma-algebra-of-subsets-of-x-generated-by-a-set-mathcala-is-the-s/54179#54179 is worth mentioning here. – Asaf Karagila Nov 07 '15 at 23:06
  • @AsafKaragila Indeed it is worth mentioning. I hadn't seen it before. Your answer there supplies several details I hand-waved about here. "How to construct a $\sigma$-algebra in $\omega_1$ easy steps", as someone commented, charmingly. I suspect this is an eternally-returning question. – BrianO Nov 07 '15 at 23:13