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Let $X$ be a set and $\{A_\alpha\}_{\alpha \in \mathcal{A}}$ be a partition of $X$. Define $\mathcal{S}=\{\bigcup_{i \in I} A_i \subseteq X \mid I \subseteq \mathcal{A}\}$. Show that $\mathcal{S}$ is a sigma-algebra

To show this I pick $F \in \mathcal{S}$. Now $F = \bigcup_{i} A_i$. So it’s union of some elements that partition $X$. Now to show that this is a sigma-algebra I need to show that $\emptyset \in \mathcal{S}$, $F^c \in \mathcal{S}$ and if I take more elements $F_1, F_2, \dots$, then $\bigcup_{k=1}^\infty F_k \in \mathcal{S}$.

The complement is $F^c = \left(\bigcup_{i}A_i \right)^c = \bigcap_{i} A_i^c$, but how do I know that the intersection of the complements is in $\mathcal{S}$?

Also if $F_k \in \mathcal{S} $, then $\bigcup_{k} F_k = \bigcup_{k} \left( \bigcup_{i} A_i \right)$, but this also doesn’t seem to be easy to work with? Is there another way to show that a set is sigma-algebra?

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1 Answer 1

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Hints:

$$\left(\bigcup_{i \in I} A_i\right)^c = \bigcup_{i \notin I} A_i$$

Think a minute why in your case this equality holds! It uses an assumption of your problem.

Also note that $$\bigcup_{n=1}^\infty \bigcup_{i \in I_n} A_i = \bigcup_{i \in \bigcup_{n=1}^\infty I_n} A_i$$

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  • $\begingroup$ Could the first equality hold since $I \subseteq \mathcal{A}$? I’m trying to figure out what it is saying. $\endgroup$
    – Jiming Le
    Sep 25, 2021 at 14:54
  • $\begingroup$ Yes with $i \notin I$ I mean $i \in I^c = \{x \in \mathcal{A}: x \notin I\}$ $\endgroup$
    – J. De Ro
    Sep 25, 2021 at 14:55

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