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Let $\{A_{i}\}_{i = 1}^{n}$ be a family of pairwise disjoint subsets of $X$. It is said that $$\mathcal{F}:=\left \{ \bigcup_{i\in I}A_{i}:I\subseteq \{1,\dots, n\} \right \}$$ is a $\sigma$-algebra.

I wanted to check if it's true. The only problem I am stuck with is about showing that it's closed under formation of (countable) unions mathematically: Let $\{I_{k}\}_{i\in\mathbb{N}}$ be a family of subsets of $\{1,\dots,n\}$. I wish to show that

$$\bigcup_{k\in \mathbb{N}}\left ( \bigcup_{i\in I_{k}}A_{{i}} \right )\in\mathcal{F}.$$

But since $|\mathcal{P}(\{1,\dots,n\})|=2^{n}$, then there are $2^{n}$ different among sets $I_{k}$ for $k\in \mathbb{N}$ and hence there are $2^{n}$ different among sets $\bigcup_{i\in I_{k}}A_{{i}}$ for $k\in\mathbb{N}$. So we have $$\bigcup_{k\in \mathbb{N}}\left ( \bigcup_{i\in I_{k}}A_{{i}} \right )=\bigcup_{k=1}^{N}\left ( \bigcup_{i\in I_{k}}A_{{i}} \right )$$ where $|\{1,\dots, N\}|=2^{n}$.

Is this approach correct?

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  • $\begingroup$ It goes in the correct way, but fails when, for example $I_k = \emptyset$ for $k\leq N$ and $I_k = \{2\}$ otherwise. $\endgroup$ – Antoine Aug 15 '15 at 21:41
  • $\begingroup$ Let $A_i = \{i\}$, then $\mathcal{F}$ is the collection of all finite sets of natural numbers. This is not a $\sigma$-algebra. It is not closed under countable unions. $\endgroup$ – Paul Sinclair Aug 15 '15 at 21:42
  • $\begingroup$ @PaulSinclair We have finitely many $A_i$'s. $\endgroup$ – Antoine Aug 15 '15 at 21:43
  • $\begingroup$ @Antoine - I see. I missed that. $\endgroup$ – Paul Sinclair Aug 15 '15 at 21:44
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$\mathcal{F}$ is defined as the set of all unions of $n$ fixed disjoint sets.

This means that taking union of those sets you get a set in $\mathcal{F}$.

Taking sections cannot have as a result a set $F\cup A'_i$ for $F\in \mathcal{F}$ and $A'_i\subset A_i$ (if $F$ does not contain $A_i), because either the whole is contained in both or none of it.

Since there are finitely many $A_i$ that makes it a $\sigma$-algebra.

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