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Let $\mathscr{A}$ be an algebra of sets over $X$. Let $\mathscr{A}_\sigma$ contain $\mathscr{A}$ as well as any countable union of any sequence of members of $\mathscr{A}$. I'm trying to figure out whether or not $\mathscr{A}_\sigma$ is a $\sigma$-algebra over $X$. If it were, we would expect it to be closed under complementation. Then

$\fbox{Let}$ $E = \bigcup_{i=1}^{\infty} X_i$ s.t. all of the $X_i \in \mathscr{A}$.

$\fbox{And suppose}$ $E \notin \mathscr{A}$

$\fbox{Then}$ $E^c \in \mathscr{A}_\sigma$ iff $E^c \in \mathscr{A}$ or $E^c = \bigcup_{i=1}^{\infty} X_i$ for some $\{X_i\}$ in $\mathscr{A}$.

Clearly $E^c \notin \mathscr{A}$ since if it were, its complement $E$ would be in $\mathscr{A}$ which is a contradiction. Now I'm investigating whether $E^c = \bigcup_{i=1}^{\infty} X_i$ for some $\{X_i\} \subseteq \mathscr{A}$ is true. I don't think it's true but I'm trying to think of a counter-example or a way show it needn't hold in general.

Any suggestions?

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up vote 2 down vote accepted

Let $X=\mathbb{R}$ and let $$\mathscr{A}=\{E\subset\mathbb{R}:E\mbox{ is a finite set or } E^c\mbox{ is a finite set}\}.$$ It is easy to see that $\mathscr{A}$ is an algebra. Now for every $i\in\mathbb{N}$, let $X_{i}=\{i\}\in\mathscr{A}$, and let $E=\cup_{i\in\mathbb{N}}X_i=\mathbb{N}$. Then $E^c$ cannot be a countable union of sets in $\mathscr{A}$.

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