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Give an example of two $\sigma$ algebras in a set $X$ whose union is not an algebra.

I've considered the sets $\{A|\text{A is countable or $A^c$ is countable}\}\subset2^\mathbb{R}$, which is a $\sigma$ algebra. I also tried to generated a $\sigma$ algebra from a collection of $\sigma$ algebra, but I've been unfruitful. So far I know $2^X,\{\emptyset, X\},$ and the measurable sets $\mathcal{L}$ are $\sigma$ algebras, but they haven't helped me too much. Thank you!

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The trivial $\sigma$-algebra on $X$ is $\{\emptyset, X\}$. The next simplest $\sigma$-algebras on $X$ are $\{\emptyset, A, A^c, X\}$ where $A \in \mathcal{P}(X)\setminus\{\emptyset, X\}$. Can you construct a counterexample using such $\sigma$-algebras?

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Let $\mathbb{N}$ be the sample space. Then look at the $\sigma$-algebra generated by $1$ and the $\sigma$-algebra generated by $2$. Is their union a $\sigma$-algebra?

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  • $\begingroup$ Hi, what does 'generated by 1' mean? Thank you. $\endgroup$ – user70520 Sep 25 '13 at 2:59
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    $\begingroup$ @user70520 $\sigma$-algebra generated by $1$ is $\sigma_1 = \{\emptyset, \{1\}, \{1\}^c, \mathbb{N}\}$. Similarly, $\sigma$-algebra generated by $2$ is $\sigma_2 = \{\emptyset, \{2\}, \{2\}^c, \mathbb{N}\}$. Now $\sigma_1 \cup \sigma_2 = \{\emptyset, \{1\}, \{2\}, \{1\}^c, \{2\}^c, \mathbb{N}\}$. Is the last one a $\sigma$-algebra? $\endgroup$ – user17762 Sep 25 '13 at 3:00
  • $\begingroup$ Hi user17762! I hope you can answer me back. I was trying to use your example, and I think that the problem relies in the countable union of sets in $\sigma_1\cup\sigma_2$, for ex., $\{\{1\},\{2\}\}\in\sigma_1\cup\sigma_2$ but $\{\{1\},\{2\}\}^c\not\in\sigma_1\cup\sigma_2$, is this right? thank you $\endgroup$ – Ana Galois Nov 20 '13 at 20:16

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