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I saw this example given as a $\sigma$-algebra in various places. It goes like this: Let $X$ be a set and assume that the collection $\{A_1,\dots, A_N\}$ is a partition of $X$. Then the collection $\mathcal{F}$ of all unions of sets $A_j$ forms a $\sigma$-algebra.

I have no problems with checking the steps of $\sigma$-algebra definition for this particular case. I am having a hard time understanding $\emptyset\in\mathcal{F}$. I tried to show this in the following way: For any $A_j\in\mathcal{F}$ it is immediate to conclude $A_j^c\in\mathcal{F}$. It is also true that if $A_j\in\mathcal{F}$ and $A_k\in\mathcal{F},j\neq k$ then we $A_j^c\cap A_k^c$. Now to show $\emptyset \in\mathcal{F}$, I used $A_1\cup\dots\cup A_N =X\in\mathcal{F}$ and $X^c=\emptyset$. Then, however,

$$ A_1^c\cap A_2^c\cap \dots\cap A_{N-1}^c\cap A_N^c=A_N\cap A_N^c=\emptyset $$ which is not a union but intersection and I don't see how to conclude this is indeed a member of $\mathcal{F}$.

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  • $\begingroup$ It seems to me like $\emptyset$ should be added as a special case, since obviously you cannot get the empty set from the union of non-empty sets. $\endgroup$ Feb 5, 2016 at 10:56

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"The collection $\mathcal{F}$ of all unions of sets $A_j$" should really mean the following: for any subset $S\subseteq\{A_1,\dots,A_N\}$, the union of all the elements of $S$ is in $\mathcal{F}$, and every element of $\mathcal{F}$ is obtained in this way. To get that $\emptyset\in\mathcal{F}$, you then just take $S=\emptyset$: the union of the empty collection of sets is the empty set.

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