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I am reading a tutorial on measure theory and it states: "Given an interval $E = [a, b]$ and a set $S$ of subsets of $E$ which is closed under countable unions, we define the following..."

I was wondering what 'closed under countable unions' meant in this case. Does it mean that $S$ contains every element of $E$?

So if $E = [0,5]$, $S$ could be something like $\{[0,2], [1,3], [2,3], [2,5]\}$ since the union of those would be $[0,5]$... but it would impossible for $S$ to be something like $\{[0,1], [1,4], [2,3]\}$ since the union of those would be $[0,4]$.

Am I correct? Thanks!

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I have edited your question so that the math is played in LaTeX. You can do this yourself by writing math within \$ \$ (see the edit for example). – Stefan Hansen Oct 7 '12 at 14:45
But $S$ is a set of subsets of $E$. So it must also include $[1,5]$. – debayon Mar 23 '13 at 15:17
up vote 5 down vote accepted

$S$ is a collection of subsets of the interval. To say that $S$ is closed under countable unions means that whenever $A_n\in S$, for all $n\in\mathbb N$, then $\bigcup_{n} A_n\in S$ as well.

Note that $S$ does not contain any element of $E$, but rather subsets of $E$.

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It should also mean that $\bigcup\emptyset =\emptyset\in S$, but many authors are not that careful. – Michael Greinecker Oct 7 '12 at 14:45
@MichaelGreinecker: True, depends if $0\in\mathbb N$ or not. – Asaf Karagila Oct 7 '12 at 14:47
Why does that matter. The set $\{E\}$ is closed under unions of sequences, whether the sequences are ordered with nonegative integers or positive integers. – Michael Greinecker Oct 7 '12 at 14:49
@Michael: Oh, yeah. You can also define countable union as if $\mathcal A\in[E]^{\leq\omega}$ then $\bigcup\cal A$ is in the collection. I was thinking about that... – Asaf Karagila Oct 7 '12 at 14:51
Ok thank you for that!! – BYS2 Oct 8 '12 at 12:56

That $S$ is closed under countable unions means that for every countable collection $(A_i)_{i\in\mathbb{N}}$ of sets from $S$, we have that $\bigcup_{i\in\mathbb{N}} A_i \in S$. In particular this must hold for every finite collection of sets, so in your case where $S=\{[0,2], [1,3],[2,3],[2,5]\}$, we have that $[1,3]\cup [2,5]=[1,5]$ which is not in $S$. So this $S$ is not closed under countable union.

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Thank you for your response!! – BYS2 Oct 8 '12 at 12:55

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