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A sigma algebra is closed under countable intersections.

This is a property of a sigma algebra, but I don't understand what this means in plain English. Can someone explain this to me in terms a n00b like me would understand? The part I don't understand the most is what "closed" means in this quote.

Addendum: The collection of subsets: {0},{x},{y,z},{x,y,z}

Is a sigma algebra of: {0},{x},{y},{z},{x,y},{x,z},{y,z},{x,y,z}


  1. The collection of subsets is not empty -> The collection contains subsets.
  2. Closed under complementation -> The collection of subsets contains {x} and {x,y,z}/{x} = {y,z}
  3. Closed under countable unions -> {x} and {y,z} are contained in our sigma algebra, so is {x} U {y,z} = {x,y,z}
  4. Close under countable intersections -> {x} and {y,z} are contained in our sigma algebra, so is {0}

For point 2. above do you just take the complement of the set of all elements or {x,y,z} (that's what it seems to be according to the Wikipedia page)?

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up vote 4 down vote accepted

In general, "closed under operation * on the set $\Sigma$" means that, if you apply operation * (in this case, taking countable intersections) to elements of the set $\Sigma$ (in this case, the set is the $\sigma$-algebra of subsets of a given set $X$), then the result is still an element of $\Sigma$.

Another example which may be more familiar to you: a vector space is closed under addition (addition of vectors is still a vector)

More precisely, in this case, if $\Sigma$ is a $\sigma$-algebra of subsets of $X$, then if $\{A_i\}_{i=1}^\infty \subset \Sigma$, i.e. $A_1, A_2,\dots, A_n, \dots \in \Sigma$ then $\bigcap_{i=1}^\infty A_i = A_1 \cap A_2 \cap \dots \cap A_n \cap \dots \in \Sigma$.

As I said in a comment above, by allowing countable intersections you get finite intersections too. If $A_1,\dots,A_n \in \Sigma$, then because $X\in \Sigma$ too by definition of $\sigma$-algebra, then using the fact $\Sigma$ is closed under countable intersections, $A_1\cap \dots \cap A_n = A_1 \cap \dots \cap A_n \cap X \cap X \cap X \cap \dots \in \Sigma$.

To answer to your addendum:

There seems to be a bit of confusion. If $X=\{x,y,z\}$, then $\mathcal{P}(X)=\{\emptyset,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$ is the power set of $X$, and a $\sigma$-algebra is a subset of the power set with some additional properties.

So, if $\Sigma=\{\emptyset,\{x\},\{y,z\},\{x,y,z\}\} \subset \mathcal{P}(X)$, then it is a $\sigma$-algebra of subsets of $X$.

To check item 2, what "closed under complementation" means is that, whenever $A\in \Sigma$, we have $A^c=X\setminus A \in \Sigma$. So, we check that for all four elements of $\Sigma$:

$\emptyset \in \Sigma \Rightarrow \{x,y,z\} \in \Sigma$.

$\{x\} \in \Sigma \Rightarrow \{y,z\} \in \Sigma$.

$\{y,z\} \in \Sigma \Rightarrow \{x\} \in \Sigma$.

$\{x,y,z\} \in \Sigma \Rightarrow \emptyset \in \Sigma$.

All these sentences are true, so item 2 is checked.

Also, as a side note, be careful, do not confuse $\emptyset$ with $\{\emptyset\}$.

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So a sigma algebra is a collection of subsets of a set, in this case the set is X or {x,y,z}? That seems to be consistent with everything else. – Soo Jan 28 '11 at 1:35
Exactly. A $\sigma$-algebra is a subset of the set of all subsets of a given set $X$ (in the preceding example, $X=\{x,y,z\}$). So, the elements of a $\sigma$-algebra consist of subsets of $X$. – Bruno Stonek Jan 28 '11 at 1:38
Wow, I think I finally understand sigma algebras :D Thanks! (also thanks for the note about 0 and {0} – Soo Jan 28 '11 at 1:46

A sigma algebra is closed under Countable Intersections means that, if suppose $\tau$ is a $\sigma$-algebra on some set $X$ and if $A_{1},A_{2},A_{3}, \cdots \in \tau$ then $\cap A_{i}$ will also be in $\tau$. Note that you can only take the countable intersections. For more information please read the wikipedia page:

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Can you comment on my addendum to the OP (particularly my question about 2.) – Soo Jan 28 '11 at 1:17

It means that if $A_1, A_2, \dots \in \tau$ then $A_1 \cap A_2 \cap \dots \in \tau$ where $\tau$ is a $\sigma$-algebra on a set $X$.

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No, this is a finite intersection, and you allow countably infinite intersections too. Of course, taking $A_n=X$ for all $n>n_0$ (which is in the sigma algebra by definition), you get finite intersections, but not conversely. – Bruno Stonek Jan 28 '11 at 0:40
Can you two comment on my addendum to the OP (particularly my question about 2.) – Soo Jan 28 '11 at 1:18

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