# Homework Help: Sigma Algebra elements

1. Sep 26, 2012

### IniquiTrance

1. The problem statement, all variables and given/known data
Suppose $\Omega = \mathbb{R}$. Let $A_n = [0,\frac{1}{n}]$. Let $\mathcal{A}=\{A_n:n\in\mathbb{N}\}$.

Is $\{0\}$ an element of $\mathcal{F}=\sigma(\mathcal{A})$?

2. Relevant equations

3. The attempt at a solution
Clearly $\lim_{n\rightarrow\infty} A_n = \{0\}$. Not sure how to show that it need be in $\mathcal{F}$ though.

2. Sep 26, 2012

### jbunniii

Think about the kinds of operations under which a sigma algebra is closed. Also, think about what you mean by the "limit" of the sequence of sets. What operation are you really performing to assert that the "limit" is {0}?

3. Sep 26, 2012

### IniquiTrance

Hmm, I know we need countable additivity. Other than that, not quite sure how to proceed.

4. Sep 26, 2012

### Dick

'Countable additivity' sounds like a property of a measure on a sigma algebra. What do you know about set theoretic properties of sigma algebras, things like complements, unions and intersections??

5. Sep 27, 2012

### non-logical

Hello!
My question will be somehow related to the topic.
I'm reading Kallenberg's Foundations now and at the very beginning I've found out that he didn't define limits before saying the following:
>> Note that any sigma-field is closed under monotone limits.
Sigma algebra A has been introduced as a non-empty subset of the set of all subsets of some set omega. The A was to be closed under certain set-theoretical operations as union, intersection and complement.
How am I to understand the limit?
My intuition (as this seems what I'm supposed to rely on) doesn't work. Could one work out the "meaning" of the limit from the information above? And, please, how?

6. Sep 27, 2012

### Ray Vickson

Additivity (countable or otherwise) has nothing to do with it. You need closure under complementation and countable unions. From that you can deduce that the empty set and the whole space belong to the σ-algebra, and do countable intersections. Of course, "countable" includes "finite".

RGV

7. Sep 27, 2012

### jbunniii

If $A_1 \supset A_2 \supset A_3 \supset \ldots$, then you can reasonably define the "limit" as the set of points that appear in every $A_n$, which is simply $\cap_{n=1}^{\infty} A_n$.

Similarly, if $A_1 \subset A_2 \subset A_3 \subset \ldots$, then you can define the "limit" as the set of points that appear in at least one $A_n$, in other words $\cup_{n = 1}^{\infty} A_n$.

This notion shows up, for example, in the monotone class theorem:

http://en.wikipedia.org/wiki/Monotone_class_theorem

I don't like this nomenclature and prefer to call the intersection and the union what they are.

8. Sep 27, 2012

### IniquiTrance

Don't you need Cantor's intersection theorem for that?

9. Sep 27, 2012

### jbunniii

For what? The intersection $\cap_{n = 1}^\infty A_n$ is well defined for any sets $A_n$. Cantor's intersection theorem ensures that under certain conditions, the intersection is nonempty. But the empty set is a perfectly good set, contained in any sigma algebra.

10. Sep 27, 2012

### IniquiTrance

Hmm, I'm just trying to make it rigorous... Why do we know that that intersection contains only {0}, when it could conceivably be,

* Empty
* Non-empty with an infinite amount of points

11. Sep 29, 2012

### non-logical

Thank you, jbunniii.
By the way, despite the definitions are reasonable in cases of $A_1 \subset A_2 \subset A_3 \subset \ldots$ and $A_1 \supset A_2 \supset A_3 \supset \ldots$, I think we can define the limits regardless to those conditions. That is for any countable set of sets we define both intersection and union limits. The limit can be then either null set or the whole Ω.

12. Sep 29, 2012

### jbunniii

That is true in general, but in this case you know exactly what the sets are: $A_n = [0, 1/n]$. Clearly, $0 \in A_n$ for all $n$, so the intersection contains 0. And for any positive number $r$, there is an $N$ such that $1/N < r$, so $r \not\in A_N$ and hence $r \not\in \cap A_n$. So this establishes that $\cap A_n = \{0\}$.

Also note that since $A_n \in \mathcal{A} \subset \mathcal{F}$ for each $n$, and $\mathcal{F}$ is a sigma algebra, it follows that $\cap A_n \in \mathcal{F}$ regardless of whether the intersection is empty, or contains one point, or contains many points.

Last edited: Sep 29, 2012
13. Sep 29, 2012

### IniquiTrance

Thank you. I understand it much better now.