1. ## measure theory

Hey,

Assume that $\displaystyle \mathcal{G}$ and $\displaystyle \mathcal{A}$ is a $\displaystyle \sigma$-algebra and that $\displaystyle \mathcal{G}\subset \mathcal{A}$. Then $\displaystyle \sigma(\mathcal{G}) \subset \mathcal{A}$.

Why is the above true?
Is it because $\displaystyle \sigma(\mathcal{G})$ is the smallest $\displaystyle \sigma$-algebra (i.e. the intersections of all $\displaystyle \sigma$-algebras containing $\displaystyle \mathcal{G}$) ?

2. ## Re: measure theory

Your statement of the problem is ambiguous. You start with "G and A" but then use the singular "is". If G is to be a sigma-algebra then this s trivial.

3. ## Re: measure theory

Originally Posted by detalosi
Assume that $\displaystyle \mathcal{G}$ and $\displaystyle \mathcal{A}$ is a $\displaystyle \sigma$-algebra and that $\displaystyle \mathcal{G}\subset \mathcal{A}$. Then $\displaystyle \sigma(\mathcal{G}) \subset \mathcal{A}$.

Why is the above true?
Is it because $\displaystyle \sigma(\mathcal{G})$ is the smallest $\displaystyle \sigma$-algebra (i.e. the intersections of all $\displaystyle \sigma$-algebras containing $\displaystyle \mathcal{G}$) ?
@detalosi, Please rewrite the question to make it clear as to what you mean.

4. ## Re: measure theory

@HallsofIvy, I appreciate your reply and I see my grammatical error. Being a new and very young student of Measure Theory (and mathematics in general) I have taken the advice of E. T. Bell to heart ("obvious" is the most dangerous word in mathematics). Below I have rewritten my question. Maybe you, Plato or someone else could shed some light on the matter.

@Plato, essentially what I would like to show is the following: If $\displaystyle \mathcal{G}$ is a $\displaystyle \sigma$-algebra then $\displaystyle \mathcal{G}=\sigma(\mathcal{G})$.

I have done the following:

To show that $\displaystyle \mathcal{G}=\sigma(\mathcal{G})$ then we must show 1) $\displaystyle \mathcal{G} \subset \sigma(\mathcal{G})$ and 2) $\displaystyle \sigma(\mathcal{G})\subset \mathcal{G}$

Part 1:

$\displaystyle \sigma(\mathcal{G})$ is the smallest algebra containing $\displaystyle \mathcal{G}$ (i.e. the intersections of all $\displaystyle \sigma$-algebras contining $\displaystyle \mathcal{G}$). Since $\displaystyle \sigma(\mathcal{G})$, by definition, contains $\displaystyle \mathcal{G}$, then $\displaystyle \mathcal{G} \subset \sigma(\mathcal{G})$.

Part 2:

I would like to use that if $\displaystyle \mathcal{G}$ and $\displaystyle \mathcal{A}$ are $\displaystyle \sigma$-algebras and $\displaystyle \mathcal{G} \subset \mathcal{A}$ then $\displaystyle \sigma(\mathcal{G})\subset \mathcal{A}$ (*)

If the above is true then since $\displaystyle \mathcal{G}$ is a $\displaystyle \sigma$-algebra and $\displaystyle \mathcal{G}\subseteq \mathcal{G}$. Then by (*) $\displaystyle \sigma(\mathcal{G})\subset \mathcal{G}$.

I would like to understand why (*) is true.

Thanks.

5. ## Re: measure theory

Originally Posted by detalosi
@HallsofIvy, I appreciate your reply and I see my grammatical error. Being a new and very young student of Measure Theory (and mathematics in general) I have taken the advice of E. T. Bell to heart ("obvious" is the most dangerous word in mathematics). Below I have rewritten my question. Maybe you, Plato or someone else could shed some light on the matter.

@Plato, essentially what I would like to show is the following: If $\displaystyle \mathcal{G}$ is a $\displaystyle \sigma$-algebra then $\displaystyle \mathcal{G}=\sigma(\mathcal{G})$
Did you read reply #2? If $\displaystyle \mathcal{G}$ is a $\displaystyle \sigma$-algebra then the question is trivial.
By definition $\displaystyle \mathcal{G}\subset\sigma(\mathcal{G})$ because $\displaystyle \sigma(\mathcal{G})$ is the intersection of all $\displaystyle \sigma$-algebras containing $\displaystyle \mathcal{G}$
If $\displaystyle \mathcal{G}$ is itself a $\displaystyle \sigma$-algebra then $\displaystyle \sigma(\mathcal{G})\subseteq\mathcal{G})$.

6. ## Re: measure theory

Originally Posted by detalosi
I would like to understand why (*) is true.

Thanks.
I think you are overthinking the problem. Let $\mathcal{G}\subseteq \mathcal{A}$. Then $\mathcal{G}\cap \mathcal{A} = \mathcal{G}$ is a very basic concept in set theory. You should be able to easily see the proof without any work at all.

Suppose you have a collection of sigma-algebras, each containing $\mathcal{G}$. In fact, $\mathcal{G}$ is one of the sigma-algebras in your collection. So, no matter what other sigma-algebra you intersect with it, you get back the same set. It is basic set theory.