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Thread: measure theory

  1. #1
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    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 $\displaystyle \mathcal{G}$) ?
    Last edited by detalosi; Sep 26th 2018 at 01:46 AM.
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  2. #2
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    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.
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  3. #3
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    Re: measure theory

    Quote Originally Posted by detalosi View Post
    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.
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  4. #4
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    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.
    Last edited by detalosi; Sep 26th 2018 at 11:11 AM.
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  5. #5
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    Re: measure theory

    Quote Originally Posted by detalosi View Post
    @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 $\displaystyle \sigma(\mathcal{G})\subseteq\mathcal{G})$.
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  6. #6
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    Re: measure theory

    Quote Originally Posted by detalosi View Post
    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.
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