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Im doing a course on measure theory and I'm stuck on one of the exercises.

Take $\{Y_{\gamma}:\gamma \in C\}$ as an arbitrary collection of random variables and $\{X_{n}: n \in N\}$ to be a countable collection of random variables

I now want to show that $$\sigma \{Y_{\gamma} : \gamma \in C\}=\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$$

So i need to show that both $\sigma \{Y_{\gamma} : \gamma \in C\}\subset\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$ and $\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}\subset\sigma \{Y_{\gamma} : \gamma \in C\}$ hold.

I know that for a single random variable X you have $\sigma(X)=\{X^{-1}(B): B \in Borel\}$. How do I expand this for the collection of random variables?

I then also have to show that if $Fn=\sigma\{X_{1},...,X_{n}\}$ and $A=\cup_{n=1}^{\infty}F_{n}$ then $$\sigma(A)=\sigma\{X_{n}: n \in N\}$$

Can anyone help me with this?

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Crossed on MO: mathoverflow.net/questions/107161/… –  Asaf Karagila Sep 14 '12 at 9:50
    
If this is homework, please add the homewok tag. Don't worry, people will still help you. –  Michael Greinecker Sep 14 '12 at 9:51
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But WITHOUT the homework tag, someone will post a complete solution almost instantly. So include the homework tag only if you want help in learning, not merely if you want us to do your work for you. –  GEdgar Sep 14 '12 at 14:08
    
The homework tag was not included, the OP's work was done for them. Ite missa est. –  Did Nov 11 '12 at 16:50

1 Answer 1

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$\def\Bor{\mathrm{Borel}}$ Write $\mathcal A = \sigma\{Y_\gamma^{-1}(B) : \gamma \in C, B \in \Bor\}$. Then every $Y_\gamma$ is $(\mathcal A, \Bor)$-measurable (as $Y_\gamma^{-1}(B) \in \mathcal A$ for each $B\in\Bor$), hence $\sigma\{Y_\gamma : \gamma \in C\} \subseteq \mathcal A$, since the former is the smallest $\sigma$-algebra making the $Y_\gamma$ measurable. On the other hand, all $Y_\gamma$ are $(\sigma\{Y_\gamma:\gamma\in C\}, \Bor)$-measurable, then for $B \in \Bor$ we have $Y_\gamma^{-1}(B) \in \sigma\{Y_\gamma:\gamma \in C\}$. This shows that $\mathcal A \subseteq \mathcal \sigma\{Y_\gamma : \gamma \in C\}$, and hence equality.

For the second claim, let $\mathcal B = \{X_n^{-1}(B) : n \ge 1, B \in \Bor\}$. We now from the first part that $\sigma (\mathcal B) = \sigma\{X_n : n \ge 1\}$. But as $X_n$ is $F_n$-measurable \[ \mathcal B = \bigcup_{n \ge 1} \{X_n^{-1}(B) : B \in \Bor\} \subseteq \bigcup_{n \ge 1} F_n =A \subseteq \sigma\{X_n : n \ge 1 \}. \] This implies \[\sigma\{X_n : n\ge 1 \} = \sigma\mathcal B \subseteq \sigma A \subseteq \sigma\{X_n : n \ge 1\}. \]

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How do you know all $Y_{\gamma}$ are $(\sigma\{Y_{\gamma}:\gamma \in C\}$,Borel)-measurable? And why is $X_{n}$, $F_{n}$-measurable?? –  Math Girl Sep 16 '12 at 19:38
    
By the very definition of $\sigma\{Y_\gamma: \gamma \in C\} = $ "the smallest $\sigma$-algebra $\Sigma$ such that all $Y_\gamma$ are $(\Sigma, \mathrm{Borel})$-measurable. Same for $F_n$ and $X_n$. –  martini Sep 17 '12 at 9:18

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