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Consider $A = \text{the sets of the form {X $\le$ x}}$. The goal is to prove that $\sigma(A) = \sigma(X)$.

The question seems obvious to me but I just don't know how to prove it. I also have difficulty understanding the definition of $\sigma(X)$. Is $\sigma(X)$ defined as the sigma field generated by the sets that the random variable X refers to? if So, isn't it obvious that $\sigma(A) = \sigma(X)$ should be equal?

I appreciate your help.

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$\sigma(X)$ is the smallest $\sigma$-field on which $X$ is measurable. It is straightforward to prove that $$\sigma(X) = X^{-1}\mathcal{B}(\Bbb{R}) = = \{ \{ X \in B \}\subset \Omega : B \in \mathcal{B}(\Bbb{R})\},$$ where $\mathcal{B}(\Bbb{R})$ is the Borel $\sigma$-field on $\Bbb{R}$. It means that you have to prove that the family $A$ generates all the sets of the form $\{X \in B\}$ for $B \in \mathcal{B}(\Bbb{R})$. – Sangchul Lee Nov 25 '12 at 5:52
Hi @sos440, thanks for your answer...very helpful. – Sam Nov 25 '12 at 6:14
Why the exclamation mark? Is it a random variable factorial? Is it a really exciting random variable? – Rahul Nov 25 '12 at 6:23
Just a habit...nothing special ! – Sam Nov 25 '12 at 6:25
Note that there is nothing special with the word "special" above ;) – Sam Nov 25 '12 at 6:26
up vote 3 down vote accepted

A typical definition of $\sigma(X)$ would be the smallest $\sigma$-algebra such that $X$ is measureable, i.e. $\sigma(X)$ satisfies $\sigma(X) \subset \mathcal A$ whenever $\mathcal A$ is a $\sigma$-algebra such that $X$ is $\mathcal A$ measurable.

The idea should be something like this. Let $\mathcal B$ be the $\sigma$-algebra generated by sets of the form $\{\omega: X(\omega) \le x\}$, and $\mathcal A$ be any $\sigma$-algebra such that $X$ is $\mathcal A$ measurable. To prove the result we must show (1) that $X$ is $\mathcal B$ measurable and (2) that $\mathcal B \subset \mathcal A$. That $X$ is $\mathcal B$ measurable follows from the fact that sets of the form $[-\infty, x]$ generate the Borel $\sigma$-algebra, while $\mathcal B$ is precisely the (edit: $\sigma$-algebra generated by the) inverse image under $X$ of the sets of this form. Argue that this implies $X$ is $\mathcal B$ measurable. To argue that $\mathcal B \subset \mathcal A$, argue that $\mathcal A$ must contain the sets of the form $\{\omega: X(\omega) \le x\}$ and so must also contain the $\sigma$-algebra generated by these sets.

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Hi @guy, thanks for your answer. I'm trying to follow your answer to get it solved... – Sam Nov 25 '12 at 6:15
I guess with your solution, we can prove that $\sigma(X)\subset \sigma(A)$. However, in order to show the equality, we also need to show that $\sigma(A)\subset \sigma(X)$; I wonder how to show this part? – Sam Nov 25 '12 at 6:23
I show that $\sigma(A)$ is contained in every $\sigma$-algebra that $X$ is measurable with respect to. This constitutes step (2). $\sigma(X)$ is a special case of this. – guy Nov 25 '12 at 16:34
I guess you are using different notation (your A is different from mine in my question) so then it causes a little confusion. Regarding your comment, I agree, $\sigma(X)$ is a special case and it means $\sigma(X) \subset \sigma(A)$. However, your argument does not show the other way that $\sigma(A) \subset \sigma(X)$. Right? – Sam Nov 25 '12 at 17:02
In your notation, $\sigma(A) \subset \sigma(X)$ is implied by (2). My notation is $\sigma(A) = \mathcal B$, and I argue that $\mathcal B$ is a subset of every $\sigma$-algebra that $X$ is measurable with respect to; in particular, it is a subset of $\sigma(X)$. Hence $\sigma(A) \subset \sigma(X)$. – guy Nov 25 '12 at 18:40

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