Sigma field generated by a random variable! 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.
 A: 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. 
