Given $(\Omega, \mathcal{F}, \mathbb{P})$, why does $\{\omega \in \Omega: X(\omega) \leq x\} \in \mathcal{F}$ imply $X$ is measurable? Given $(\Omega, \mathcal{F}, \mathbb{P})$, the definition of a random variable is a function $X$ from $\Omega$ to the real numbers $\mathbb{R}$, such that: 
$$
\{\omega \in \Omega: X(\omega) \leq x\} \in \mathcal{F}, \ x \in \mathbb{R}
$$
I understand that this is a technical requirement equal to saying that the function $X$ must be measurable. However, I am not sure why it implies $X$ is measurable since it deals with outcomes being in the sigma field.
I had a few questions I was wondering if anyone had insight into:
1) My understanding is that $\{\omega \in \Omega: X(\omega) \leq x\}$ denotes the collection of ALL outcomes in $\Omega$ which cause $X(\omega)$ to be non-infinite, is this correct? 
2) Could $X(\omega)$ be $-\infty$?
3) It seems that the statement $\{\omega \in \Omega: X(\omega) \leq x\} \in \mathcal{F}$ is supposed to imply $X$ is measurable. However, it seems odd as all we are doing is ensuring a collection of outcomes in $\Omega$ is in the sigma field, and that doesn't seem to directly involve $X$ other than the "constraint" $\{X(\omega) \leq x\}$. What would be an example where $X$ is not measurable? Would it involve some $\omega$ not being in the sigma field?
Thank you!
 A: By definition, the measurable function's domain is the measure space (sigma field). That's what you're measuring! 
1) $\{\omega \in Omega : X(\omega)<x\}$, better put, in my opinion, as the set $X^{-1}\big((-\infty, x]\big)$, is the collection of all outcomes that evaluate less than or equal to the real number $x$. The goal isn't to bar it from taking on an infinite value (it cannot take on an infinite valuation or even $-\infty$ since it is only defined to be in the reals). 
2) It could have a LIMIT of $-\infty$ but no $\omega \in \Omega$ is such that $X(\omega)=-\infty$ by definition of the range of $X$ as $\mathbb{R}.$
3) The objects you're considering are sets - a nonmeasurable function would bring some real number $c$ back to a set not in your sigma field. An example of such a set is the Vitali set on the real number line, or any old set not in your sigma field for your particular measure space. Call this nonmeasurable set $V.$ Take $X(v)=1$ if $v \in V$ and $X(v)=0$ otherwise. Then $X^{-1}(1)=V$ a nonmeasurale set, so $X$ is not measurable.  
There are several equivalent "constraint" options: $X<x; X \leq x; X>x; X \geq x.$ Each of the four imply that for each EXTENDED real number (reals $\pm \infty$), $\{\omega \in \Omega : f(\omega)=x\}$ is in the sigma field (CAUTION: this doesn't work the other way around!!!!). 
The value of all this is that you can build intervals out of these four types of rays, and from them you can take unions and intersections and so forth, and get to the real point: $X$ is measurable if and only if $X^{-1}(Q)$ is in the sigma field for every open set $Q$ of the reals. Works for Borel sets, too! 
