Definition of random variable, Borel $\sigma$-algebra

I´m a beginner in more advanced probability and measure theory and there is this definition that I simply can´t understand. It says, a random variable is a function $X\colon\Sigma\to \mathbb R$ with the property that the set $\{\sigma \subseteq \Sigma": X(\sigma)\in B\}$ belongs to $\mathcal F$ for each Borel set $B$. ($\mathcal F$ is a $\sigma$-algebra).

Does it mean that $X(\sigma)$ should be contained in every possible Borel set for all values of sigma? Or is it the other way around, that we start by "looking" at each Borel set to find out which values of sigma that makes $X(\sigma)$ belong to each one of the Borel sets, and then finally "look" if all these sigmas belong to the sigma algebra $\mathcal F$?

I've had a really hard time trying to find out what this definition really means so your answers would be much appreciated. Thanks in advance! (Sorry, but I´m not used to Latex).

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You can, and should, use $\LaTeX$ to make your question readable. (Also linebreaks!) – Asaf Karagila Apr 10 '12 at 22:21
@erict Could you check if the edits in $latex$ are correct? – azarel Apr 10 '12 at 22:54
While it is entirely equivalent, I think it is easier to think of $X^{-1}(B) \in F$, $\forall B$ Borel. Very loosely it says that events in $\mathbb R$ (Borel sets) are 'compatible' with the sigma algebra $F$ when 'viewed' through $X^{-1}$. – copper.hat Apr 10 '12 at 23:19
Don't forget that Borel sets are generated by open intervals. A common equivalent definition of measurability is that $X^{-1}(-\infty,a]=\{\omega: \ X(\omega)\leq a\}$ belongs to the sigma algebra for every $a$. So if you can check that the preimage on such sets is in your sigma algebra, then your r.v. is measurable. – Alex R. Apr 11 '12 at 5:05
Presumably $\sigma\subseteq\Sigma$ was intended to be $\sigma\in\Sigma$. – Andreas Blass Jun 18 '13 at 17:27

The definition says that $X$ is $F$-measureable, which (as Alex pointed out) amounts to requiring the sublevel sets $\{\omega:X(\omega)\le \alpha\}$ to be in $F$ for every real number $\alpha$. (Equivalently, one can ask for this property with $<\alpha$, $\ge\alpha$, or $>\alpha$).
To see why we might care about such a thing, consider the special case of $X$ taking values $0$ and $1$ only. Then the expected value of $X$ is the measure of the set $\{\omega:X(\omega)=1\}$ which is not defined unless this set is in $F$. In general, measurability is needed to compute the expected value, variance, and other quantities describing $X$.