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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!)2012-04-10
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    @erict Could you check if the edits in $latex$ are correct?2012-04-10
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    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 $X^{-1}$.2012-04-10
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    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.2012-04-11
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    Presumably $\sigma\subseteq\Sigma$ was intended to be $\sigma\in\Sigma$.2013-06-18

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