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Suppose I have a random variable $ X: \Omega \longrightarrow R$. Where $\Omega$ is the sample space.

Suppose then that I have another random variable $Y$ which is some function of $X$, $Y=f(X)$.

What is the sample space of $Y$? Is it $\Omega$ or $R$?

If the sample space is still $\Omega$ can I also say that $A_1: X<0 , A_2: X \geq 0$ is a partition of $\Omega$ ?, or is it only a partition of $R$ since $X$ is real valued?

I am asking this question because I want to apply the law of total expectations on $Y$, i.e. I want to express $E[Y]$ as

$E[Y]=E[Y|X<0]P(X<0) + E[Y|X \geq 0]P(X \geq 0)$

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The sample space is normally still $\Omega$ in that $Y:=f\circ X :\Omega \to \mathbb{R}$. Furthermore $A_1$ and $A_2$ are disjoint subsets of $\Omega$ in that $$ A_1 = (X<0) = X^{-1}((-\infty,0)) = \{\omega\in\Omega: X(\omega)<0\} \subseteq \Omega, $$ and $$ A_1 = (X\geq 0) =X^{-1}([0,\infty)) = \{\omega\in\Omega: X(\omega)\geq 0\} \subseteq \Omega, $$ for which it holds that $$ A_1\cap A_2 = \{\omega\in\Omega:X(\omega)<0, X(\omega)\geq 0\} = \emptyset. $$ They also cover the entire of our sample space, that is $$ A_1 \cup A_2 = \{\omega\in\Omega: X(\omega)\in \mathbb{R}\} = \Omega. $$ Thus this is a partition of $\Omega$ if we don't require a partition not to include the empty set. Otherwise you have to include more information about $X$ not being strictly negative or non-negative.

To use the formula you use you need $P(A_1),P(A_2)>0$, implying that they indeed are non-empty. However given they have positive probability you can indeed use the law of total expectation.

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