Let $(\Omega,\mathcal{F}, P)$ a probability space with $P: \mathcal{F} \to [0,1] $. In my understanding, measuring the probability of an event (any kind event) is equivalent to measure the "size" of a set in $\mathcal F$. Also, in my understanding, thanks to the Radon-Nikodym theorem we can convert the probability measure $P$ into the Lebesgue measure over $\mathbb R$. In other works, for any set $A\in \mathcal F$ we have $$ P(A)= \int_{A}dP=\int_{X(A)}f_X(x)dx $$ By keeping in mind the conversion "probability measure" $\to$ "Lebesgue measure on the real-line", from the above equality one can easily grasp that $X(A)$ is a subset of $\mathbb R$. Therefore, I find more natural to define $X : \mathcal F \to C \subset 2^{\mathbb R}$, where $C$ is appropriately chosen, rather than $X : \Omega \to \mathbb R$. That way, it is easier to grasp that $X$ maps events into intervals of the real-line. Does it make sense?

Furthermore, why $f_X$ is a distribution? My intuition would suggest that $f_X$ is a distribution because in reality (considering the standard definition of random variable) it should be written $P(A)=\int_{X(A)}f_X(X(\omega))dX(\omega)$, but I am not sure if it this is actually the reason. If so, should the "x-axis" of the plot of $f_X(x)$ represent $X(\omega)$?

  • 2
    $\begingroup$ "In other words, for any set $A\in\mathcal F$ we have". Forget it.If $X:\Omega\rightarrow\mathbb{R}$ is a function, measurable w.r.t. the Borel $\sigma$-algebra $\mathcal{B}$ then it induces a probability $\nu$ measure on $\left(\mathbb{R},\mathcal{B}\right)$ by $\nu A:=P\left(\left\{ \omega\in\Omega:X\left(\omega\right)\in A\right\} \right)$. If $\nu$ is absolutely continuous then the Radon-Nikodym theorem tells us that there is a density $f_{X}$ that satisfies $P\left(\left\{ X\in A\right\} \right)=\nu A=\int_{A}f_{X}\left(x\right)dx$. Compare this with what you write in your question. $\endgroup$
    – drhab
    Nov 27 '15 at 11:05

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