# Making sense of measure-theoretic definition of random variable

I have some understanding of measure theory / real analysis, and some understanding of probability theory, but I'm having some trouble putting the two together.

According to Wikipedia:

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $(E, \mathcal{E})$ a measurable space. Then an $(E, \mathcal{E})$-valued random variable is a function $X : \Omega \to \mathcal{E}$ which is $(\mathcal{F}, \mathcal{E})$-measurable.

Now for example, let's take $X$ be a standard Gaussian random variable, $X \sim \mathcal{N}(0, 1)$.

• I think $E = \mathbb{R}$ since $X$ takes values in $\mathbb{R}$.
• Also, we should have $\mathcal{E} = \mathscr{B}(\mathbb{R})$ the Borel $\sigma$-field of $\mathbb{R}$.
• But, what should $(\Omega, \mathcal{F}, P)$ be?

Furthermore, let's try to calculate $\mathbb{E}[X]$ the mean of $X$. By Wikipedia's definition,

$$\mathbb{E}[X] = \int_\Omega X\, dP = \int_\Omega X(\omega)\, P(d\omega).$$

This raises some questions.

• How does this relate to the elementary computation: $$\mathbb{E}[X] = \int_{-\infty}^{\infty} x\cdot f_X(x)\, dx$$ How does $f_X : \mathbb{R} \to \mathbb{R}^{\geq 0}$ relate to the measure-theoretic definition of $X$?
• What is the meaning of $P(d\omega)$? $P$ is a measure so it makes sense to integrate $dP$, but what is $d\omega$?
• You don't usually care about the underlying probability space. What matters is the distribution function of your random variable. But if you need to see an example, just take $\Omega = [0,1]$, $P$ the Lebesgue measure and define $X(\omega) = \Phi^{-1}(\omega)$ where $\Phi$ is the distribution function of a standard normal random variable. Commented Feb 20, 2017 at 12:25
• Not only you don't care about the underlying probability space, but you don't want to care. Consider $X$ the random variable that gives you the result of a dice roll. What is the underlying probability space? Should it be the space of all possible positions of all atoms in the universe ? Something simplier? something more complex? Commented Feb 20, 2017 at 13:03
• As of today, [1] has a description of these five symbols given "in more intuitive terms." [1] en.wikipedia.org/wiki/… Commented Aug 31, 2019 at 23:56

But, what should be $(Ω,F,P)$?

Depends on your application. If want to look at arbitrary random variables, then it is simple arbitrary. If you have a specific example in mind, it is not.

How does this relate to the elementary computation [...]

In this case you assumed that $X \sim \mathcal N[0,1]$. In particular $X$ is a continuous variable. In measure theory we say that the distribution of $X$ is absolutely continuous w.r.t. to Lebesgue measure. The Radon-Nikodym theorem then guarentees the existence of an $f_X$ with the property you have stated, so that we can apply the change of variable formula to make the computation of the expectation easier. Without the change of variable formula, we would have to compute the expectation with the definitions of expectations for indicator functions then simple functions then (?simple integrable then $L_1$ then?) nonnegative functions and then measurable functions.. But, again, this is a particular example, where $X$ is continuous. The measure-theoretic definition of expectation is much more general.

What is the meaning of $P(dω)$?

It doesn't mean anything. It is a notational crutch, like writing $\lim\limits_{n\to \infty} a_n$ instead of $\lim a$. It becomes useful, if you have multiple nested integrals / integrate w.r.t. to product measures.

• Stefan Perko, suggestion: so that we ... 'can apply the change of variable formula to make the computation of the expectation easier. Without the change of variable formula, we would have to compute the expectation with the definitions of expectations for indicator functions then simple functions then nonnegative functions and then measurable functions. ' ?
– BCLC
Commented Mar 14, 2018 at 8:33
• Related question? Computing expectation without change of variable formula
– BCLC
Commented Mar 14, 2018 at 8:34
• @BCLC Knock yourself out. Btw. you can also go 'integrable simple functions' $\to$ '$L_1$-functions' $\to$ 'non-negative functions' if you like. Commented Mar 14, 2018 at 10:18
• Thanks, Stefan Perko! ^-^
– BCLC
Commented Mar 14, 2018 at 11:17
• Stefan Perko, wait, what's $L_1$? integrable?
– BCLC
Commented Mar 14, 2018 at 11:18

The probability space is abstract (We know, or suppose it exists but we don't care how it looks). By Radom-Nikodym Theorem, if a measure $\nu$ is absolutely continuous with respect to other $\mu$, this is $\forall A\in\mathcal{F},\mu(A)=0\implies\nu(A)=0$, if both are $\sigma$-finite, there is a measurable function $f$ such that $$\forall A\in\mathcal{F},\nu(A)=\int_Af\mathrm{d}\mu$$ Besides you prove that in this case for all measurable function $g$ we have that $$\int g\mathrm{d}\nu=\int fg\mathrm{d}\mu$$ in this case $\nu(\dot{})=\int_{\dot{}}f_X\mathrm{d}x,\mu=\mathrm{d}x=$lebesgue measure, obviously $f=f_X$ and $g$ is the random variable $X$. $P(dw)$ is just notation.

You can use the following derivations:

$$\int X dP = \int t dPX^{-1}(dt) = \int t dF = \int t f_{X} dx$$ where F is the distribution function for $$X$$ and $$PX^{-1}$$ is the pull back measure given by $$PX^{-1}(B)=P(X^{-1}(B))$$.