I was staring at one of my questions at SE and realized that I do not really understand what I mean by $dP(\omega)$ when I write:

$$EX = \int_{\Omega} X(\omega) \, dP (\omega)$$

where $X: \Omega \to \mathbb{R}$ is a real-valued random variable defined on a probability space $(\Omega, \mathcal{F}, P)$. It seems $P(\omega)$ does not make much sense since $P: \mathcal{F} \to [0, 1]$. I started looking at what Durrett writes in his book, and, when it comes to integrals with respect to measures, there is no any argument next to $P$:

$$EX = \int X \, dP = \int_\Omega X(\omega) \, dP.$$

Question: What does one actually mean by $dP$ in this context?

If it is $P(\{ \omega \})$ then again I cannot make heads or tails of it since $P(\{ \omega \})$ is zero assuming that there are no atoms, and $\{ \omega \}$, $\forall \omega \in \Omega$, should be in $\mathcal{F}$.

Thank you!

Regards, Ivan


1 Answer 1


The notation $dP(\omega)$ doesn't mean that you are plugging $\omega$ in to something called $dP$.

The notation $dP$ is generally used to indicate that your integral is actually a measure-theoretic integral with respect to the measure $P$. The notation $dP(\omega)$ is simply used to indicate that $\omega$ is the "variable" that is allowed to range over your space $\Omega$ that is measured by $P$.

This is fairly common in measure-theoretic integration outside of a probability context as well; you may see something like $\int f(x)\,d\mu(x)$, for instance, to indicate that $x$ is your "variable of integration".

In most contexts, this is clear, and you can simply write $\int X(\omega)\,dP$ or $\int X\,dP$. However, when you start dealing with multiple variables, or when you are considering expectations of integrals or vice versa, this notation is very helpful. For instance, when you are trying to prove that $$ \mathbb{E}[X]=\int_0^{\infty}P(X\geq x)\,dx $$ for a non-negative random variable $X$, the typical first step is to write $$ \int_0^{\infty}P(X\geq x)\,dx=\int_0^{\infty}\int_{\Omega}1_{X(\omega)\geq x}\,dP(\omega)\,dx; $$ here, with multiple variables, it is visually helpful to know which one is associated with which measure. The $dx$ notation makes it perfectly clear where $x$ is coming from; $dP(\omega)$ lets us do the same for $\omega$.

  • $\begingroup$ Thank you for your quick reply. So, $dP$ and $dP(\omega)$ are merely symbols used to denote this type of integration, and we cannot spell them out explicitly to show exactly what $P$ is doing (that is, what set it intakes)? What if we shift to the probability space induced by $X$ and write $\int_\mathbf{R} x \, dP \circ X^{-1}(x)$? $\endgroup$
    – Ivan
    Commented Jan 6, 2014 at 17:04
  • $\begingroup$ I think you'd be advised to avoid that notation; usually, for this, we say something like $\mu=\text{law}(X)$ (that is, $\mu$ is the Borel measure on $\mathbb{R}$ induced by the random variable $X$) and then integrate with respect to $d\mu$. $\endgroup$ Commented Jan 6, 2014 at 17:06

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