# $\int_a^b \mathbb{P}(\mathrm{d}x) = \int_a^b \mathbb{P}(x)\mathrm{d}x$?

Putting $\mathrm{d}x$ inside a probability, is the same to write $\mathbb{P}(x)\mathrm{d}x$ like in the following example?

$\int_a^b \mathbb{P}(\mathrm{d}x) = \int_a^b \mathbb{P}(x)\mathrm{d}x$

## 2 Answers

In general, the $\mathbb{P}$-integral of a measurable function $X$ on $\left(\Omega, \mathcal{A}, \mathbb{P} \right)$ is given by

$$\int_{\Omega} X(\omega) \ d \mathbb{P}(\omega)$$

which is also written as

$$\int_{\Omega} X(\omega) \ \mathbb{P}(d\omega)$$

In probability theory this is just the expectation of $X$ under the measure $\mathbb{P}$. $\int_\Omega \mathbb{P}(x)\mathrm{d}x$ on the other hand, I have never seen anywhere - it is not even clear if this is an abstract Lebesgue integral or a Riemann integral.

Just started to write a similar question before already finding it asked before on SO. Your suspicion is probably correct, at least as the notation $Pr(dx,dy)$ instead of $\mathbb P$ is concerned. See this SO answer to a question regarding an equation in the book The elements of statistical learning.

@Jerry wrote:

For the notation of $Pr(dx,dy)$, it is equal to $g(x,y)dxdy$, where $g(x,y)$ is the joint pdf of $x$ and $y$.