Notation: Expectation of empirical measure In general, if $X$ is a random variable defined on a probability space $(Ω, Σ, P)$, then the expected value of $X$ is defined as
\begin{align}
\int_\Omega X \, \mathrm{d}P = \int_\Omega X(\omega) P(\mathrm{d}\omega) 
\end{align}
Let $X_1, X_2, \dots$ be a sequence of independent random variables identically distributed with probability measure $P$.
The empirical measure $P_n$ is given by
\begin{align}
P_n =\frac{1}{n}\sum_{i=1}^n \delta_{X_i}
\end{align}
Is the following the correct notation for the expectation with respect of the empirical measure?
\begin{align}
\int_\Omega X \, \mathrm{d}P_n = \int_\Omega X(\omega) P_n(\mathrm{d}\omega) = \frac{1}{n}\sum_{i=1}^n X_i
\end{align}
Edit:
What is important for me, is that $P_n$ is an approximation for $P$.
I think $X$ and $X_1,\dotsc,X_n$ should map from $(Ω, Σ)$ to some measureable space $(\mathcal{F},\mathscr{F})$.
 A: It's not correct.
The empirical measure isn't a measure on the sample space $\Omega$, it's a (random) measure on $\mathbb{R}$.  Notationally, I think most people reserve letters like $P, P_n$, etc, for measures on $\Omega$, using letters like $\mu, \nu$ for measures on other spaces.
So I'd call your empirical measure $\mu_n$ and then write its mean as 
$$\int_{\mathbb{R}} x\,\mu_n(dx) = \frac{1}{n} \sum_{i=1}^n X_i.$$
Note that the left-hand side denotes the integral over $\mathbb{R}$, with respect to the measure $\mu_n$, of the identity function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = x$.  The lower-case $x$ is intentional and not a typo.
A: Let us formalize this a bit. Let $(\Omega_0,\Sigma_0,P_0)$ be a probability space and take $X_1,X_2,\ldots$ be a sequence of independent random variables on $\Omega_0$ with values in $\mathbb R,$ we consider the random measure $P_n$ on $\mathbb R$ given by
$$P_n=\frac{1}{n}\sum\limits_{i=1}^n\delta_{X_i}$$
where $\delta_{X_i}(A)=1$ when $X_i\in A,$ that is with probability $P(A).$ Fix now $\omega_0\in\Omega_0$ and consider the measure $P_n(\omega_0).$ Additionally, take some random variable $Y$ on $\mathbb{R}.$ Then
$$\int\limits_{\mathbb{R}}Y\mathrm{d}P_n(\omega_0)=\frac{1}{n}\sum\limits_{i=1}^n\int\limits_{\mathbb{R}}Y\mathrm{d}\delta_{X_i(\omega_0)}.$$
Since $$f(x)=\int\limits_{\mathbb{R}}f\mathrm{d}\delta_x,$$
we obtain
$$\int\limits_{\mathbb{R}}Y\mathrm{d}P_n(\omega_0)=\frac{1}{n}\sum\limits_{i=1}^n(Y\circ X_i)(\omega_0).$$
In other worlds,
$$\frac{1}{n}\sum\limits_{i=1}^n(Y\circ X_i)=\int\limits_{\mathbb{R}}Y\mathrm{d}P_n:\Omega_0\to\mathbb{R}$$
is a random variable on $\Omega_0.$
The random measure $P_n$ is a good approximation of $P$ in the following sense.
Pick some $A\in\mathcal{B}(\mathbb R).$ Then $(\delta_{X_i}(A))_{i\in\mathbb{N}}$ are iid random variables on $\Omega_0$ with expectation $P(A).$ The strong law of large number implies
$$P_n(A)=\frac{1}{n}\sum\limits_{i=1}^n\delta_{X_i}(A)\xrightarrow{n\to\infty}\mathbb{E}[\delta_{X_i}(A)]=P(A)$$
almost surely, so that $P_n\to P$ pointwise, almost surely.
From the comment below, it even can be shown that this convergence is uniform.
