Question: 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 $(\mathcal{F},\mathscr{F})$.

Answer: 
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.
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.