Rigorous definition of characteristic function of random variable as Lebesgue integral

Let $(\Omega, \mathcal{A}, P)$ be a probability space and $(\Omega', \mathcal{A}')$ be a measurable space. Furthermore, let $X :\Omega \rightarrow \Omega'$ be a random variable.

If $(\Omega', \mathcal{A}') = (\mathbb{R}, \mathcal{B}_{\mathbb{R}})$, then the expected value of $X$ is defined (if it exists) as the Lebesgue integral of $X$ over the measure space $(\Omega, \mathcal{A}, P)$ :

$$E(X):= \int_{\Omega} X \ dP.$$

I am currently learning about characteristic functions of random variables. The usual definition for the characteristic function $\phi_X : \mathbb{R} \rightarrow \mathbb{R}$ of a random variable is:

$$\phi(t) := E(e^{itX}) .$$

Usually, no further explanation is given. I have problems bringing this definition together with the definition of the expected value as a Lebesgue integral.

1.) The function $e^{itX}$ is complex-valued and the Lebesgue-integral is only defined for real-valued functions. How can this be resolved?

2.) Is there a general way to define a Lebesgue-integral for complex-valued measurable functions $g : \Omega \rightarrow \mathbb{C}$. If yes, how does it work and what is the usual $\sigma$-algebra on $\mathbb{C}$?

I am happy to hear any thoughts on this to get some more clarity about this obscure definition. I am interested to understand the definition of a characteristic function in the most rigorous way possible. Thanks in advance for any help.

For a given complex-valued function $f;\Omega \to \mathbb{C}$ we can use the decomposition

$$f(x) = \text{Re} \, f(x) + i \, \text{Im} \, f(x)$$

to define the integral of $f$. Indeed, since the real part and the imaginary part are both real-valued functions, we know how to integrate them, and therefore the following definition makes sense:

$$\int f(x) \, \mu(dx) := \int \text{Re} f(x) \, \mu(dx) + i \int \text{Im} \, f(x) \, \mu(dx) \tag{1}$$

where $\mu$ is some measure on the measurable space $(\Omega,\mathcal{A})$ (e.g. the Lebesgue measure or the distribution of a random variable). Obviously, we need to ensure that the integrals on the right-hand side of $(1)$ are well-defined, i.e. we need $\text{Re} f \in L^1(\mu)$ and $\text{Im} \, f \in L^1(\mu)$.

Note that the so-defined integral is linear - this is one of the most important properties of the integral. Moreover, if $f$ is a real-valued function, then we recover the "classical" integral for real-valued functions.

The most common $\sigma$-algebra on $\mathbb{C}$ is the Borel-$\sigma$-algebra $\mathcal{B}(\mathbb{C})$, i.e. the $\sigma$-algebra induced by the metric

$$d(z_1,z_2) := |z_1-z_2| := \sqrt{(\text{Re}(z_1-z_2))^2 + (\text{Im}(z_1-z_2))^2}.$$

It is possible to show that a function $f: (\Omega,\mathcal{A}) \to (\mathbb{C},\mathcal{B}(\mathbb{C}))$ is measurable if, and only if, $\text{Re}\, f: (\Omega,\mathcal{A}) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $\text{Im} \, f: (\Omega,\mathcal{A}) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ are measurable. This stems from the fact that the bijective mapping

$$\iota: \mathbb{C} \to \mathbb{R}^2, z \mapsto (\text{Re} \, z, \text{Im} \, z)$$

is bi-measurable, i.e. both $\iota$ and its inverse are (Borel-)measurable.