Consider the sketched system below. $x_c(t)$ is an arbitrary, continuous-time signal at the input and $s(t)$ is an impulse train, defined as $s(t)=\sum_{n=-\infty}^{\infty} \delta(t-nT)$, where T is the sampling period. Hence:
$$ x_s(t) = x_c(t)s(t)=\sum_{n=-\infty}^{\infty} x_c(nT)\delta(t-nT) $$
Now I calculate and compare the fourier transform of $x_s(t)$ and $x_c(nT)$.
$$ X_s(j\Omega)=\int_{-\infty}^{\infty} x_s(t)e^{-j\Omega t}dt=\int_{-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x_c(nT)\delta(t-nT)e^{-j\Omega t}dt=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-j\Omega nT} $$
$$ X(e^{j\omega})=\sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n} $$
The results look quite similar. In fact, they are equal if $\omega=\Omega T$.
The calculation is no problem, but the issue I have is understanding what this really means. So obviously it is a scaling. If I define the angular frequency as $\Omega_s=\frac{2\pi}{T}$ and plug it into the above formula for $\omega$, I get $\omega_s=2\pi$. So it seems that through the process of discretization the actual frequency information is lost and the fourier transform goes from 0 to $2\pi$. In order to retrieve the information, I need to multiply $\omega$ with the sampling frequency. Is this statement correct? Why does this actually happen? What meaning is behind this? Probably I am missing an obvious thing, I feel like missing the forest for the trees.
Answer
You've shown that the (continuous-time) Fourier transform (CTFT) of a sampled continuous-time signal equals the discrete-time Fourier transform (DTFT) of the corresponding discrete-time signal. In both cases, the spectrum is periodic. The independent variables are related by
$$\omega=\Omega T\tag{1}$$
where $\Omega$ is the angular frequency ($\Omega=2\pi f$), $T=1/f_s$ is the sampling period ($f_s$ is the sampling frequency), and $\omega$ is the normalized angular frequency, which is often used as the independent frequency variable when describing spectra of discrete-time signals. Note that you could also use $\Omega$ as the independent variable for discrete-time signals, but it always appears together with the sampling interval $T$. This is an important property of sampled or discrete-time signals, and that's also why the two variables $\Omega$ and $T$ are commonly combined into a single variable $\omega=\Omega T=2\pi f/f_s$.
Note that sampling results in a periodization of the spectrum, and this is why information can be lost. This is true for both spectra $X_s(j\Omega)$ and $X(e^{j\omega})$. The periodicity of the spectra implies that the fundamental frequency interval $f\in [-f_s/2,f_s/2]$, or, equivalently, $\Omega\in [-\pi/T,\pi/T]$ or $\omega\in [-\pi,\pi]$ contains all information of the sampled signal, and all frequencies outside that interval are redundant. This is an inherent property of sampled (or discrete-time) signals, and it is the cause of aliasing, i.e. the mapping of different continuous-time signals to the same sampled signal.
