We give a simple version of the theorem, usually sufficient for the engineer.
**Sampling (Nyquist-Shannon)** Let $x\in L^2(\mathbb{R})$ a band limited analog signal un signal, such that the frequency band is such that $\nu\in[-\nu_0,\nu_0]$. Then $x$ can be recovered without error from the samples $x(t_n)$ taken at the times $t_n=\frac{n}{2\nu_0} = n T_e$ (with $T_e = \frac{1}{2\nu_0}$). $\nu_f = 2\nu_0$ is called the Nyquist-Shannon sampling frequency.
Elements of the proof are the following. $x$ being band limited, we can construct a periodised version of $\hat x$, denoted by $\hat x^{P}$, such that $$ \hat x^{P}(\nu) = \hat x(\nu)\quad\text{if } \nu\in[-\nu_0,\nu_0] $$ and for all $\nu$ $$ \hat x^{P}(\nu + 2\nu_0) = \hat x^{P}(\nu) $$
$\hat x^{P}$ being a periodic signal, we can develop it in Fourier serie $$ \hat x^{P}(\nu) = \sum_{n=-\infty}^{+\infty} c_n(\hat x^{P})e^{-i2\pi\frac{n}{2\nu_0}\nu} $$ where $c_n(\hat x)$ are the complex Fourier coefficient of $\hat x^{P}$: $$ \begin{aligned} c_n(\hat x^{P}) & = \frac{1}{2\nu_0} \int_{-\nu_0}^{\nu_0} \hat x^{P}(\nu) e^{i2\pi\frac{n}{2\nu_0}\nu} \mathrm{d}\nu\\ & = \frac{1}{2\nu_0} \int_{-\nu_0}^{\nu_0} \hat x(\nu) e^{i2\pi\frac{n}{2\nu_0}\nu} \mathrm{d}\nu \end{aligned} $$
One can remark that the inverse Fourier transform of $\hat x$ at the time $t_n = \frac{n}{2\nu_0}$ is given by $$ x(t_n) = \int_{-\nu_0}^{\nu_0} \hat x(\nu) e^{i2\pi t_n \nu} \mathrm{d}\nu $$ then the Fourier coefficients read $$ c_n(\hat x^{P}) = 2\nu_0 x(t_n) $$
The Fourier serie of $\hat x^{P}$ can then be written as $$ \hat x^{P}(\nu) = \sum_{n=-\infty}^{+\infty} 2\nu_0 x(t_n) 1_{[-\nu_0,\nu_0]} e^{-i2\pi\frac{n}{2\nu_0}\nu} $$ Moreover, we have for all $\nu$ $\hat x(\nu) = \hat x^P(\nu)1_{[-\nu_0,\nu_0]}(\nu) $, i.e. $$ \hat x(\nu) = \sum_{n=-\infty}^{+\infty} 2\nu_0 x(t_n) 1_{[-\nu_0,\nu_0]}(\nu) e^{-i2\pi\frac{n}{2\nu_0}\nu} $$
Finaly, taking the inverse Fourier transform of $\hat x$, we get for all $t$ the Whittaker–Shannon interpolation formula: $$ x(t) = \sum_{n=-\infty}^{+\infty} x(t_n) \mathrm{sinc}(2\nu_0 (t-t_n)) $$ with $\mathrm{sinc}(t) = \frac{\sin(\pi t)}{\pi t}\ $
Under-sampling If the sampling frequency $\nu_f$ is such that $\nu_f<\nu_0$, then the perfect reconstruction is not possible in general without further hypothesis.
Over-sampling if the sampling frequency $\nu_f$ is such that $\nu_f>\nu_0$, then the perfect reconstruction is still possible.