What does $[-\pi \; \pi]$ band-limited signal mean?
Now first of all concept of bandlimited signal exists in the context of analog signals not digital ones. Digital signals are always (by their definition and existance) bandlimited. 

Second, frequency in analog signals is expressed either as $f$ in Hertz (cycle per second, repetation per second) or as $\Omega$ in radians per second, where they are related as $$\Omega = 2 \pi f$$ So for example a frequency of $f=1$ kHz would equivalently make a radian-frequency of $\Omega=2 \pi 1k \approx 6283$ rad/sec.

When a bandlimited analog signal $x_a(t)$ is sampled by a period of $T_s$ (and digitized afterwards in practice), the resulting discrete-time sequence $x[n]$ will have a discrete Fourier transform expression of 
$$ H(e^{j\omega}) = \frac{1}{T_s} \sum_k X_a( \frac{\omega + 2\pi k}{T_s})$$

such that it will be frequency normalized to the discrete-time frequency range of $-\pi \leq \omega &lt; \pi$. Where $\omega$ is the discrete-time frequency in radians per sample. Where $X_a$ is the continuous-time Fourier transform (frequency content) of the analog signal $x_a(t)$.

for $k=0$ the base period of $H(e^{j \omega})$ becomes $$ H(e^{j \omega}) = \frac{1}{T_s} X_a( \frac{\omega}{T_s}) $$ where the frequency normalization becomes more apparent as setting $ \Omega = \frac{\omega}{T_s} $ yields the those analog radian-frequencies for $X_a(\Omega)$ when $\omega$ of $H(e^{j \omega})$ is selected from $\omega \in [-\pi,\pi)$