How can spectrum of DTFT of sampled band-limited analog signal be in $[-\pi,\pi]$ Say the analog signal $x(t)$ and its spectrum $X_a(\Omega)$. After sampling with frequency $F_s$ or sampling period $T$ we get 

$$x[n] \triangleq x(nT) = x\left(\frac{n}{F_s}\right)$$

and its spectrum $X_d(\omega)$.

The relationship between the two spectrums is

$$X_d(\omega) =\frac1T \sum ^{\infty}_{n=-\infty}X_a\left(\frac{\omega + 2\pi n}{T}\right)$$
$$\omega=T\Omega$$

For band-limited analog signal having spectrum in range $[-B,B]$. $X_a(\Omega)=0$ for $|\Omega| \geq B$ if $F_s$ is above Nyquist rate then
$$X_d(\omega) =\frac1T X_a(\Omega)$$

For digital signal, its frequency is restrict in $[-\pi,\pi]$ but not the case of analog signal. The equation above means that $X_d(\omega)$ is not restricted in $[-\pi,\pi]$
The DTFT $X_d(\omega)$ of a discrete-time signal is periodic with period $2\pi$, so it's sufficient to consider the interval $[-\pi,\pi]$. "Restricted" to this interval means that due to the inherent periodicity there can be no new frequency components outside the interval $[-\pi,\pi]$. Of course, $\pi$ corresponds to half the sampling frequency.

If $X_a(\Omega)$ is the spectrum of a band-limited continuous-time signal, and if the sampling frequency is greater than twice the band-width, then the equality

$$X_d(\omega)=\frac{1}{T}X_a\left(\frac{\omega}{T}\right)$$

holds in the interval $[-\pi,\pi]$. Outside that interval, $X_a(\Omega)$ is zero, and $X_d(\omega)$ is continued periodically.