Behaviour of Voltage on Base in a fixed BIAS Common Emitter Amplifier
The transistor does not just go from hard on to hard off.  Rather, the emitter current of the transistor follows the diode equation.  Roughly, \$i_e = I_{SS}\left(e^\frac{v_{be}}{V_T} - 1\right)\$ (Google for the diode equation for the definitions of \$I_{SS}\$ and \$V_T\$).  At room temperature, \$V_T \simeq 26\mathrm{mV}\$.

So in your simple circuit, a small (like, less than 1mV or so peak-peak) signal will act to increase and decrease the emitter current in an amount proportional to the signal voltage.