MOSFET cut-off region

In cut-off there’s no channel and no current, so the device is off. The condition:

$V_{GS}<V_t$

gate-to-source voltage below the Threshold voltage. With $V_{GS}$ below $V_t$ the gate field is too weak to invert the surface, so no channel forms. No channel means no conducting path between source and drain, and $i_D=0$ regardless of $V_{DS}$.

What it means in a circuit

Cut-off is the open-switch state. In a Common-source amplifier or any resistor-loaded stage, no drain current means no drop across the drain resistor $R_D$, so the output sits all the way up at the supply:

$V_{DS}=V_{DD}(V_{GS}<V_t)$

This is the flat top-left plateau of the Voltage-transfer characteristic: as long as the input is below threshold the output is pinned at $V_{DD}$ and nothing happens. Driving the input further below $V_t$ changes nothing, the device is already fully off. For amplification you have to bias the input above $V_t$ and into the saturation region. Cut-off is the wrong side of the curve for linear signals, but it’s exactly the off-state you want when using the MOSFET as a switch.

Cut-off is one of the three MOSFET regions of operation, alongside the MOSFET triode region and MOSFET saturation region. For a p-MOSFET the off condition is the magnitude version: cut-off when $|V_{GS}|<|V_{tp}|$ (with $V_{tp}<0$).