voltage leak to MOSFET gate? Fried?
You say you are probing the gate, but you are watching the drain on the scope? How is this possible without a second probe on the drain?  

Assuming this is actually what you have done, then since the gate is extremely high impedance and floating, then pretty much anything moving near it or touching it will cause the voltage at the gate to vary, and hence the drain-source current and drain voltage also. Touching the probe to it will certainly do this.  

This is why you should never leave a FET gate floating (unless you are building a static detector) Add a pulldown resistor to ground (e.g. 10-100k&Omega;) to keep it off until you apply a turn on voltage.

EDIT - another possibility is that the drain has a signal present at it, and you are seeing this at the gate due to capacitive coupling between them - in this case Dave's answer fits.

EDIT 2 - It seems you are trying to implement something like this:



This circuit will stay on for a while after touching. On touching the finger bridges the +V and the MOSFET gate, charging the cap and turning it on. The high value resistor then gradually discharges the cap to turn it off again after a while.
You can change the resistor/cap values for a longer/shorter on time. 

EDIT 3 - here is a simple Set/Reset circuit using 2 transistors:



To Set, apply the supply voltage to the S input (e.g. button between base and Vs. Vs could be e.g. 5V - 12V), to Reset, do the same for the R input. Output can be taken from either O point (one is inverted) and used to drive your alarm circuit. Transistors can be any general purpose NPN (BC337, 2N2222, etc)

Here are the waveforms:



Also, here is a simpler version, where you ground the inputs to switch states (R1,R2 = 1k&Omega;, R3,R4 = 10k&Omega;):