Drive a 12v indicator lamp/LED from AC mains
A moderate size capacitor will provide an AC current.
Consider 0.047 uF.
Notice this current will NOT be isolated.
For 6 milliAmps, you need 1/0.006 = 160 ohms per volt.
If your input voltage is 160 volts peak (about right for 120 Vrms), your impedance should be 160 * 160 = 256 * 10 * 10 = 25,600 ohms.
A 1uF capacitor at 1MHz has reactance (impedance) of -j 0.159 ohms. This is just 1/(2 * pi).
At 100 Hz, the reactance is 10,000X higher, or 1,590 ohms.
At 60Hz, about 2,000 ohms. (getting closer)
Thus 0.1 uF produces 10X the reactance, or 20,000 ohms.
And I'll let you chose the final capacitor value. A 0.068 or 0.047 should be good. I don't know if 0.068 uF is stocked in HIGH VOLTAGE models.
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Note the comment of Tony_Stewart about SURVIVING LIGHTENING. Thus a simple capacitor, rated 400 volts, is not adequate.

I've seen 470 ohms at large sizes (1 watt size, for arc_over prevention) used in series with the capacitor.

And a spark_gap laid out on the PCB.


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regarding the comment/question of switching at peak of voltage, there is INDUCTANCE in the wiring. 100uH (about 100 meters of wire, or some unknown transformer secondary inductance) has this current:

100uH, 100volts

Using V = L * dI/dT and re-arranging, you get the equation

dI/dT = V / L = 100 volts / 100 uH = 1 amp/uS risetime

How fast will the capacitor charge? Using Q = C * V = I * T, we have
T = C * V / I = 0.047 uF * 100v / 1amp = 4.7 uS to charge the capacitor
This linear model ignores any ringing of the 100uH and the 0.047uF.