An ideal voltmeter (which does not exist in real life) has infinite resistance and no current flows when a measurement is made. In that case there would be no current flow in the series resistor, so no voltage drop, so the metter would still read full battery voltage.
A real voltmeter has finite resistance. A cheap meter may have a 1 megohm resistance, maybe even 100 kohm - but there is no good reason for a digital meter to be under 1 megohm. When connected current will flow in the meter. This will have minimal effect on the battery voltage, but the current will cause voltage drop in the series resistor and the meter will read low by this amount. As the same current flows in meter and in the series R and as V= I x R , the voltages across resistor and meter will be in proportion to their resistances.
Say Rtotal = Rmeter + Rseries.
All the voltage will drop across Rtotal (of course).
So the proportion across the resistor will be Vbattery x Rseries/Rotal.
And the proportion across the meter will be Vbattery x Rmeter/Rotal.
If Rmeter = 1,000,000 ohms and Rseries = 10,000 ohms then
Rmeter/Rtotal = Rmeter / (Rseries + Rmeter)
= 1,000,000 / (1,000 + 1,000,000) = 1,000,000/1,001,000
= 0.999000999 ~= 0.9990 of original
ie for small Rseries compared to Rmeter the drop in voltage will be proportional to the series resistance compared to the meter resistance.
So a 100 k resistor with a 1 megOhm meter will cause about a 10% drop.
A 20k resistor will cause a ~= 2% drop etc.
So, in most cases, things like contact resistance and wiring resistance will have minimal effects on voltage measurements as long as there are no other currents flowing to affect the reading.
Best Answer
Of course you can.
Of course, a certain percentage of the times you do that you will die, but that's not what you asked.
Your circuit, as shown, is guaranteed to kill the LED, since on the negative halves of the power cycle as much as 310 volts will appear across the LED and destroy it. If you are lucky, the LED will fail short, and this will protect you. Murphy's Law ("Anything which can go wrong, will.") suggests that the LED will instead fail open, so touching your point will apply 220 volts to you. Of course, the 100k resistor will limit that current, but the result will still not be good.
As KH has answered, at your stage of knowledge, you should make an ironclad rule NEVER to deal with mains voltages directly. You simply don't know enough to deal with the hazards in a safe manner. Know your limitations.