LED with built-in resistor and voltage drop
Your battery, LED and voltmeter are all in parallel.



simulate this circuit &ndash; Schematic created using CircuitLab
Figure 1. You're not measuring the voltage &quot;after&quot; (or &quot;before&quot;) the LED. You are measuring the voltage across the LED-resistor combination.
In this circuit it is the same as measuring the supply voltage.
To measure the LED voltage you would need to do this:



simulate this circuit
Figure 2. To measure the LED voltage in your 12 V LED would require some delicate surgery.

Now I measured the voltage and get after the LED still a voltage of 11.3V, when the LED is supplied with 12.5V from a power supply.

That means that your 12 V supply is drooping. To confirm this connect your voltmeter to the battery and monitor while you switch the LED on and off.



simulate this circuit
Figure 3. Measuring the voltage in series with the LED gives no useful information.
In this situation the voltmeter presents a series resistance of (typically) 10 MΩ in series with the 1.2 kΩ of the LED. Since the meter's resistance is 10,000 times that of the LED resistor almost all the voltage is dropped across it rather than the LED.
If you wish to do something useful then switch your meter to mA and use Figure 3 to measure the current through the LED. Remember to switch back to V when finished. If you forget and connect it up as shown in Figure 1 you will pass a high current through the meter and blow the fuse if it has one and blow the meter if it hasn't.