If it's a DC signal, then, yes, there should be no change in the measured signal. If, however, you have other signals, such as, for example, a 1 kHz signal, this would be significantly attenuated. If you want a bit more of the math here, try to think of your signal in the Fourier domain. The Fourier transform of a signal breaks a signal into a number of sinusoids at many different frequencies that, when added together, recreate your signal. The zero-th term corresponds to your DC signal. Higher terms represent AC components of the signal, starting at the fundamental frequency, a sinusoidal signal with period equal to your measurement duration. These higher terms will also be present in the output signal, until the frequency of the term is higher than the cutoff frequency of your low-pass filter. Higher frequencies are attenuated.
As for "mean value", you need to be careful of your word choice. The mean value of a signal is its average value. More commonly encountered is the root mean square (RMS) value. So, for a signal V(t):
$$V_{RMS}= \sqrt{{1\over{T}}\int_0^TV(t)^2dt}$$
This is a bit different from your zero-th term DC signal.
As Andy mentions in a comment, you need a signal source that you can tune to make a frequency that you measure the filter with.
Luckily you say it is an Audio filter, so you can use an audio source.
Find a program that you can use in the freeware (or sample-ware) world to make signals around the cut-off frequency.
Use your scope to measure the signal going into the filter and going out. The scope can also probably even do some maths (since it's a digital one) and show "Input" - "Output" for you to easily then calculate the factor.
Since you designed the filter you should probably know about the log() scale of Decibels, but since I'm in a decent-ish mood today I'll continue to explain that the "cut-off" frequency is the 3dB point for many filter designs.
The 3dB point is where the power is reduced by a factor of two, but that means for the voltage that it should be a factor of sqrt(2) =~ 1.41 lower. Half the voltage output would be 6dB.
This website has a nice table to use if you don't feel like maths.
Many digital scopes can also (or have a purchasable upgrade for it) calculate the dB change in voltage. All you'd need to do is tweak the sine-wave output by the sound card until it says "-3dB". These digital toys have really taken the art out of it :-).
In fact, if it has "infinite persistence" you can make your soundcard sweep the frequency over 10seconds and then set the scope to a timebase that gets 10 seconds for a full screen and you can get a nice 2-line frequency plot and just use a ruler to instantly find a 3dB, 6dB or 20dB point.
Best Answer
Input voltage is DC and this is dead easy to measure with a DC volt meter. Current into the power supply will be dc plus an amount of AC ripple. The AC part is irrelevant and any old meter reading dc amps will measure average.
The result is: -
It is incorrect to measure RMS current because this will give the wrong answer when the input voltage is DC.
Same with the output - if the output voltage is largely DC i.e. it has very little ripple voltage then, output power is: -
Applying a filter makes no sense if you are trying to measure power when one of the consituents of power (namely voltage) is constant dc.
As for using a power meter with a filter, you'll probably find that it won't make much difference which setting you use. Try it and prove it.