Electronic – Converting a high frequency ac signal into square wave without creating a low pass filter by accident

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.

Frankly, I suspect you have drive problems. I'd suggest monitoring the voltage across R2 as a measure of the current through the LED. If it's not perfectly clean (and I suspect it's not) that's your problem. I'd guess that your low phase has a major spike about a millisecond before nominal transition to low, and the high phase of the drive signal shows a major drop about 5 msec in, down to about a volt or less with a rather low rate of recovery.

If the drive signal is clean, try a new optocoupler. Every once in a while devices do very strange things.