Okay, but in that case you would have to include In your definition both clipped and unclipped ISOs, which to me would be confusing.
Let's take for example a sampled wave for which the peaks do not coincide with samples. If we were to perfectly reconstruct that wave, it would have peaks that go beyond the samples.
Now, let's normalize those samples to 0dBFS. The peaks of that wave, if perfectly reconstructed, would exceed the 0dBFS threshold.
Let's say the peaks go over the sample points by 3dB.
Let's pretend to run that signal into an actual oversampling DAC with no headroom. Now, instead of a peak between samples we'd have a train of 0dBFS samples.
Let's now skip to the output of the DAC after the analog low-pass filter. The reconstructed wave would now have peaks that exceed the 0dBFS threshold due to the Gibbs phenomenon, but since the wave was digitally clipped they obviously wouldn't be at +3dBFS.
Let's say those peaks are now at +1dBFS (I'm making up a number).
By your definition, you would have to call those peaks "+1dBFS ISOs or ISPs", which is fine, but it wouldn't really give us an idea of how much information was lost during the conversion due to clipping.
If I were to say "this DAC can't accurately reproduce +3dBFS ISOs above 0dBFS" it wouldn't make any sense to you, because you would say that the ISOs only measure 1dB above 0dBFS, even though the original wave had 3dB worth of information that was lost.
Furthermore, if we were to change the values of the analog low-pass filter in the DAC, that +1dBFS value could become +0.5dBFS, or +0.2dBFS (again, I'm making up numbers.
@JIW will correct me with the math). The values would also change if we increased the headroom of the interpolator. If we instead stayed in the digital realm, every clipped ISO would measure 0dB above 0dBFS.
This to me would be confusing. I'd rather say that our wave has +3dBFS ISOs or ISPs. That stays true regardless of what the DAC is doing to the signal.
Does this make any sense?