Aliasing in wave-scanning oscillator
i am a little confused by the first sentence. i dunno how a wavetable oscillator scans a sound file at all. some algorithm can scan a file and extract either grains (or little "wavelets", which is a term i have misused in the past) or single-cycle wavetable "snapshots". but the wavetable oscillator starts with wavetable(s) and outputs a periodic or quasiperiodic tone.   

about aliasing, if you extract wavetables from a real tone (and let's assume you have a decent pitch detector, so you know how long each period is to a precision of fraction of sample, and decent interpolation, so you you can resample the period to a, say, 256 or 1024 point wavetable) and you up-pitch the tone by more than a couple of semitones, there is a danger of aliasing unless you remove some of the higher harmonics. you can remove these higher harmonics by performing a DFT (a.k.a. FFT) on the wavetable and zero some of the bins centered around bin N/2 . then inverse DFT back to a wavetable.

actually, the number of bins centered around bin N/2 that are already zero is dependent on the degree of upsampling in the interpolation when the period in the soundfile is converted into a wavetable.  consider this example:  suppose the sample rate of the sound file is 44100 Hz and you have a note that is at middle C, with a fundamental at 261.6255653 Hz.  this means that your period length is 168.5615 samples and a really good pitch detector can return a period value with that precision (even though i doubt it would be that accurate, precision and accuracy are different things).  now a 128-sample wavetable would be insufficient to represent that waveform, or at least possibly insufficient to represent all of the harmonics of that waveform.  there are conceptually 84 harmonics.  a 256-samplewavetable would be sufficient to represent all of the harmonics, but let's say, for playback reasons, you resample that 168.5615 sample period to a 1024-sample wavetable, $x[n]$.  that means you have a step size (the reciprocal of the sample-rate-conversion ratio) of 168.5615/1024 = 0.16461084 .

now, if you were to FFT that 1024-sample wavetable, you would possibly have non-zero bins for $X[0]$ up to $X[84]$ (and also the mirror image of $X[1024-84]$ up to $X[1023]$ .  all of the bins from $X[85]$ to $X[1024-85]$ should be zero if the interpolation kernel was any good.

now, when you play that wavetable back at the same note, middle C, then the sample stride would be the reciprocal of the step size above, or 6.074934 wavetable samples per output sample (assuming the same 44100 Hz sample rate).  and you would have no aliasing because all of those harmonics above the 84th harmonic are zero.  but if you start pitching that note higher, your stride will be longer.  if you pitched it 7 semitones higher (to G), then the stride would be 9.1021167 wavetable samples per output sample.  but then your 84th harmonic gets bumped up to around 33 kHz which is above Nyquist and you will have aliasing.

so what you need to do, if you plan to pitch that note up 7 semitones (i would not pitch a sampled piano up that far, because it will sound "plinky"), you must lose all of those harmonics above the 56th harmonic.  ($56 \approx 84 \times 2^{-7/12} $).  you do that with the FFT, zeroing bins, and then inverse FFT