Does the Discrete Fourier Transform assume a periodic signal, or one that dies off?

Question

I keep hearing that the DFT assumes a periodic signal. E.g. the first answer in this MATLAB Q&A site. This doesn't make any sense to me. According to the derivations I've seen of the DFT one approximates the continuous FT by a Riemann sum as in $$H(f_n)=\int_{-\infty}^{\infty} h(t) e^{2\pi i f_n t} dt \approx \Delta \sum_{k=0}^{N-1} h_k e^{2\pi i k n }{N}$$

Which to me pretty much implies that the signal dies off and isn't repeating itself indefinitely.

So my question is: does the DFT assume signals that die off? Does it assume a continuous signal (if yes, why?) and reflect this assumption in $H(f_n)$?

If it dies off then why would the whole discussion on aliasing occur? Namely, it wouldn't make much sense for that discussion to arise since what determines the precision of the DFT would be doing a good Riemann sum, not the sampling rate and its relation to the bandwith of the signal.

Thanks!

Answer 1

It shouldn't really matter. The DFT approximates the function by a finite number of points, and you can't tell from a finite number of points whether a function is periodic or decays.