Define:
Function '$f$' defined at all (-oo, oo).
$U(to)$ is a step function that equals 1 when $t \geq to$
$\sum_{n \epsilon Z}\delta(t - nT)$ is an impulse train that is non-zero for all $t = nT|{n \epsilon Z}$ for a given $T$
$I[to, t1] = U(to) - U(t1)$
Problem:
Given a sequence of numbers $f[0], f[T], f[2T],..., f[(N-1)T]$, for a given $T$
prove that the DTFT of the sequence is given by
$F[\omega] = \frac{1}{NT} \sum_{n = 0}^{N-1}f[nT]e^{-j\omega nT}$,
where $F$ is the Fourier representation/ frequency representation of $f$.
Proof:
The Fourier representation of $x(t)$ is given by,
$X(\omega) = \int_{-oo}^{oo} x(t) e^{-j\omega t} dt$
If the assumption of periodic is true, $f[NT] = f[0], f[(N+1)T] = f[T], ...$ and as the sequence is periodic finding the Fourier series coefficients suffices. Therefore, using the complex exponential form of Fourier series,
$x(t) = \sum_{n\epsilon Z}C_{n} e^{j\omega t}$,
where,
$C_n = \frac{1}{NT}\int_{0}^{nT}x(t)e^{-j\omega t}dt$,
As $x(t)$ in our case is a sequence, the sequence can be written as a product of the function with an impulse train as shown in the expression below.
$C_{n} = \frac{1}{NT} \int_{0}^{nT}f(t)[\sum_{n\epsilon Z}\delta(t - nT)]e^{-j\omega t} dt$
Interchanging the integration and summation and using the sift property of impulses,
$C_{n} = F(\omega) =\frac{1}{NT}\sum_{n = 0}^{N-1}f[nT]e^{-j\omega nT}$ and hence the proof.
In case, the assumption of periodic is not true, as the function is no more periodic we need to find the Fourier transform.
$x(t) = f(t)I[0, NT]\sum_{n \epsilon Z}\delta(t-nT)$
and the Fourier transform the above function is
$F(\omega)*sinc(\omega)*\sum_{n\epsilon Z}\delta(\frac{\omega}{T}-n)$,
where, $*$ is a convolution operator and the expression clearly doesn't match the true expression and hence DTFT assumes the sequence to be periodic.