why total energy of a finite duration continuous signal becomes infinite after sampling
If you multiply a continuous-time finite energy signal $f(t)$ with an impulse train you get
$$\tilde{f}(t)=\sum_{n=-\infty}^{\infty}f(nT)\delta(t-nT)\tag{1}$$
where $T$ is the sampling interval and $\delta(t)$ is the Dirac delta impulse. Note that the &quot;signal&quot; $\tilde{f}(t)$ is a mathematical fiction, it cannot exist in practice, and it cannot even be evaluated for any value of $t$. Of course it is pretty safe to say that $\tilde{f}(t)$ is zero for $t\neq nT$, but you can't assign any value at $t=nT$. This comes from the fact that the Dirac delta impulse is no ordinary function but a generalized function or distribution, which only makes sense under an integral. So we can do calculations with $(1)$, we can take its Fourier transform, etc., but we cannot evaluate $(1)$ directly.
As for the energy of $\tilde{f}(t)$, since the square of $(1)$ is undefined (because the square of a distribution is undefined), we cannot compute the energy in the time domain. As you know, the spectrum of $(1)$ is the sum of shifted spectra of $f(t)$. Consequently, if you compute the energy in the frequency domain you get a diverging integral. So you could state that the energy of $(1)$ is definitely not finite. But the question remains if that has any meaning, given that $(1)$ is not even a function or signal in the ordinary sense.
Of course, the actual discrete-time signal $f[n]=f(nT)$, which is a sequence of finite numbers (the samples of $f(t)$), does have finite energy. If $f(t)$ is band-limited with maximum frequency $f_{max}&lt;\frac{1}{2T}$ the energy of the sampled signal is equal to the energy of the continuous-time signal $f(t)$ (up to a constant):
$$E_f=\int_{-\infty}^{\infty}|f(t)|^2dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}|F(\omega)|^2d\omega\tag{2}$$
$$\begin{align}\sum_{n=-\infty}^{\infty}|f(nT)|^2&amp;=\frac{T}{2\pi}\int_{-\pi/T}^{\pi/T}\left|\frac{1}{T}\sum_{k=-\infty}^{\infty}F\left(\omega-\frac{2\pi k}{T}\right)\right|^2d\omega\\&amp;=\frac{1}{2\pi T}\int_{-\pi/T}^{\pi/T}|F(\omega)|^2d\omega=\frac{E_f}{T},\qquad\textrm{if } F(\omega)=0\textrm{ for }|\omega|&gt;\pi/T\end{align}$$
A different way of showing this can be found in this answer.