Discrete Fourier Transform of PMF and PDF I got this explenation:

"It is well-known that an image’s histogram is essentially
the probability mass function (pmf) of the image (only
differing by a scalar). Multiplying each component of the
pmf by a correspondingly shifted unit impulse results in the probability density function (pdf). Obviously, in the context
of discrete Fourier transform (DFT), the unit impulses can be
ignored, implying that we can treat pmf and pdf
exchangeable. Thus, the pdf can be thought as the
normalized version of a histogram."

and I am trying to fully understand it.

It is clear to me that histogram of an image are probability mass functions. How can we treat is as probability density function since it is discrete signal, not continuous?
Is the only reason that DFT of both PMF and PDF would give us same result so it doesn't really matter in this scenario?

And how can we even get PDF from PMF?
I don't get this part: "multiplying each component of the
pmf by a correspondingly shifted unit impulse results in the probability density function"
and I cannot find any good example or explenation.
A random variable (RV) doesn't need to be continuous in order to have a PDF. The PDF of a discrete RV, i.e. a RV that takes on only discrete values, is of course not a continuous function but is only different from zero at the possible discrete values of the RV.

An example would be the outcome of a fair die. It takes on the values 1 to 6 with equal probability $1/6$. It's PDF is given by

$$p(x)=\frac16\sum_{i=1}^6\delta(x-i)\tag{1}$$

where $\delta(x)$ is the Dirac delta impulse. The Dirac delta impulses are necessary in the PDF (as opposed to the PMF) because probabilities are computed from the PDF by integration. Since for any interval $[a,b]$ containing the value $0$

$$\int_{a}^b\delta(x)dx=1$$

the PDF given by (1) integrates to $1$, as desired. And the probability of the RV $X$ being inside an arbitrary interval $[a,b]$ is

$$P[a\le X\le b]=\int_a^b p(x)dx$$