Understanding where the constant $2/N$ comes from in Fourier transformation

I'm implementing Fourier transformation in my analysis and I wanted dig a bit deeper on the reasons why the absolute value of Fourier transformation is usually multiplied by the constant $2/N$ to get the peak amplitude value of a sinewave with certain frequency.

In the book Understanding Digital Signal Processing by Lyons, the author states the following relationship between the peak amplitude $A$ of a sinewave and the output magnitude $M_r$ of the discrete Fourier transformation (DFT) for that particular sinewave is:

$$M_r=AN/2,\;\;\;\;\;\; \tag{1}$$

where the $r$ stands for real input values to DFT and $N$ is the number of input values to DFT. From this relationship, I trivially get the amplitude I want to as $A=2M_r/N$, which I see is many times done in many fft-examples in Matlab found throughout the web.

Now my big question was, why is the relationship in $(1)$ true? I started to read more from the book Fourier Analysis and Its Applications by Folland and I found the following in his book (in section about DFT):

$$\widehat{f}\left(\frac{2\pi m}{\Omega}\right)\approx \frac{\Omega}{N}\widehat{a}_m,\;\;\;m=0,1,...,N-1\;\;\;\;\;\;\tag{2}$$

where $\widehat{f}$ is the amplitude function, $\widehat{a}_m$ is the $m^{th}$ output of DFT, $N$ is again the number of inputs and $\Omega$ is the length of the time interval $[0, \Omega]:$

$$\widehat{f}\left(\frac{2\pi m}{\Omega}\right)=\int_0^\Omega e^{-2\pi i m t/\Omega}\;f(t)\;dt,$$

where $f$ is the wave function. Now when I look at $(1)$ and $(2)$, there seems to be a connection between them:

$${\color{red}{\widehat{a}_m}} \approx{\color{blue}{\frac{N}{\Omega}}}{\color{green}{\widehat{f}\left(\frac{2\pi m}{\Omega}\right)}},\;\;\;\;\;\;{\color{red}{M_r}}={\color{blue}{\frac{N}{2}}}{\color{green}{A}}.$$

These two results are almost satisfying but I wondered why it seems to be the case that $\Omega=2$?

My questions: Where does this $2$ come from? Why in Lyons's book there is $2$ instead of $\Omega$?

I thought could it be somehow related to the symmetry of the DFT output? One time unit to left and right: $[-1,1]$ so the length of the interval would be $\Omega=2$? A bit vague this last part but could I be onto something here?

UPDATE:

The definition for DFT in book Understanding Digital Signal Processing is given as:

$$X(m)=\sum_{n=0}^{N-1} x(n) e^{-2\pi i nm/N},$$

where $x(n)$ is some continuous time-domain signal. In the book Fourier Analysis and Its Applications the corresponding definition is:

$$\widehat{a}_m = \sum_{n=0}^{N-1}a_n e^{-2\pi i mn/N}\;\;\;(0\leq m

where $a_n = f\left(\frac{n\Omega}{N}\right)$.