Q: It is said that "to maintain the same quality in the two cases, we require that the power spectral densities remain the same". Why is this a measure of the same quality? Why is not the integral of each power spectral density over the corresponding Nyquist interval not used, instead?
Why is not the integral of power spectral densities over the Nyquist interval not used, instead?
Depends on what you mean by Nyquist interval. Since you have two sample rates you have two Nyquist intervals. Assuming you mean the Nyquist interval in the original sample rate, than it's the same thing. The noise power in the band of interest is in either case the integral of the noise spectral density over the original Nyquist interval. Since the interval is the same, the integral is the same of the densities are the same.
The oversampled version has more noise but that's outside the desired bandwidth and can be filtered out.
Let $x(t)$ be your continuous-time bandlimited signal, modeled as a WSS random process with a PSD of $S_{xx}(\Omega)$ in the band $\Omega \in[-W,W]$, $\Omega$ in radians per second.
Sampling $x(t)$ at the critical rate $T_s = \frac{\pi}{W}$, yields the unquantized, discrete-time sequence $x[n] = x(nT_s)$. Let the associted PSD with $x[n]$ is
$$\Phi_{xx}(e^{j \omega}) = \frac{1}{T_s} S_{xx}(\frac{\omega}{T_s}) ~~~,~~~\omega \in [-\pi, \pi] \tag{1}$$
Assume some $N$ bits of quantization is used with a step-size $\Delta$. The process of quantization converts the unquantized sequence $x[n]$ into quantized (digital) $\hat{x}[n]$ with the quantization error denoted as $e[n]$ such that: $$\hat{x}[n] = x[n] + e[n] \tag{2}$$
For simplification reasons, the quantization error $e[n]$ is assumed to be an i.i.d., white-noise of independent (from $x[n]$ and itself at nonzero lags) WSS random process uniformly distributed in $[-\frac{\Delta}{2} ~,~ \frac{\Delta}{2}]$. Which, therefore has the PSD given by :
$$ S_{ee}(e^{j\omega}) = \frac{\Delta ^2}{12} ~~~,~~~\pi < \omega < \pi \tag{3}$$
Notice, how I modeled the quantization noise process, directly in discrete-time, with a power spectral density independent of sampling period $T_s$. This means that whatever sampling rate is used to obtain the quantized samples $\hat{x}[n]$, the assoiced noise power at the discrete-time will be the same as $\sigma_e^2 = \Delta^2/12$ in the range $-\pi < \omega < \pi$. It only depends on the number of bits; hence the step-size $\Delta$.
However, the same is not true for the PSD of the signal $x[n]$. As the sampling rate changes, its PSD will scale both in amplitude and in the frequency band as given by the relation. Assume we employ $M$ times oversampling so that $T' = T_s/M$, then:
$$ x[n] = x(n T_s) \longleftrightarrow \Phi_{xx}(e^{j\omega}) \implies \\ x_o[n]=x(n \frac{T_s}{M}) \longleftrightarrow \begin{cases}{ M \cdot \Phi_{xx}(e^{j\omega/M}) ~~~,~~~|\omega| < \frac{\pi}{M} \\ 0 ~~~,~~~ \frac{\pi}{M} < |\omega| < \pi }\end{cases} \tag{4}$$
Now, if you lowpass filter the quantized oversampled sequence $x_o[n]$ and decimate the result back into its critical rate, both PSDs of $x_o[n]$ and $e[n]$ will be reduced by $M$, however, this will only revert PSD of $x[n]$ into its original, while reducing the equivalent noise power of quantization into $1/M$ of its original value, thus enabling one to use more bits to represent the original signal samples $x(nT_s)$.
