Understanding the Probability Distribution of Intensity Values
Let $R$ be the random variable "intensity" with probability density function $f_R(r)$ ($=p_r$). That implies that $R$ also has a cumulative density function $F_R(r) = P(R \le r) = \int_{-\infty}^r f_R(u)\,\mathrm du$.

Let $S = T(R)$ be another random variable, the transformed version of $R$, with the transformation mapping $s=T(R)$ smooth and monotonous¹. We're looking for the pdf of $S$, $f_S(s)$ ($=p_s$).

We take the convenient detour through $f_S(s) = \frac{\mathrm d}{\mathrm ds}F_S(s)$:

\begin{align}
\frac{\mathrm d}{\mathrm ds} F_S(s) &amp;= \frac{\mathrm d}{\mathrm ds}P(S\le s)\\
&amp;= \frac{\mathrm d}{\mathrm ds}P(S\le T(r))\\
&amp;= \frac{\mathrm d}{\mathrm ds}P(T^{-1}(S)\le r)\\
&amp;= \frac{\mathrm d}{\mathrm ds}F_R(T^{-1}(S)) &amp; \text{chain rule of deriv.}\\
&amp;= \frac{\mathrm d}{\mathrm d \left(T^{-1}(s)\right) }F_R(T^{-1}(S))\cdot \underbrace{\frac{\mathrm d}{\mathrm dS}T^{-1}(S)}_\text{deriv. of inv. function}\\
&amp;=f_R(T^{-1}(S))\cdot \overbrace{\frac1{\frac{\mathrm d}{\mathrm dr}T(r)}}\\
&amp;=f_R(R)\cdot \frac1{\frac{\mathrm d}{\mathrm dr}s}\\
&amp;=f_R(R)\frac{\mathrm dr}{\mathrm ds} &amp; \text{in original notation:}\\
&amp;=p_r(r)\frac{\mathrm dr}{\mathrm ds}
\end{align}

I feel like I've forgotten the formal step that introduces the absolute value, but in case of $T(r)$, I required local invertability, so that kind of goes with that. If someone has a good notation for that, please do not hesitate to edit my answer. It's kinda late!


¹ that's an addition to the original problem, but for this way of derivation I need $T$ to be invertible. We can go a lot more measure-theoretic at this, but I don't really think that helps, so I'm going with this restriction, which I don't think is too harsh, as other mappings would lead to questionable results, just from a picture perspective.