Is integration by substitution a special case of Radon–Nikodym theorem?

I was wondering

1. if Integration by substitution is a method only for Riemann integral?
2. if Integration by substitution is a special case of Radon–Nikodym theorem, and why?

Thanks and regards!

1. There are measure-theoretic versions of integration by substitution, a couple of which can be found on the page you linked to (search the page for "Lebesgue"). Another version is an exercise in Royden's Real analysis (page 107 of the 2nd edition) which says that if $g$ is a monotone increasing, absolutely continuous function such that $g([a,b])=[c,d]$, and if $f$ is a Lebesgue integrable function on $[c,d]$, then $\displaystyle{\int_c^d f(y)dy=\int_a^bf(g(x))g'(x)dx}$. This is also an exercise in Wheeden and Zygmund's Measure and integral (page 124). One of the versions on the Wikipedia page generalizes this to subsets of $\mathbb{R}^n$ in the case where the change of variables is bi-Lipschitz.
2. I don't see how it would be. The Radon-Nikodym theorem says that if $\nu$ and $\mu$ are measures on $X$ such that $\nu$ is absolutely continuous with respect to $\mu$, then there is a $\mu$-integrable function $g$ such that $\int_X fd\nu=\int_Xfgd\mu$ for all $\nu$-integrable $f$. Both integrals are over the same set, with no change of variables. Maybe I'm not seeing what you have in mind.
However, you can at least derive the formula for linear change of variables for Lebesgue measure using the Radon-Nikodym theorem, and maybe there's more to this than I initially thought. If $T:\mathbb{R}^n\to\mathbb{R}^n$ is an invertible linear map and $m$ is Lebesgue measure, then $\int_{\mathbb{R}^n}fdm=|\det(T)|\int_{\mathbb{R}^n}f\circ Tdm$ for all integrable $f$. A proof of this (without Radon-Nikodym) is given in a 1998 article by Dierolf and Schmidt, and they mention that in the proof they could also have used the Radon-Nikodym theorem. They don't pursue this, but the idea is that $f\mapsto\int_{\mathbb{R}^n}f\circ Tdm$ corresponds to an absolutely continuous measure on $\mathbb{R}^n$, so there is a $g$ such that $\int_{\mathbb{R}^n}f\circ Tdm=\int_{\mathbb{R}^n}fgdm$. In particular, considering $f=\chi_E$ shows that $m(T^{-1}(E))=\int_Egdm$ for all measurable $E$. From this you can show that $g$ must be constant, and the constant must be the measure of the image of the unit $n$-cube under $T^{-1}$, which is $|\det(T^{-1})|=\frac{1}{|\det(T)|}$.