In the setting of Riemann integration, we have the following change of variables formula:
Let $[a,b]$ be a closed interval, and let $\phi:[a,b]\to[\phi(a),\phi(b)]$ be a continuous monotone increasing function. Let $f:[\phi(a),\phi(b)]\to{\Bbb R}$ be a Riemann integrable function on $[\phi(a),\phi(b)]$. The $f\circ\phi:[a,b]\to{\Bbb R}$ is Riemann-Stieltjes integrable with respect to $\phi$ on $[a,b]$ and
$$
\int_{[a,b]}f\circ\phi\ d\phi=\int_{[\phi(a),\phi(b)]}f. \tag{1}
$$
In the setting of Lebesgue integration, we have the following:
Let $(X,{\mathcal B},\mu)$ be a measure space, and let $\phi:X\to Y$ be a measurable morphism from $(X,{\mathcal B})$ to another measurable space $(Y,{\mathcal C})$. Define the pushforward $\phi_*\mu:{\mathcal C}\to[0,+\infty]$ of $\mu$ by $\phi$ by the formula
$\phi_*\mu(E):=\mu(\phi^{-1}(E))$. If $f:Y\to[0,+\infty]$ is measurable, then
$$
\int_Y f\ d\phi_*\mu=\int_X(f\circ\phi)\ d\mu. \tag{2}
$$
My question is: how can I interpret (1) in terms of (2)?
Best Answer
The key is to understand what $d\phi$ is; formally, it is the Lebesgue-Stieltjes measure associated to the increasing, right-continuous function $\phi$ (if you're concerned that $\phi$ is not defined on all of $\mathbb{R}$ as in the usual treatment, you can extend $\phi$ in the canonical way). In particular, $$d\phi( (a,b)) = \phi(b) - \phi(a) $$
Looking at the Lebesgue change-of-variables theorem, if we take $\mu$ to be the above measure, then we obtain $$ \int_{[a,b]} f \circ \phi \, d\phi = \int_{[ \phi(a), \phi(b)]} f d( \phi_{\ast}d\phi) $$ Well, what is the measure $\phi_{\ast} d\phi$? If $E$ is an interval $E = (\phi(c_1), \phi(c_2))$, then by the definition of the pushforward and the fact that $\phi$ is increasing, we have $$ \phi_{\ast} d\phi(E) = d\phi((c_1, c_2)) = \phi(c_2) - \phi(c_1) $$ Thus $\phi_{\ast} d\phi$ takes intervals and returns exactly their length! Then, completing this measure to a Borel measure, etc, it must be the Lebesgue measure, and so it is $dx$ as desired.