In measure theory we learn that
$$
\int_\Omega g \circ f d\mu = \int_{f(\Omega)}g d(\mu \circ f^{-1})
$$
where $(\Omega, \mathscr F, \mu)$ is a measure space and $f$ and $g$ are measurable.
Now in calculus we have that
$$
\int_{\phi(a)}^{\phi(b)} h(x)dx = \int_a^b h(\phi(t))\phi'(t)dt
$$
for a suitable substitution $x = \phi(t)$. From this substitution we have $dx = \phi'(t)dt$ with $\phi'(t)$ being the Jacobian, but I want to understand this in terms of the measure theoretic formulation.
Clearly I'll have $g = h$ and $f = \phi$, so this ought to be equivalent to
$$
\int_{[a,b]} h \circ \phi d\mu \stackrel ?= \int_{\phi([a,b])} h \ d(\mu \circ \phi^{-1}).
$$
Let $\lambda$ be the Lebesgue measure. It seems fair to assume that $\mu \ll \lambda$ so the Radon-Nikodym derivative $\frac{d\mu}{d\lambda}$ exists. This means that
$$
\int_{[a,b]}h \circ \phi d\mu = \int_{[a,b]} h \circ \phi \frac{d\mu}{d\lambda}d\lambda
$$
so we have $\phi' = d\mu / d\lambda$?
Now how do we reconcile this with the $\mu \circ \phi^{-1}?$
Best Answer
Symbolically we have $\phi'(t) \, dt = d\phi$ which gives $$ \int_a^b (h \circ \phi)(t) \, \phi'(t) dt = \int_{[a,b]} (h \circ \phi)(t) \, d\phi(t) = \int_{\phi([a,b])} h(x) \, d\phi(\phi^{-1}(x)) = \int_{\phi(a)}^{\phi(b)} h(x) \, dx $$
The measure $\mu$ in your post is defined by $\mu(E) = \int_E \phi'(t) \, dt = \int_E \phi'(t) \, d\lambda(t)$ so $$\frac{d\mu}{d\lambda}(t) = \phi'(t).$$