In my course we have proven the following result
Let $g$ is a non-negative measurable function on $G$ with measure $\nu$. Suppose $f: E \rightarrow G$ is a measurable function and that $\mu$ is a measure on $(E, \mathcal{E})$ such that $$ \nu=\mu \circ f^{-1} . $$ Then $$ \int_G g \mathrm{~d} \nu=\int_E g \circ f \mathrm{~d} \mu . $$
Separetly then, we proved (via a different argument)
Let $\phi:[a, b] \rightarrow \mathbb{R}$ be continuously differentiable and increasing. Then for any bounded Borel function $g$, we have $$ \int_{\phi(a)}^{\phi(b)} g(y) \mathrm{d} y=\int_a^b g(\phi(x)) \phi^{\prime}(x) \mathrm{d} x . $$
For $g$ non negative measurable I should be able to derive the second result from the first since it is more general. However, I am confused with what is going on. Here is how I am thinking about this formulation:
First let us convert the second equation into a measure theoretic formulation: \begin{align*} \int_{\mathbb{R} }^{} 1_{(\phi (a), \phi(b))} g \, \mathrm{d} \lambda = \int_{\mathbb{R} }^{} g \circ \phi * 1_{(a,b)} * \phi' \, \mathrm{d} \lambda \end{align*} where $\lambda $ is the Lebesgue measure on $\mathbb{R} $. My issue now is that I do not know how to apply the previous part. I need to seek $\mu $ such that \begin{align*} \lambda (A) = \mu (\phi^{-1} (A)). \end{align*} But how do I find it?
Question: Is it that we can't just derive the second from the first because it is hard to find such a $\mu $ moreover even if we find it it might not be the lebesgue measure as we desired?
Additional: If so, when do we use the first? I am aware that one can derive formulas for lebesgue to lebesgue measure, however, the above attempted proof has managed to confuse me.
Could someone help me understand what is going on?
