Expectation of composition of functions with density as R-N derivative

Expectation of composition of functions with density as R-N derivative

In prior probability courses, I've always seen and used the fact that, for
a continuous random variable X and a function $\phi$, $E[\phi(X)]=\int_{
\mathbb{R}}\phi(x) f_X(x)dx,$ but I cannot find a rigorous statement of
such a theorem listing all of the necessary assumptions on $X$ or $\phi$
in any of my probability references. I'm currently trying to reconcile my
previous probability knowledge and what I've learned in my measure theory
course, using Rudin's "Real and Complex Analysis" and would like to prove
this result. My questions are:
Is my understanding of the set up correct, and is the proof of my result
below correct?
Can the theorem below be strengthened?
Theorem: Let $(X, \mathcal{M}, \mu)$ be a positive measure space with
$\mu(X)=1$, and let $g \in L^1(\mu)$ be a real "absolutely continuous"
function with density $f:\mathbb{R} \to \mathbb{R}$. If $\phi: \mathbb{R}
\to \mathbb{R}$ is absolutely continuous on $[a,b]$ for every $a,b \in
\mathbb{R}$ and $\lim_{t \to -\infty} \phi(t)=0$ with $\phi' \in
L^1(\mathbb{R})$, then $$\int_X \phi \circ g d\mu = \int_{\mathbb{R}}\phi
f dm.$$
SET UP: Define $F: \mathbb{R} \to \mathbb{R}$ by $F(x) = \mu(g \leq x)$,
the distribution function of $g$. Keeping things general, let $F_-$ be the
left continuous function agreeing with $F$ on all points of continuity
(this function won't matter once absolute continuity is defined, but it is
necessary in general). Define $\Lambda: C_c(\mathbb{R}) \to \mathbb{C}$ by
$\Lambda(h)=\int_{\mathbb{R}} h dF_-$ (the Riemann-Stieltjes integral).
Since $F$ is bounded and monotone, $F \in BV$, so $\Lambda$ is well
defined and forms a positive linear functional. The Riesz representation
theorem implies the existance of a measure $F_*$ on $\mathbb{R}$ so that
$F_-(b)-F_-(a) = F_*([a,b))$ for all real $a < b$. (as in Rudin's exercise
7.12). We say $g$ is "absolutely continuous" if $F_* << m$, in which case
the density of $g$ is $f:= dF_*/dm.$ Note that if $g$ is absolutely
continuous, it follows that $F_-=F.$
Proof I first show that $\int_X \phi \circ f d\mu=\int_{\mathbb{R}}\mu(g >
t) \phi'(t)dt$, and the proof closely follows that of Rudin's theorem 8.16
about distribution functions.
Define $E=\{(x,t): g(x) > t\} \subset X \times \mathbb{R}$, which Rudin
argues is measurable for positive $g$, so taking the positive and negative
parts of $g$ implies that $E$ is measurable. Also, observe that $\int_X
\int_{\mathbb{R}} |\chi_E(x,t) \phi'(t)|dt d\mu(x) \leq \int_X
||\phi'||_{L^1(\mathbb{R})} d\mu < \infty.$ Therefore, Fubini's theorem
implies that $$\int_{\mathbb{R}}\mu(g>t)\phi'(t)dt
=\int_{\mathbb{R}}\int_X \chi_E(x,t)\phi'(t) d\mu(x) dt =\int_X
\int_{\mathbb{R}} \chi_E(x,t) \phi'(t)dt d\mu(x) =$$ $$\int_X
\int_{-\infty}^{g(x)} \phi'(t) dt d\mu(x) .$$
Since $\phi' \in L^1(\mathbb{R})$, the dominated convergence theorem and
absolute continuity imply that this equals $\int_X
\phi(g(x))-\phi(-\infty) d\mu(x)= \int_X \phi \circ g d\mu.$
Also,
$$\int_{\mathbb{R}}\mu(g > t) \phi'(t)dt = \int_{\mathbb{R}}(1-F(t))
\phi'(t)dt = \int_{\mathbb{R}}\int_{t}^{\infty}f(x)\phi'(t)dx dt.$$
A similar argument as before implies that Fubini's theorem applies to this
integral, and therefore $$\int_X\phi \circ g d\mu=
\int_{\mathbb{R}}\int_{-\infty}^{x}f(x) \phi'(t) dt dx = \int_{\mathbb{R}}
f(x)(\phi(x)-\phi(-\infty))dx=\int_{\mathbb{R}}\phi f dm.$$
Note that the second to last equality holds again by the dominated
convergence theorem, since $\phi$ is absolutely continuous and $\phi' \in
L^1(\mathbb{R})$.