Intuition for $\int_X g\,d\phi = \int_X gf\,d\mu$? Is there any intuition for the following result?
Theorem: Let $(X, \mathcal{A}, \mu)$ be a measure space. Let $f: X \to [0,\infty]$ be measurable. Let $\phi$ be the measure defined by $\phi(E) = \int_E f \,d\mu$ for each $E \in \mathcal{A}$. Then for each measurable $g: X \to [0,\infty]$, we have $$\int_X g\,d\phi = \int_X gf\,d\mu.$$
Reference: Theorem 1.29 of Rudin's Real and Complex Analysis
 A: It may be useful to seek intuition in the finite, discrete case. Suppose, for simplicity, that $X=\{1,2,\dots ,n\}$ and $\mu$ is the counting measure. Pick any set of positive numbers $a_1,\dots,a_n$, Let $f:X\to [0,\infty)$ be defined by $f(i)=a_i$, and for $E\subset X$ put
$$\phi(E)=\sum_{i\in E}a_i=\int_Ef\,d\mu$$
Let $g$ be any function defined on $X$, assuming non-negative values, i.e. $g:X\to [0,\infty)$ is a measurable function. Then
$$\int_X g\,d\phi=\sum_{i=1}^na_ig(i)=\sum_{i=1}^nf(i)g(i)=\int_Xgf\,d\mu$$
The general case is the "continuous" version of this.
A: As said in epsilon-emperor's comment, intiutively you can think it as $ d\phi = f d\mu $. Thus,
$$ d\phi = f d\mu \quad \implies \quad g d\phi = fg d\mu \quad \implies \quad \int_X g d\phi = \int_X fg d\mu $$
The above is just a formal calculation, a priori. But actually it can be justified rigorously using the notion of Radon-Nikodym derivative.
A non-negative function $f: X \to [0, \infty]$ is said to be a Radon-Nikodym derivative of a measure $\phi$ with respect to another measure $\mu$ if for any measurable set $E$, we have
$$ \phi(E) = \int_E f d\mu $$
Any two functions that satisfy this conditions are equal $\mu$-almost everywhere and we can denote it as
$$ \frac{d\phi}{d\mu} = f \quad \mu\text{-a.e.}
\quad \text{or} \quad d\phi = f d\mu $$
Now, if $f$ is such a function, then a theorem says that for any measurable function $g$ such that either $g \geq 0$ or $g$ is integrable, we have
$$ \quad \int_X g d\phi = \int_X fg d\mu $$
The proof uses the standard argument of simple approximation, which is a four-step procedure. First, you prove the identity for an indicator function $g = \mathbb{1}_E$. Next, you extend the identity linearly for any simple functions $g$. Then it comes the most important step: approximate $g \geq 0$ by an increasing sequence of simple functions and use monotone convergence theorem to prove the identity for any non-negative function $g$. Finally, conclude the result for any integrable function $g$ by splitting $g$ into positive and negative parts, and into real and imaginary parts in case $g$ is complex-valued.
In fact, doing the routine proof procedure described above would also give you some intuition of why the identity is true, so you should try to do it once. I leave it as an exercise.
