(Hint Needed) Real-Analysis Exam Let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite measure space, and let $\mathcal{A}_0$ be a sub-$\sigma$-algebra of $\mathcal{A}$ so that $(X,\mathcal{A}_0,\mu)$ is $\sigma$-finite. Given a nonnegative $\mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $\mathcal{A}_0$-measurable function $f_0$ on $X$, such that
$$\int_Xfg\,d\mu=\int_Xf_0g\,d\mu $$
for every nonnegative $\mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?
 A: Notice that the above problem is false if we do not require $(X,\mathcal{A}_0,\mu)$ to be $\sigma$-finite. As a counter-example, take the usual Lebesgue measure on $\mathbb{R}$, and $\mathcal{A}_0=\{∅,\mathbb{R}\}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $\infty ($contradiction $)$. Thus we will assume $(X, \mathcal{A}_0, \mu)$ is σ-finite in this proof.
We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
$$\int_Ef\ d\mu=\int_E\ f_0\ d\mu,\mbox{ for any }E\in\mathcal{A}_0\tag{1}$$
Consider the $\sigma$-finite measure space $(X,\mathcal{A}_0,\mu)$. Let $v$ be a measure defined on this space by $$v(E)=\int_Ef\ d\mu\mbox{ for my }E\in \mathcal{A}_0$$
It is clear that $v$ is a $\sigma$-finite measure and $v<<\mu$. Radon-Nikodym theorem allows us to
construct an $\mathcal{A}_0$-measurable function $f_0=\dfrac{dv}{d\mu}\ge0$.
In the case when $g\ge0$ is $\mathcal{A}_0$-measurable, we can find an increasing sequence of $\mathcal{A}_0$-simple functions $\psi_n$ converges to $g$ pointwise. Then $f\psi _n$ and $f_0\psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$\int_Xfg\ d \mu=\lim\int_Xf\psi _n\ d\mu=\lim\int_Xf_0\psi _n\ d\mu=\int_Xf_0g\ d\mu$$
Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $\mu\upharpoonright \mathcal{A}_0$
