Let $(\Omega,\mathcal{F},\mu)$ a complete measure space, with $\mu(\Omega)=1$. Consider $ \mathcal{G}\subseteq \mathcal{F}$ sub-$\sigma-$algebra.
Let $f:\Omega \longrightarrow\mathbb{R}_+$ a $\mathcal{F}$-medible function such that $f\in L^2(\Omega)$. Consider the function defined on $\mathcal{G}$ given by $$\nu(A)=\int_Afd\mu\hspace{0.5cm}A\in\mathcal{G}$$ $\nu$ is a finite measure with $\nu<<\mu$ . So exists only one function $g:\Omega \longrightarrow\mathbb{R}_+\hspace{0.2cm}\mu-$a.e. and $\mathcal{G}$-medible such that $$\nu(A)=\int_Agd\mu=\int_Afd\mu\hspace{0.5cm}A\in\mathcal{G}$$ If $\mathcal{G}=\sigma((A_i)^n_{i=1})$ is the $\sigma-$algebra generated by a $\mathcal{F}-$medible partition $(A_i)^n_{i=1}$ of $\Omega$
Why in this case $g$ is constant in each $A_i$? How can I calculate $g$ explicitly? Also, Why in this case it's not true that $f=g$ (or $f=g \hspace{0.2cm}\mu-$a.e.)? Any example?
$\sigma$-algebra $\mathcal G$ is generated by a partition $(A_i)_{i=1}^n$.
This implies that: $$\mathcal G=\left\{\bigcup_{i\in J}A_i\mid J\subseteq\{1,\dots,n\}\right\}$$
Consequently a function $g:\Omega\to\mathbb R$ is measurable wrt $\mathcal G$ if and only if $g$ is constant on each of the sets $A_1,\dots,A_n$.
That means that we can write: $$g=\sum_{i=1}^nr_i\mathbf1_{A_i}$$where $r_1,\dots,r_n\in\mathbb R$.
So $g$ is determined by these $r_i$ and for every $i$ we have:$$\int_{A_i}f\;d\mu=\int_{A_i}g\;d\mu=r_i\mu(A_i)$$This enables you to find the $r_i$.
Note that $f=g$ can only be true if $f$ is measurable wrt $\mathcal G$, but that does not have to be the case.