Given a probability space $(X , \mathcal{M} , m)$ and $\mathcal{A}$ is a $\sigma$ sub algebra of $\mathcal{M}$. Let $\mathbb{E}$ be the condition expectation given $\mathcal{A}$.
Given $f$ is an real valued integrable function on $X$ .
Show that $\mathbb{E}(gf)=g\mathbb{E}(f)$ if $g$ is bounded and $\mathcal{A}$-measurable
The approach that I read:
The statement is true for $g=\chi_{B}$ for any $B \in \mathcal{A}$.
Since $\int_{A} \chi_{B}\mathbb{E}(f) dm=\int_{A\cap B} \mathbb{E}(f) dm = \int_{A\cap B} f dm=\int_{A} \chi_{B}f dm$ for all $A\in\mathcal{A}$
Then, it is true for simple function$\sum c_i\chi_{A_{i}}$
We can approximate $g$ by simple function.
Hence, the result follow.
My question is "Why the limit can move in the integral ?"
i.e. Why $\int_{A} g_n\mathbb{E}(f) dm=\int_{A} g_nf dm$ imply $\int_{A} g\mathbb{E}(f) dm=\int_{A} gf dm$ for $g_n\rightarrow g$ poinwisely and $g_n$ is simple
$g$ bounded, so $g_n$ can be chosen bounded. $f$ integrable, so dominated convergence applies to get $\int g_n f dm \to \int gf dm$.