(The freezing lemma) Let $(\Omega, \mathscr{F}$, P) be a probability space and $\mathscr{G}$ and $\mathscr{D}$ independent sub- $\sigma$-algebras of $\mathscr{F}$. Let $X$ be a $\mathscr{D}$-measurable r.v, taking values in the measurable space $(E, \mathcal{E})$ and $\psi: E \times \Omega \rightarrow \mathbb{R}$ an $\mathcal{E} \otimes \mathscr{G}$-measurable function such that $\omega \mapsto \psi(X(\omega), \omega)$ is integrable. Then $$ \mathrm{E}[\psi(X, \cdot) \mid \mathscr{D}]=\Phi(X) \qquad(4.9) $$ where $\Phi(x)=\mathrm{E}[\psi(x, \cdot)]$
Proof: Let us assume first that $\psi$ is of the form $\psi(x, \omega)=f(x) Z(\omega)$, where $Z$ is $\mathscr{G}$-measurable. In this case, $\Phi(x)=f(x) \mathrm{E}[Z]$ and (4.9) becomes $$ \mathrm{E}[f(X) Z \mid \mathscr{D}]=f(X) \mathrm{E}[Z], $$ which is immediately verified. The lemma is therefore proved for all functions $\psi$ of the type described above and, of course, for their linear combinations. One obtains the general case with the help of Theorem 1.5.
I have tried the first part of the proof as above. However, I am not sure how to continue on extending this simple case to a more general case? Is there any theorem supporting such approximation?