I want to show that $\mathbb {E}(\mathbb {E}(X\mid \mathcal {F}))=\mathbb {E}X$. My thought is that
\begin{align*}\mathbb {E}(\mathbb {E}(X\mid \mathcal {F}))=\sum_{\Omega_{i}} \frac {\mathbb {E}(X;\Omega_{i})}{\mathbb {P}(\Omega_{i})}\mathbb {P}(\Omega_{i})=\mathbb{E}X, \end{align*} where $\Omega_{i}'s$ are either finite or infinitely countable partition of $\Omega$ into disjoint sets, and each one has positive probability. I wonder if it's always true that there exists such $\Omega_{i}'s$ such that $\mathcal {F}=\sigma(\Omega_{1},\Omega_{2},\ldots)$ with the desired properties? Thanks.
Not every sigma field is countably generated and not many sigma fields are generated by countable partitions.
$E(E(X|\mathcal F) I_A)=EXI_A$ for all $A \in \mathcal F$ by the definition of $E(X|\mathcal F)$. Just taking $A =\Omega$ gives $E(E(X|\mathcal F)) =EX$