Here is the statement that I wish to prove (Bayes formula):
Suppose $\mathbb{P}$ and $\mathbb{Q}$ are equivalent probability measures on $(\Omega, ~ \mathcal{F})$, let density: $Z = \frac{d\mathbb{Q}}{d\mathbb{P}}$，and $\mathcal{G} \subset \mathcal{F}$ to be a $\sigma-$field, then: $$ \mathbb{E}^{\mathbb{Q}}\big[ X ~ \big| ~ \mathcal{G} \big] = \frac{\mathbb{E}^{\mathbb{P}}[ZX ~ | ~ \mathcal{G}]}{\mathbb{E}^{\mathbb{P}}[Z ~ | ~ \mathcal{G}]} $$
I derive a proof as follows, but I think there's obviously something wrong with it (or it may not even make any sense).
For $\forall G \in \mathcal{G}$，since $\mathcal{G} \subset \mathcal{F}$，and hence $G \in \mathcal{F}$, then for $\forall G \in \mathcal{G}$: $$ \begin{align*} \mathbb{E}^{\mathbb{Q}}\big[ X ~ \big| ~ \mathcal{G} \big] & = \frac{\mathbb{E}^{\mathbb{Q}}[ X \cdot \mathbb{I}_{G} ]}{\mathbb{Q}(G)} \\ & = \frac{\int ~ X \cdot \mathbb{I}_{G} ~ d\mathbb{Q}}{\int ~ \mathbb{I}_{G} ~ d\mathbb{Q}} \\ & = \frac{\int ~ ZX \cdot \mathbb{I}_{G} ~ d\mathbb{P}}{\int ~ Z \cdot \mathbb{I}_{G} ~ d\mathbb{P}} \qquad \text{(Since $d\mathbb{Q} = Z \cdot d\mathbb{P}$)} \\ & = \frac{\mathbb{E}^{\mathbb{P}}[ ZX \cdot \mathbb{I}_{G} ]}{\mathbb{E}^{\mathbb{P}}[Z \cdot \mathbb{I}_{G}]} \\ & = \frac{\frac{\mathbb{E}^{\mathbb{P}}[ ZX \cdot \mathbb{I}_{G} ]}{\mathbb{P}(G)}}{\frac{\mathbb{E}^{\mathbb{P}}[Z \cdot \mathbb{I}_{G}]}{\mathbb{P}(G)}} \\ & = \frac{\mathbb{E}^{\mathbb{P}}\big[ ZX ~ \big| ~ \mathcal{G} \big]}{\mathbb{E}^{\mathbb{P}}\big[ Z ~ \big| ~ \mathcal{G} \big]} \end{align*} $$
Would you please identify the mistake that I made in the above?
Work with the definition of conditional expectations by the Radon-Nikodym integral equations: A random variable $Y$ is the conditional expected value of $X$ given a $\sigma$-algebra $\mathcal G$ with respect to a probability measure $\mathbb P$ if for all $G\in\mathcal G$ we have $$\int_GXd\mathbb P=\int_GYd\mathbb P$$
Now in this setting we have for every $G\in\mathcal G$: $$\int_G Xd\mathbb Q = \int_G \mathbb E^\mathbb Q[X|\mathcal G]d\mathbb Q$$ as well as $$\int_G Xd\mathbb P=\int_G\mathbb E^\mathbb P[X|\mathcal G]d\mathbb P$$ Then by $Z=d\mathbb Q/d\mathbb P$ we get $$ \int_GZ\cdot Xd\mathbb P=\int_GXd\mathbb Q= \int_G \mathbb E^\mathbb Q[X|\mathcal G]d\mathbb Q = \int_G Z\cdot \mathbb E^\mathbb Q[X|\mathcal G]d\mathbb P$$ It follows that $\mathbb E^\mathbb P\big[Z\cdot\mathbb E^\mathbb Q[X|\mathcal G]\big|\mathcal G\big]=\mathbb E^\mathbb P[X|\mathcal G]$
Now you need to know a bit of basic theory of conditional expectations to rewrite the left-hand side and obtain the desired result.