I'm reading definition of conditional expectation from a lecture note:
the the author presents a proposition:
I'm able to prove that $\sum_{i \in \mathbb{N}} \frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \textbf{1}_{\Lambda_{i}}$ satisfies CE1, but failed to prove that it satisfies CE2. Below is my attempt. Let $A \in \mathcal G$, we have
$$\begin{aligned} & \int_A \sum_{i \in \mathbb{N}} \frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \textbf{1}_{\Lambda_{i}} \, \mathrm{d} \mathbb P &&= \sum_{i \in \mathbb{N}} \int_A \frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \textbf{1}_{\Lambda_{i}} \, \mathrm{d} \mathbb P\\ = & \sum_{i \in \mathbb{N}} \int_A \frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \textbf{1}_{\Lambda_{i}} \, \mathrm{d} \mathbb P &&= \sum_{i \in \mathbb{N}} \int_{A \cap \Lambda_{i}} \frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \, \mathrm{d} \mathbb P \\ =& \sum_{i \in \mathbb{N}}\frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \int_{A \cap \Lambda_{i}} \, \mathrm{d} \mathbb P &&= \sum_{i \in \mathbb{N}}\frac{\mathbb{E}\left[X \textbf{1}_{\Lambda_{i}}\right]}{\mathbb{P}\left[\Lambda_{i}\right]} \mathbb P [A \cap \Lambda_{i}] \\ =& \sum_{i \in \mathbb{N}} \frac{\mathbb P [A \cap \Lambda_{i}]}{\mathbb{P}\left[\Lambda_{i}\right]} \int_{\Lambda_{i}} X \, \mathrm{d} \mathbb P \end{aligned}$$
Could you please help me to proceed? Thank you so much for your help!
Now you have to use the fact that $\mathcal G$ is generated by a partition: the set $A$ is of the form $\bigcup_{j\in J}\Lambda_j$ for some subset $J$ of $\mathbb N$. Then $\mathbb P[A\cap \Lambda_i]/\mathbb P[\Lambda_i]$ is either zero or one, according to the case where $i\in J$ or not. We end up with $\sum_{i\in J}\int_{\Lambda_j}X\mathrm dP$.