Lets $\sigma(Y)=\sigma(\{A_1,\cdots A_n\})$ where $A_1,\cdots , A_n$ is a partition on $\Omega$. Show
$$E(X\mid Y)=E(X\mid A_1)1_{A_1}+\cdots E(X\mid A_n)1_{A_n}$$.
My try:
By definition $Y$ is a simple function, that is $Y=\sum c_i 1_{A_i}$. $E(X\mid Y)$ is a function of $Y$ so
\begin{eqnarray} E(X|Y)=\left\{ \begin{array}{cccc} a_1 & \omega \in A_1 \\ a_2 & \omega \in A_1 \\ . & . \\ a_n & \omega \in A_n \end{array} \right. \end{eqnarray} and it is enough to calculate $a_i$. By definition of conditional expectation
$$E(X1_B)=E(E(X\mid Y)1_B), \text{for all} B\in \sigma(Y)$$ so
$$E(X 1_{A_i})=E(E(X\mid Y)1_{A_i})=E(a_i 1_{A_i})=a_i E(1_{A_i})$$ so $a_i=\frac{E(X 1_{A_i})}{ E(1_{A_i})}= E(X\mid A_i).$
Q_1) Is this proof valid?
Q_2) Is this valid for a countable partition on $\Omega$. It means , If $\{A_n , n\geq 1\}$ be a countable partition on $\Omega$ and $\sigma(Y)=\sigma(A_1,A_2,\cdots )$ $E(X\mid Y)=\sum_{n\geq 1} E(X\mid A_n)$ ,that is, (countable case).
Q_2) Is this valid for an uncountable case. For $\{A_i , i\in I\}$ be a uncountable partition on $\Omega$, where $I$ is an uncountable index set, if $E(X\mid Y)=\sum_{i\in I} E(X\mid A_n)$ ? For example $\Omega=[0,1]$ , $\{A_x=\{ x\} ,x\in [0,1]\}$
Thanks in advance for any help you are able to provide or any clarification.
1) Yes, your proof is valid. Though personally I would avoid the expression where $\mathbb E\mathsf1_{A_i}$ serves as denominator. The equality $\mathbb{E}X\mathsf{1}_{A_{i}}=a_{i}\mathbb{E}\mathsf{1}_{A_{i}}=a_{i}P\left(A_{i}\right)$ can directly be interpreted as the statement that $a_i=\mathbb E[X\mid A_i]$. Also it is not necessary to invoke in between that $\mathbb E[X\mid Y]$ is a function of $Y$. It is immediate that a function measurable wrt $\sigma\left(\left\{ A_{1},\dots,A_{n}\right\} \right)$ where the $A_i$ form a partition has the shape $\sum_{i=1}^{n}a_{i}\mathsf{1}_{A_{i}}$.
2) Yes, it will also work for countable case where automatically a function measurable wrt $\sigma\left(\left\{ A_{1},A_{2},\dots\right\} \right)$ will have the shape $\sum_{i=1}^{\infty}a_{i}\mathsf{1}_{A_{i}}$ (because the function can only be constant on the $A_i$).
3) No in the uncountable case. If $I$ is uncountable then sums of the form $\sum_{i\in I}a_i$ are not even defined.