Csdrhrt

Consider a probability space $(Omega,mathcal{F},P)$ and a bounded random variable $X$. Let $(A_n)_{nge 1}$ be a countable partition of $Omega$ and define $mathcal{A}:=sigma({A_n:nge 1})$. Then it holds:
$$E[X|mathcal{A}]=sum_{nge 1, P(A_n)>0}E[X|A_n]cdotmathbb{1}_{A_n}$$

One can prove it by using the uniqueness of $E[X|mathcal{A}]$, i.e. I should be done, if I can show that the expression on the right is $mathcal{A}$-measurable, and that for any $Binmathcal{A}$, it holds

$$EBig[E[X|mathcal{A}]cdot 1_BBig]=EBig[Big(sum_{nge 1, P(A_n)>0}E[X|A_n]cdotmathbb{1}_{A_n}Big)cdot 1_BBig]$$

EDIT:

First property: For any $nge 1$: $E[X|A_n]$ and $1_{A_n}$ are $sigma(1_{A_n})$-measurable, therefore $mathcal{A}$-measurable. Now using the fact that products and sums of measurable functions are measurable, we have that
$$sum_{nge 1, P(A_n)>0}E[X|A_n]cdotmathbb{1}_{A_n}$$
is $mathcal{A}$-measurable.

Second property:
Now let $A_ninmathcal{A}$ with $P(A_n)=0$. Then we know that

$$E[Xcdot1_{A_n}]=sum_{omegain A_n}X(omega)cdot underbrace{P({omega})}_{=0}=0.$$

Therefore we have for any $Binmathcal{A}$, i.e. $B=cup_{min M} A_m$ with a countable subset $Msubsetmathbb{N}$

begin{align}
EBig[E[X|mathcal{A}] 1_B Big]
&=E[X 1_B]\
&=E[Xsum_{min M}1_{A_m}]\
&=E[sum_{min M}X 1_{A_m}]\
&=sum_{min M}E[X 1_{A_m}],text{ since $X$ is bounded we can use Fubini here}\
&=sum_{min M,P(A_m)>0}E[X 1_{A_m}],text{ with the statement above}\
&=sum_{min M,P(A_m)>0}EBig[E[X|A_m] 1_{A_m}Big]\
&=sum_{min M,P(A_m)>0}EBig[E[X|A_m] 1_{A_m}1_BBig],text{ since }1_Bcdot 1_{A_m}=1_{A_m}.\
&=sum_{nge 1,P(A_n)>0}EBig[E[X|A_n] 1_{A_n}1_BBig],text{ since }1_Bcdot 1_{A_n}=0text{ for }nnotin M.\
&=EBig[Big(sum_{nge 1,P(A_n)>0}E[X|A_n] 1_{A_n}Big)1_BBig],text{ again Fubini}\
end{align}

For the Fubini statement see here

Assume for simplicity that $mathsf{P}(A_i)>0$ for all $i$. Every element of $mathcal{A}$ is an at most countable union of sets in $(A_n)$. Thus, for $B=bigcup_{i}A_i$,
$$
mathsf{E}X1_B=sum_{i}mathsf{E}[X1_{A_i}]=sum_{i}mathsf{E}[mathsf{E}[Xmid A_i]1_{A_i}]=mathsf{E}left[sum_imathsf{E}[Xmid A_i]1_{A_i}1_Bright]
$$
because $(A_n)$ are disjoint and $X$ is bounded (although, it suffices to assume that $mathsf{E}|X|<infty$).