Introduction
We have a finite population $U$ with $N$ individuals, namely $U:=\{1,\dots,N\}$. Each individual stores a secret fixed value, so let's write $y_i$ for the value corresponding to the $i$-th individual.
Let's say we randomly knock the door of some subset $s\subset U$, so they reveal their value to us. By doing this, we get a finite sequence of key-value pairs of the form $(i,y_i)$. We may call $X$ the set of all the possible sampling results.
Notice that we drew $s$ randomly, so we know the probability $p(s)$ of selecting each sample. We don't know, however, the probability $q(x)$ for each sampling result $x\in X$, as the secret values $\theta:=(y_1,\dots,y_N)$ are unknown to us.
Example
Let $U=\{1,2\}$ and only two possible samples $s_1:=\{1\}$ and $s_2:=\{2\}$ with $p(s_i)=\frac{1}{2}$. Let say the secret value must be in $\{0,1\}$. A possible $x$ may be $x=((1,0))$, but we don't know the probability of this sampling result, because it depends on $\theta$. Notice that if $\theta=(0,*)$, then $q(x)=\frac{1}{2}$, but if $\theta=(1,*)$ then $q(x)=0$.
Recall
So we have a "sample space" $X$ and a probability measure $q_\theta$ over it, so we have all the ingredients to define the concept of sufficiency.
Let $T:X\to Y$ be an statistic. We say that $T$ is sufficient for $\theta$ if, for all measurable $A\subset X$ and all $y\in Y$, the conditional probability $q_{\theta}(A|T=y)$ is the same for all values of $\theta$.
Problem
Each statistic $T$ induces a partition in the sample space $X$ via the equivalence relation $\sim_T$, where $x\sim_{T}y$ if and only if $T(x)=T(y)$.
Our goal is to prove the following:
If the partition $\{C_y\}_{y\in Y}$ induced by $T$ is such that every point in $C_y$ is associated with the same sample $s$, then $T$ is sufficient.
The proof is omitted here.
My approach
By definition $q_{\theta}(A|T)=E_{\theta}(1_{A}|T)$, which is a constant function in each $C_y$, so \begin{equation} \int_{C_y}E_{\theta}(1_{A}|T)dq_{\theta}=E_{\theta}(1_{A}|T=y)q_{\theta}(C_y)\text{. } \end{equation} Also, by definition of conditional expectation \begin{equation} \int_{C_y}E_{\theta}(1_{A}|T)dq_{\theta}=\int_{C_y}1_Adq_{\theta}=q_{\theta}(A\cap C_{y}) \end{equation} hence \begin{equation} E_{\theta}(1_{A}|T=y)q_{\theta}(C_y)=q_{\theta}(A\cap C_{y})\text{, } \end{equation} and then I'm stuck. Any idea?