Solution
= Solution
Conditional expectation is unique up to <almost sure equality>. If $Y$ and $Z$ both satisfy the definition, then for every $G\in\mathcal G$,
$$
\int_G(Y-Z)\,d\mathbb P=0.
$$
The events $\{Y>Z\}$ and $\{Z>Y\}$ belong to $\mathcal G$. Testing on them, or first on $\{Y-Z\geq1/k\}$ and $\{Z-Y\geq1/k\}$, shows that both have probability zero. Hence $Y=Z$ almost surely.