 Solution

Conditional expectation is unique up to almost sure equality. If $Y$ and $Z$ both satisfy the definition, then for every $G\in \mathcal{G}$,The events $\{Y>Z\}$ and $\{Z>Y\}$ belong to $\mathcal{G}$. Testing on them, or first on $\{Y-Z\ge 1/k\}$ and $\{Z-Y\ge 1/k\}$, shows that both have probability zero. Hence $Y=Z$ almost surely.