Figure 2.

Top: probability mass diagram for $\mathcal{X}\times Y$. Bottom: probability mass diagram for $\mathcal{X}\times Z$. Note that the events $y_1$ and $z_1$ can induce different exclusions in $P\left(X\right)$ and yet still yield the same conditional distributions $P\left(X\right|y_1)=P\left(X\right|z_1)$ and hence provide the same amount of information $i(x_1;y_1)=i(x_1;z_1)$ about the event $x_1$.