I want to show that $d(\xi, \eta) = H_{\mu}(\xi|\eta) + H_{\mu}(\eta|\xi)$ defines a metric on the space of all partitions (considered up to sets of measure zero) of a probability space $(X, \mathscr{B}, \mu)$ with finite entropy, where $H_{\mu}$ is the measure-theoretic entropy
The harder part is to show the triangular inequality. I think it is equivalent to $H_{\mu}(\xi \vee \zeta)\le H_{\mu}(\xi | \eta)+H_{\mu}(\eta \vee \zeta)$, but I don't know how to continue.
You want to show that $H_{\mu}(\xi\vee\zeta)\leq H_{\mu}(\xi|\eta)+H_{\mu}(\eta\vee \zeta)$.
First, we will write $$H_{\mu}(\xi\vee\zeta)=H_{\mu}(\zeta)+H_{\mu}(\xi|\zeta)$$ and $$H_{\mu}(\eta\vee\zeta)=H_{\mu}(\zeta)+H_{\mu}(\eta|\zeta).$$ So we are left to show that $$H_{\mu}(\xi|\zeta)\leq H_{\mu}(\xi|\eta)+H_{\mu}(\eta|\zeta).$$ Well, this is quite immediate, \begin{align*} H_{\mu}(\xi|\eta)+H_{\mu}(\eta|\zeta)&\geq H_{\mu}(\eta|\zeta)+H_{\mu}(\xi|\zeta\vee\eta)\\ &=H_{\mu}(\eta\vee\xi|\zeta)\\ &\geq H_{\mu}(\xi|\zeta). \end{align*}