If $\mathcal{C}$ is a collection of measurable subsets of $(X,\mu)$, let us denote by $\mathcal{B}(\mathcal{C})$ the smallest $\sigma$-algebra containing $\mathcal{C}$. Note that one can choose $\mathcal{C}$ to be in particular a partition or any other collection of measurable subsets.
For $\alpha_\infty$ a measurable a.e. partition and $\alpha_\bullet$ a sequence of measurable a.e. partitions of $(X,\mathscr{B},\mu)$, let us write $\alpha_\bullet\uparrow \alpha_\infty$ if
- $\mathcal{B}(\alpha_1)\leq \mathcal{B}(\alpha_2)\leq \cdots \leq \mathcal{B}(\alpha_n)\leq\cdots \leq\mathscr{B}$, and
- $\mathcal{B}(\alpha_\infty)=\mathcal{B}\left(\bigcup_{n\in\mathbb{Z}_{\geq1}}\mathcal{\alpha_n}\right)$.
Note that then $\alpha_\bullet\uparrow \alpha_\infty$ iff
- For any $n\in\mathbb{Z}_{\geq1}$, $\alpha_{n+1}$ refines $\alpha_n$ (i.e. $\alpha_1\leq \alpha_2\leq \cdots \leq \alpha_n \leq \cdots$ in symbols commonly in use), and
- $\alpha_\infty$ is the coarsest measurable a.e. partition that refines any $\alpha_n$ (i.e. if $\alpha_n\leq \alpha$ for all $n$, then $\alpha_\infty\leq \alpha$).
Hint: If $\alpha$ is a measurable a.e. partition with finite entropy, there is always a sequence $\alpha_\bullet$ of finite a.e. partitions such that $\alpha_\bullet\uparrow \alpha$ (provided choosing is allowed). Then it is straightforward that $\lim_{n\to \infty} d_\mu(\alpha_n,\alpha)=0$, where $d_\mu(\zeta,\xi)= H_\mu(\zeta\,|\, \xi)+ H_\mu(\xi\,|\, \zeta)$ is the Rohlin metric, which is precisely the RHS of the inequality of wrote.
(Succinctly, the reason why entropy can be calculated using only finite a.e. partitions is that finite partitions are dense in the space of all partitions.)
For more background and references please see this answer of mine to an older question: https://math.stackexchange.com/a/4281578/169085.
