Let $T:X\to X$ a ergodic transformation in a measure space $(X,\mu)$ and let $A\subseteq X$ with positive measure. If $\mathcal{A}=\{\alpha\subseteq 2^X\mid\alpha\text{ is a finite partition with }X\backslash A\in\alpha$}, is it true that $$ h(T) = \sup_{\alpha\in \mathcal{A}}h(T,\alpha) ?$$ I think this should be true because since $T$ is ergodic the partitions $\bigvee_{n=0}^{N-1}T^{-n}\alpha$ become finer and uniform along $X$ when $N\to\infty$, and we also have the formula $h(T,\alpha) = h\left(T,\bigvee_{n=0}^{N-1}T^{-n}\alpha\right)$.
Equivalently: do $T$ ergodic implies $T^{-1}\mathcal{B}=\mathcal{B}$? (where $\mathcal{B}$ is the set of measurable subset of $X$ and the equality is but measure zero)
$T$ ergodic does not necessarily imply $T^{-1}\mathcal{B} = \mathcal{B}$. Take for example, the left shift on $\{0,1\}^{\mathbb{N}}$. Then $\{(x_n)_n \in \{0,1\}^\mathbb{N} : x_1 = 1\}$ is in $\mathcal{B}$ but not $T^{-1}\mathcal{B}$.