False: ergodicity implies strong mixing

Fix an irrational $\alpha$ and the ergodic Lebesgue system $R_{\alpha }$ of [F1]. Set $A=[0,1\mathrm{/}2)$, of measure $1\mathrm{/}2$. By [F2], define recursively $n_j$ to be the least integer greater than $n_{j-1}$, with $n_0=0$, such that $d(\{n_j\alpha \},0)<1\mathrm{/}(j+3)$. The existence is [F2] and leastness gives unique choices. Thus $n_j\to \mathrm{\infty }$ and ${\delta }_j=d(\{n_j\alpha \},0)\to 0$.

For a translation by a circle displacement with representative $t\in [-1\mathrm{/}2,1\mathrm{/}2]$, the half-circle $A$ and its inverse translate overlap in length $1\mathrm{/}2-\mathrm{\mid }t\mathrm{\mid }$: if $0\le t\le 1\mathrm{/}2$, the part in $A$ is $[0,1\mathrm{/}2-t)$, up to endpoints, and for $-1\mathrm{/}2\le t\le 0$ it is $[-t,1\mathrm{/}2)$. Consequently $\lambda (A\cap R_{\alpha }^{-n_j}A)=1\mathrm{/}2-{\delta }_j\to 1\mathrm{/}2$. Strong mixing in [F3] would require the full sequence, hence this subsequence, to tend to $\lambda (A{)}^2=1\mathrm{/}4$. Since $1\mathrm{/}2\mathrm{\ne }1\mathrm{/}4$, mixing fails. Countable choice is inherited only from the ergodic Lebesgue system in [F1]; the return-index recursion is canonical.