Doubling ergodicity depends on the invariant measure

The circle map $D(x)=\{2x\}$ fixes zero and exchanges $1\mathrm{/}3$ with $2\mathrm{/}3$. Thus ${\delta }_0(D^{-1}E)={\mathbf{1}}_E(D0)={\mathbf{1}}_E(0)={\delta }_0(E)$ for every Borel $E$. Every Borel set has ${\delta }_0$-measure either zero or one, so in particular every strictly invariant Borel set does; [F2] proves ergodicity for ${\delta }_0$. Ergodicity for $\lambda$ is [F1].

The finite weighted sum $\mu$ is a Borel probability by [F3] and the finite-sum interchange with nonnegative series. Its inverse-image mass is $\frac{1}{2}{\mathbf{1}}_E(0)+\frac{1}{4}{\mathbf{1}}_E(2\mathrm{/}3)+\frac{1}{4}{\mathbf{1}}_E(1\mathrm{/}3)=\mu (E)$, so it too is invariant. Let $Z={\bigcup }_{n\ge 0}(D^n{)}^{-1}\{0\}$. Continuity of the iterates makes this Borel. Since zero is fixed, $x$ eventually reaches zero exactly when $Dx$ does; hence $D^{-1}Z=Z$. Zero belongs to $Z$ and neither point of the two-cycle does, so $\mu (Z)=1\mathrm{/}2$. This contradicts the ergodicity criterion in [F2]. The atomic calculations are choice-free; countable choice is used only to include the Lebesgue system of [F1].