False: constant continuous invariants characterize measure ergodicity

For $D$ from [F1], $D(0)=0$, $D(1\mathrm{/}3)=2\mathrm{/}3$ and $D(2\mathrm{/}3)=1\mathrm{/}3$. The measure $\mu$ displayed in the statement is a Borel probability by [F3] and finite additivity of the weighted sum of measures; countable additivity follows by commuting a finite sum with increasing partial sums. For every Borel $E$, $\mu (D^{-1}E)=\frac{1}{2}{\mathbf{1}}_E(0)+\frac{1}{4}{\mathbf{1}}_E(2\mathrm{/}3)+\frac{1}{4}{\mathbf{1}}_E(1\mathrm{/}3)=\mu (E)$, so it is invariant. Continuity of $D$ gives its Borel measurability.

Put $Z={\bigcup }_{n\ge 0}(D^n{)}^{-1}\{0\}$. Each inverse image is Borel since $D^n$ is continuous and $\{0\}$ is closed, so $Z$ is Borel. A point lies in $Z$ exactly when it eventually maps to zero. If it does, then so does its image, since zero is fixed; conversely if its image eventually maps to zero, the point does one step later. Thus $D^{-1}Z=Z$. The point zero is in $Z$, while $1\mathrm{/}3$ and $2\mathrm{/}3$ stay in their two-cycle and never reach zero. Therefore $\mu (Z)=1\mathrm{/}2$, which violates [F2].

Now let continuous real $f$ satisfy $f\circ D=f$ at every circle point. Iteration shows $f(k\mathrm{/}{2}^n)=f(0)$ for $0\le k<2^n$, since $D^n(k\mathrm{/}{2}^n)=0$. For any $x\in [0,1)$ put $q_n=2^{-n}\lfloor 2^{n}x\rfloor$. Then $0\le x-q_n<2^{-n}$, so $q_n\to x$ in the circle metric: $2^n\ge n+1$ by induction and [F4] gives $2^{-n}\to 0$. Continuity yields $f(x)={\lim }_{n}f(q_n)=f(0)$. Hence all the stated continuous invariants are constant while the invariant probability is nonergodic. No Lebesgue measure, countable union of countable sets, or choice principle is used; the sets and sequences are explicitly defined.