Doubling is ergodic for Lebesgue measure

Let A be strictly invariant: $D^{-1}A=A$. Then $(D^n{)}^{-1}A=A$ for every n. On $I_{n,k}=[k\mathrm{/}{2}^n,(k+1)\mathrm{/}{2}^n)$ the map $D^n$ is $x\mapsto 2^{n}x-k$, so $A\cap I_{n,k}=(A+k)\mathrm{/}{2}^n$. Translation and dilation give $\lambda (A\cap I_{n,k})=2^{-n}\lambda (A)=\lambda (I_{n,k})\lambda (A)$.

If $\lambda (A)>0$, choose a density-one point $x\in A\cap (0,1)$. Let I_n be its unique half-open depth-n cell and put h_n=2^{-n}. This cell lies in the centered interval $(x-h_n,x+h_n)$ apart from possibly an endpoint of measure zero. Thus $\lambda (I_n\setminus A)\mathrm{/}{h}_n\le 2\lambda ((x-h_n,x+h_n)\setminus A)\mathrm{/}(2h_n)\to 0$. But step 1.1 makes the left side identically $1-\lambda (A)$, so $\lambda (A)=1$. Every strict invariant set is therefore null or conull, and the invariant-set criterion proves ergodicity. Countable choice is used in the measure, scaling and density suppliers.