Circle rotations preserve Lebesgue measure

Let $a=\{\alpha \}$. For any Borel $B\subseteq [0,1)$, $R_{\alpha }^{-1}B=((B\cap [a,1))-a)\cup ((B\cap [0,a))+(1-a))$. The two pieces lie respectively in $[0,1-a)$ and $[1-a,1)$ and are disjoint. Translation invariance and additivity give $\lambda (R_{\alpha }^{-1}B)=\lambda (B\cap [a,1))+\lambda (B\cap [0,a))=\lambda (B)$. For a=0 the second piece is empty.

In particular this proves the inverse-image identity on the pi-system of half-open intervals and the whole circle. These generate the circle Borel sigma-algebra: ordinary open intervals are countable unions of half-open subintervals, and the Borel sigma-algebras agree by the circle definition. The whole-space measure is one, so the generating criterion applies (or directly use the identity for every Borel B in step 1.1). The completion clause extends preservation to all completed sets. The countable-choice use is exactly the earlier Lebesgue and completion construction.

Fractional-part arithmetic gives $R_{-\alpha }{R}_{\alpha }x=x=R_{\alpha }{R}_{-\alpha }x$. The same argument applies to $-\alpha$, so the inverse is measurable for either sigma-algebra. Thus invertibility here includes measurability of the inverse.