I'm having a little trouble understanding following the example in my book as to why the rotation map $R_{\alpha}$ preserves Lebesgue measure.
We have $R_{\alpha}([x])=[x+\alpha]$ and $R_{\alpha}^{-1}=R_{-\alpha}([x])=[x-\alpha].$ WLOG, assume $\alpha\in [0,1]$ and for each set $B\subseteq [0,1]$ in the Borel-sigma algebra on $\mathbb{R}$, define $\mu(B)=\lambda(B)$. This defines a measure $\mu$ on $S^1$ with $\mu(S^1)=1$. Also, for $B-\alpha:=\{x-\alpha:x\in B\}\subset\mathbb{R}$, we have $R_{\alpha}^{-1}(B)=B-\alpha$. Thus $\mu(R_{\alpha}^{-1}(B))=\lambda(B-\alpha)=\lambda(B)=\mu(B)$, since Lebesgue measure is invariant under translations.
What I'm confused about is following the equivalence class notation. The author writes $R_{\alpha}^{-1}(B)=B-\alpha$, when I think it should be $R_{\alpha}^{-1}([B])=R_{\alpha}^{-1}([\bigcup_{x\in B}x])=R_{\alpha}^{-1}(\bigcup_{x\in B}[x])=\bigcup_{x\in B}R_{\alpha}^{-1}([x])=\bigcup_{x\in B}[x-\alpha]=[B-\alpha]$. So then $\lambda(B-\alpha)$ should be $\lambda([B-\alpha])$, i.e. $\lambda$ should be well defined on the equivalence classes. This is true because if $D\in[B-\alpha]$, then $D=\{B-\alpha+m:m\in\mathbb{Z}\}$. Then $\lambda(D)=\lambda(B-\alpha+m)=\lambda(B-\alpha)$, since $\lambda$ is translation invariant.
Is this correct?
Your question is more about formal definition of Lebesgue measure on the circle than about the dynamics, right? Technically you can define measure both on the sets (as normalised Riemmanian metric on the circle) and on the equiv classes (as a Lebesgue measure on $\mathbb{R}$ of intersection with the unit interval) and they will coincide.