My book gives the following definition of Lebesgue measure on $S^1$:
For each $B\subseteq [0,1]$ in the Borel sigma-algebra on $\mathbb{R}$, define $\mu(B)=\lambda(B)$. This defines a finite measure $\mu$ on $S^1$ with $\mu(S^1)=1$.
I'm unsure if more needs to be done. These are my thoughts:
Show $\mu$ is well-defined on $[0,1]/\sim\cong S^1$: If $B\in (0,1), [0,1),$ or $(0,1]$, then $B=[B]$ in $S^1$, so $\mu([B])=\mu(B)$. If $B$, as a subset of $\mathbb{R}$, contains both $0$ and $1$, then $[B]=B\setminus\{1\}$, so $\mu([B])=\mu(B\setminus\{1\})=\lambda(B\setminus\{1\})=\lambda(B)=\mu(B)$ since singletons are zero sets. So in every case, $\mu([B])=\mu(B)$.
Now for a homeomorphism $\phi:[0,1]/\sim\to S^1$, we can define a measure $\upsilon$ on $(S^1,\scr B$$(S^1))$, where $\scr B$$(S^1)$ is borel measure on $S^1$, by $\upsilon(V)=\mu(\phi^{-1}(V))$ for all $V\in\scr B$$(S^1)$.
Are my thoughts correct?