This is a question from my measure theory book:
Let $\mathbb{U}=\{(x,y)\in \mathbb{R^2}:x^2+y^2=1\}$ denote the unit circle. Prove the existence of a finite nonzero measure $v$ on the $\sigma$-algebra $\mathcal{B}(\mathbb{U})=\mathbb{U} \cap \mathcal{B}(\mathbb{R^2})$ which is invariant under rotations of $\mathbb{U}$. Hint: Construct $v$ as an image of $\lambda$ (Lebesgue-Borel measure).
My solution:
Let $f:[0,2\pi)\to \mathbb{U}$ be the the map defined by $f(\theta)=(\cos(\theta),\sin(\theta))$, and define $v(A)=\lambda(f^{-1}(A))$ for $A\in \mathcal{B}(\mathbb{U})$. Since $f$ is continuous, it is measurable and so $v$ is well-defined. We also see that $v$ is a finite nonzero measure on $\mathcal{B}(\mathbb{U})$. Consider the rotation map $R_\alpha:\mathbb{U}\to \mathbb{U}$ defined by
$$R_\alpha((\cos(\theta),\sin(\theta))=(\cos(\theta+\alpha),\sin(\theta+\alpha))$$
and the translation map $T_\alpha:\mathbb{R}\to \mathbb{R}$ defined by
$$T_\alpha(x)=x+\alpha$$
for $\alpha \in \mathbb{R}$. Both maps are continuous, hence measurable. Let $\alpha \in \mathbb{R}$ and $A\in \mathcal{B}(\mathbb{U})$ be given. Define $\bar{\alpha}=\alpha$ mod $2\pi$ and
$$B_1=f^{-1}(A)\cap T_{\bar{\alpha}}^{-1} [0,2\pi)$$
$$B_2=f^{-1}(A)\cap T_{\bar{\alpha}}^{-1} [2\pi,4\pi)$$
Then $B_1$,$B_2$ are measurable, disjoints, their union is $f^{-1}(A)$, and
$$f^{-1}(R_{\alpha}(A))=\big(B_1+\bar{\alpha} \big) \cup \big(B_2+\bar{\alpha}-2\pi \big)$$
with $B_1+\bar{\alpha},B_2+\bar{\alpha}-2\pi$ disjoints. Hence we get
$$ v(R_\alpha(A))=\lambda(f^{-1}(R_{\alpha}(A)))=\lambda(B_1+\bar{\alpha})+\lambda(B_2+\bar{\alpha}-2\pi)=\lambda(B_1)+\lambda(B_2)=\lambda(f^{-1}(A))=v(A)$$
where the third equality is due to the translation invariance of $\lambda$. We conclude that $v$ is invariant under translation.
Am I missing something?