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Map the torus $\mathbb T^2$ to $2 \mathbb D$ by the projection $\pi: (e^{i\theta}, e^{i\phi}) \mapsto e^{i\theta} + e^{i\phi}$. The trajectory is uniformly distributed on $\mathbb T^2$, and what you have is the image of that trajectory under the projection. Thus $\mu(E) = m(\pi^{-1}(E))$ where $m$ is normalized Lebesgue measure.