A set defined by $$\mathcal{D}_{q_2}=\left\{a\in[0,1]:\frac{q_1-A^{-1}}{q_2}<a<\frac{q_1+A^{-1}}{q_2},0\leqslant q_1\leqslant q_2\right\}.$$ I don't known whether its Lebesgue measure is $2A^{-1}$ or $\dfrac{2A^{-1}}{q_2}$. Assume $[q_1-A^{-1},q_1+A^{-1}]\subseteq [0,1]$ and $q_1$ is fixed. Please show me the reason.
2026-04-03 22:40:13.1775256013
Lebesgue measure of $\mathcal{D}_{q_2}$
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