Discussion
The Lebesgue measure of the set \(A = \left\{0 \lt x \le 1 : x\sin\left(\frac{\pi}{2x}\right) \ge 0\right\}\) is
Answer: D
The measure is \(1 - \ln\sqrt{2}\).
For \(x \gt 0\) the condition is \(\sin(\pi/2x) \ge 0\). Put \(t = \pi/2x\), which runs over \([\pi/2, \infty)\) as \(x\) runs over \((0, 1]\). The sine is non-negative on \([\pi/2, \pi]\) and on each \([2k\pi, (2k+1)\pi]\), which correspond to
\[ \begin{aligned} x &\in [1/2, 1] \text{ (length } 1/2) \\ x &\in \left[\tfrac{1}{4k+2}, \tfrac{1}{4k}\right],\ k \ge 1 \end{aligned} \]
Adding the lengths:
\[ \begin{aligned} m(A) &= \tfrac{1}{2} + \sum_{k\ge 1}\left(\tfrac{1}{4k} - \tfrac{1}{4k+2}\right) \\ &= \tfrac{1}{2} + \tfrac{1}{2}\left(\tfrac{1}{2} - \tfrac{1}{3} + \tfrac{1}{4} - \cdots\right) \\ &= \tfrac{1}{2} + \tfrac{1}{2}(1 - \ln 2) = 1 - \ln\sqrt{2} \end{aligned} \]
using \(1 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2\). The value is about 0.653, so 0 and 1 are wrong, and \(\ln 2\) (about 0.693) is close to it but does not equal the sum of the lengths.
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