Find Lebesgue measure of set $\left\lbrace0
Problem:Find Lebesgue  measure of set $\left\lbrace0<x \leq1:x \sin\left(\frac {\pi}{2x}\right) \geq0\right\rbrace$
Solution: I know the formula of Lebesgue measure of set
But I have no idea about this question
 A: Define $E=\{0<x\leq1:x\sin(\dfrac{\pi}{2x}) \geq 0\}$.
We just need to solve the inequality $x\sin(\dfrac{\pi}{2x}) \geq 0$. Since $x>0$, this yields $\sin(\dfrac{\pi}{2x}) \geq 0$. Easy to conclude that (since $x>0$)
$\dfrac{1}{2} \leq x <1$ or $2k\pi \leq \dfrac{\pi}{2x} \leq (2k+1)\pi$ for $\forall k \in \mathbb{N^+}$. 
For the latter, we have
$\dfrac{1}{4k+2} \leq x \leq \dfrac{1}{4k}$ for $\forall k \in \mathbb{N^+}$. 
That is, $E=\{[\dfrac{1}{4k+2},\dfrac{1}{4k}]|k \in \mathbb{N^+}\}\cup [\dfrac{1}{2},1)$.
Hence $m(E)=\dfrac{1}{2}+\sum_{k=1}^\infty (\dfrac{1}{4k}-\dfrac{1}{4k+2})=\dfrac{1}{2}+\dfrac{1}{2}\sum_{k=1}^\infty (\dfrac{1}{2k}-\dfrac{1}{2k+1})=1-\dfrac{\ln2}{2}$
Remark: We already know that $\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}=\ln2$, this gives $\sum_{k=1}^\infty (\dfrac{1}{2k}-\dfrac{1}{2k+1})=\sum_{k=2}^\infty \dfrac{(-1)^{k}}{k}=1-\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}=1-\ln2$.
A: Hint: The zeroes in $(0,1]$ of the (continuous) function are $\tfrac{1}{2n}$, $n = 1,2,3, \dots$.
