Find the Lebsegue measure of the set $A= \left\{ 0<x \leq 1: x \sin \left(\frac{\pi}{2x}\right) \geq 0 \right\}$.
The answer given is $1 - \ln \sqrt{2}$.
My thought: I only know that Lebsegue measure of countable set is zero (The Lebesgue measure of an uncountable set is not necessarily positive).
Kindly help me.
$\sin(x)$ is nonnegative for $x \in [2n\pi,(2n+1)\pi]$ for each integer $n$. So you should solve $\frac{\pi}{2x} \in [2n\pi,(2n+1)\pi]$ for each integer $n$. You get $x \in \left [ \frac{1}{4n+2},\frac{1}{4n} \right ]$ for $n \neq 0$ and $x \in [1/2,\infty)$ for $n=0$. (Check this yourself.) So your answer is the sum of the lengths of all these intervals when they are intersected with $[0,1]$. Only the zeroth one, which is $[1/2,\infty)$ has any piece not in $[0,1]$, so that contributes $1/2$. The rest contribute
$$\sum_{n=1}^\infty \frac{1}{4n} - \frac{1}{4n+2}.$$
Deriving the value of this sum from first principles is not a trivial matter. But since you are given the final answer, you can check it by using the Maclaurin series of $f(x)=\ln(1+x)$. Note that $\ln(\sqrt{2})=\frac{1}{2} \ln(2)$; so it is enough to show that the sum above is $\frac{1-\ln(2)}{2}$.