Find the Lebesgue Measure of the following sets

Question

Find the Lebesgue Measure of the following sets :

$A=\{0<x\leq 1:x\sin (\dfrac{1}{x})\geq 0\}$

$B=\{0<x\leq 1:\sin(\dfrac{1}{x})\geq 0\}$

In order to find Lebesgue Measure of a set we have to cover the set by open balls in $\mathbb R^2$ and take the infimum of the area of the balls covered.

But how to cover the graph of the above sets by the open balls?

Any help will be great.

Answer 1

Some initial comments:

1. You don't have to cover the graph as $A,B$ are real sets not subsets of $\mathbb R^2$.
2. If we note $\mu$ the Lebesgue measure, you have $\mu(A)=\mu(B)$ as for $0 <x \le 1$, $x \sin \frac{1}{x} \ge 0 \Longleftrightarrow \sin \frac{1}{x} \ge 0$.

Now you have $$B=\{0 < x \le 1\ : \ \sin \frac{1}{x} \ge 0\} = [\frac{1}{\pi}, 1] \cup \bigcup_{k \ge 1} [\frac{1}{2k \pi +\pi},\frac{1}{2k \pi}].$$

As those intervals are disjoint, you can find the Lebesgue measure and get $$ \begin{align*} \mu(B) &= 1 - \frac{1}{\pi} + \sum_{k \ge 1} (\frac{1}{2k \pi}-\frac{1}{2k \pi +\pi}) \\ &= 1 - \frac{1}{\pi} + \frac{1}{\pi}\sum_{k \ge 1} (\frac{1}{2k}-\frac{1}{2k+1}) \\ &= 1 + \frac{1}{\pi} \sum_{n \ge 1} \frac{(-1)^n}{n} \\ &= 1 - \frac{\ln 2}{\pi}. \end{align*} $$