Welcome to our community

Be a part of something great, join today!

Problem of the Week #250 - May 23, 2017

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,894
Here is this week's POTW:

-----
Consider a strictly increasing sequence of natural numbers $(n_k)_{k = 1}^\infty$, and suppose $X$ is the subset of $[0,2\pi]$ consisting of all $x$ such that the sequence $(\sin(n_k x))_{k = 1}^\infty$ is convergent. Prove $X$ has Lebesgue measure zero.


-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,894
Hi MHB community,

Due to illness, I'm posting a solution later than usual. No one answered this week's problem. You can read my solution below.


Let $f(x) = \lim_{k\to \infty} \sin(n_k x)$, for all $x\in X$. By the Riemann-Lebesgue lemma, $\int_X \cos(2n_k x)\, dx \to 0$ as $k \to \infty$. On the other hand, due to the identity $2\sin^2(n_k x) = 1 - \cos(2n_k x)$, the dominated convergence theorem gives $\int_X 2f(x)^2\, dx = \int_X 1\, dx$. Therefore, $f(x) = \pm \frac{\sqrt{2}}{2}$ for almost every $x\in X$. Again, by the Riemann-Lebesgue lemma and the dominated convergence theorem, $\int_X f(x)\, dx = 0$. Consequently, $m(X) = 0$.
 
Status
Not open for further replies.