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I've got the following problem... We are given a sequence of integers $n_k\to+\infty$ as $k\to\infty$. Then we define $$A:=\{x\in[0,2\pi]\,\colon\,\lim_{k\to\infty}\sin(n_kx) \text{ exists}\}.$$

I have to prove now that $\operatorname{meas}(A)=0$, where the measure is the usual Lebesgue measure on the real line.

I've proved the following relation which I assume can be useful for the sequel, namely for any measurable $B\subseteq A$, setting $$f(x):=\lim_{k\to\infty}\sin(n_kx),$$ it follows $$\int_B 2(f(x))^2\mathrm dx=\lim_{k\to\infty}\int_B1-\cos(n_kx)\mathrm dx=\operatorname{meas}(B).$$ Then $A$ is measurable (I have proved it), so I want to apply the previous result: my plan is to show that $f(x)$ must be almost everywhere $0$ but I cannot see a neat argument.

Another strategy I've thought about would be to reason by contradiction and see what happen if $A$ has strictly positive measure. At any rate this way seems nice to me because the result in this case would imply that $f$ must be equal to $0$ almost everywhere on $A$ in contrast with the previous way which uses this conclusion as the starting assumption.

In any case, I've spent quite a lot of time on this without success, so I'm asking you to help me...

thanks for your patience

-Guido-

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I think it suffices to prove that if $\lim\sin(n_k x)$ exists then it is zero. You can argue this by contradiction. Then it's fairly clear that $A$ has measure zero, since it's merely a countable set of points. If anything's unclear let me know, and I'll write it up in full below! –  Edward Hughes Jun 11 '12 at 12:49
    
but what if, for example, $n_k=4k+1$ and $x=\frac\pi2$? In this case the limit exists and it is constantly $1$ so I cannot claim that if the limit exists then it is $0$. I don't get your point sorry... –  guido giuliani Jun 11 '12 at 13:11
    
Oh yes damn, sorry that doesn't work. I think I was halfway to something more like Siminore has written below, but didn't quite think it through! –  Edward Hughes Jun 11 '12 at 13:29

1 Answer 1

up vote 1 down vote accepted

Here is a sketch of the proof, as suggested by W. Rudin in his "Real and Complex Analysis".

  1. Prove that $$\lim_{n \to +\infty} \int_E \sin nx \, dx = \lim_{n \to +\infty} \int_E \cos nx \, dx =0$$ for every measurable subset $E$ of $[0,2\pi]$. Hint: Riemann-Lebesgue.
  2. Assume that the subsequence $\{\sin n_k x\}_k$ converges. Since $2 \sin^2 \alpha = 1 -\cos 2\alpha$, we deduce from the previous step that $\sin n_k x \to \pm \frac{1}{\sqrt{2}}$ almost everywhere.

Can you complete the proof?

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you mean... $\sin(n_kx)\to\pm\frac{1}{\sqrt2}$? –  guido giuliani Jun 11 '12 at 13:27
    
@guidogiuliani: Yes, it was a misprint on the french translation of the book. –  Siminore Jun 11 '12 at 13:50

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