How one can prove $$|\{x\in[0,2\pi] : \lim_{n\rightarrow \infty} e^{inx} \ \ \ \text{exists}\}| = 0$$ where $|\cdot|$ denotes the Lebesgue measure. Any hints please?
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Possible duplicate of ${m \alpha, m \in \mathbb Z}$is dense in $[0,1]$ for $\alpha$ irrational – Oct 30 '15 at 08:03
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Suppose $\{e^{inx}\}$ converges. Then what is $$ \lim_{n\to\infty}\bigl(e^{i(n+1)x}-e^{inx}\bigr)? $$ What does it tell you about $x$?
Julián Aguirre
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1Thank you! So that limit would be $0$ and this implies that $e^{ix}=1$ therefore $x\in {2\pi n : n \in Z}$ and the latter set has measure $0$. Correct? – user285292 Oct 30 '15 at 23:39
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