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$a_n$ is a sequence of real numbers such that $a_n \to +\infty$. Show that the set $E = \{x \in \mathbb{R}| \lim_{n \to \infty} \sin(a_n x) \mbox{ exists}\}$ has zero (Lebesgue) measure.

The hint for the exercise says "Riemann-Lebesgue lemma and dominated convergence theorem". But for a start, I have no idea of how to show that $E$ is measurable (or of finite measure).

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  • $\begingroup$ My first idea, which may be rubbish, is that it might be possible to show that $E_m = \{x \in \mathbb{R} \mid \text{$\sin(a_n x)$ eventually stays within an interval of size $\frac{1}{m}$} \}$ is measurable. Then $E$ is just the intersection of all of those. $\endgroup$ May 15, 2016 at 9:26
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    $\begingroup$ Showing that $E$ is measurable is not hard: $E$ is the set of equality of the two functions $f(x) = \limsup_n(\sin(a_nx))$ and $g(x) = \liminf_n(\sin(a_nx))$; since the limsup of sequences of measurable functions is measurable (exercise) and the set of equality of measurable functions is measurable, $E$ is measurable. It is trickier to show the measure is zero. $\endgroup$
    – guest
    May 15, 2016 at 9:38
  • $\begingroup$ The answer I posted has some problems so I moved the partial solution to comment, let $A$ be the given set and $\sin(a_nx)\rightarrow f(x)$ on $A$, let $g$ be any integrable function on $A$, suppose $A$ has finite positive measure, then $$0=\lim_{n\to \infty}\int_A g\sin(a_nx)d\mu=\int_A gfd\mu$$ First equality is by Riemann-Lebesgue lemma and second equality is by dominated convergence theorem(dominated by $|g|$). If we set $g=1$ when $f>0$ and $g=-1$ when $f<0$, $g$ is clearly integrable on $A$ and last integral is just $\int_A |f|d\mu$, we conclude that $f=0$ a.e. on $A$. $\endgroup$
    – user175968
    May 16, 2016 at 5:09
  • $\begingroup$ $\sin(a_nx)\rightarrow 0$ on $B$, positive measure and $\mu(B-A)=0$. $x\in B\iff \forall\epsilon>0\exists m\in\mathbb{N}$ such that $n\geq m\Rightarrow |\sin(a_nx)|<\epsilon$. That means $B$ can be written as $$\displaystyle\bigcap_{j\in \mathbb{N}}\bigcup_{m\in \mathbb N}\bigcap_{n\geq m}\{x\in A: |\sin(a_nx)|<1/j\}$$ $\endgroup$
    – user175968
    May 16, 2016 at 5:11
  • $\begingroup$ Need to prove $B$ has measure 0 to reach contradiction, the prove in my old solutions were wrong... $\endgroup$
    – user175968
    May 16, 2016 at 5:12

2 Answers 2

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If $\lim_{n\to \infty}\sin (a_n x)$ exists on a set of positive measure, then by Egorov's theorem you can find a set $E$ with $0<m(E)<\infty$ such that $f_n(x)=\sin (a_n x)$ converges uniformly on $E$ to a measurable function $f(x)$.

Then by the Riemann-Lebesgue lemma we must have $\int _F \sin (a_n x) dx \to 0 $ for any measurable subset $F$ of $E$. But also, by uniform convergence we have $\int_F \sin (a_n x)dx \to \int_F f(x) dx$ for any measurable set $F\subset E$. Thus, $\int_F f=0$ for all $F\subset E$, which implies that $f\equiv 0$ on $E$.

Now, observe that $\sin^2 (a_nx)$ also converges uniformly to $0$ on $E$, hence $$0 = \lim_{n\to\infty} \int_E \sin^2(a_nx) dx= \lim_{n\to\infty} \int_E (1-\cos(2a_nx))/2 dx = m(E)/2$$ because $\int_E \cos(2a_nx)dx \to 0$ by the Riemann-Lebesgue lemma.

However, this is a contradiction, since $m(E)$ was assumed to be $>0$.

Note: In fact, instead of using Egorov's theorem and uniform convergence, you can just use the dominated convergence theorem, since $\sin(a_n x)$ is bounded.

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  • $\begingroup$ Nice idea to bring in $\sin^2(a_nx)$... $\endgroup$
    – user175968
    May 16, 2016 at 5:26
  • $\begingroup$ That's a trick also used in the proof of the Cantor-Lebesgue lemma, which says that if $\sum_{-\infty}^\infty c_ne^{inx}, c_n\in \mathbb C$ converges on a set of positive measure, then $c_n\to 0$. The proof is almost identical. $\endgroup$
    – Dimitris
    May 16, 2016 at 5:40
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For any sequence $f_n$ of measurable functions, the set $\{x: \lim_{n \to \infty} f_n(x) \ \text{exists}\}$ is measurable. For example, you can write it as

$$ \bigcap_{m \in \mathbb N} \bigcup_{N \in \mathbb N} \bigcap_{i,j>N} \{x: |f_i(x) - f_j(x)| < 1/m \}$$

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