# Prove the set of which sin(nx) converges has Lebesgue measure zero (from Baby Rudin Chapter 11)

I am trying to work through Rudin. This is a question from chapter 11:

Suppose that $\{n_k\}$ is an increasing sequence of positive integers and $E$ is the set of all $x$ in $(-\pi,\pi)$ at which $\sin(n_kx)$ converges. Prove that $E$ has Lebesgue measure zero. Hint: For every subset $A$ of $E$, $\int_A \sin (n_kx) dx$ tends to zero, and $2\int_{A} \sin^2 (n_k x)dx$ tends to the measure of $A$.

So, if I can use the hint, I'm pretty sure I can get the question. The thing is, I have no idea how to prove the hint.

Using the hint, here's how I think you do the question:

Define $f$ on $E$ as \begin{align*}f(x)= \lim_{k\rightarrow \infty} \sin(n_kx) \end{align*} Notice that $\sin(n_kx)\leq 1$, and $1\in L$ on $A\subset E$ for every measurable $A$ (since $E$ has finite measure), so Theorem $11.32$ in Rudin (the Lebesgue dominated convergence theorem) says \begin{align*} \int_A f(x) dx = \int_A \lim_{k\rightarrow \infty} \sin(n_kx) = \lim_{k\rightarrow \infty} \int_A \sin(n_kx) = 0 \end{align*} (by the first hint). But this was true for all $A\subseteq E$, so by one of the questions on the last assignment, $f(x)=0$ almost everywhere on $E$ and so $f^2(x)=0$ almost everywhere on $E$. Using Theorem $11.32$ again, we get \begin{align*} \mu(A) = \lim_{k \rightarrow \infty} \int_A 2\sin^2(n_kx) = 2\int_A \lim_{k\rightarrow \infty} \sin^2(n_kx)= 2 \int_A f^2(x)=0 \end{align*} Therefore $\mu(E)=0$.

Does anybody know how to prove the hint? Thanks!

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The first part of the hint follows from the Riemann-Lebesgue lemma (which I will state). We'll sketch the proof here as it is essentially the answer you are looking for. You should be able to coerce this into seeing why the second statement is true (the difference is that instead of having the cancellation of $\sin(nx)$, $\sin^2(nx)$ fills up half roughly half the space of a rectangle).

The Riemann-Lebesgue lemma: If $f \in L^1(\mathbb R)$, then

$$\int_{\mathbb R} f(x)\sin(nx)\,dx \to 0$$

as $n\to\infty$.

To prove the lemma, recall that step functions with compact support are dense in $L^1(\mathbb R)$. Thus using dominated convergence it is enough to prove the result for such functions. However for a step function, the result follows from the straightforward calculation

$$\int_a^b \sin(nx)\,dx = -\frac 1n(\cos(bn)-\cos(an))$$

The analogous statement for the second part is

$$\int_{\mathbb R} f(x)\sin^2(nx)\,dx \to \frac 12 \int_{\mathbb R} f(x)\,dx$$

Applying these results to $\chi_A$, the characteristic function of $A$, we obtain the statements of the hint. Note that replacing $n$ by $n_k$ does not change the argument at all, we simply need $n_k$ to go off to infinity.

A few ideas to take away from this:

1. Instead of attempting to prove results directly, establish them first for "nice" objects and then hope "nice" objects are dense
2. Instead of integrating $f$ over $A$, integrate $f\,\chi_A$ over the entire domain

P.S. I'd leave this as a comment, but cannot due to low reputation: there is a typo in the hint as you've typed it.

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Thank you so much! – Samuel Reid Dec 1 '12 at 2:59