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!