Exercise 11.16

Question

Suppose $\{n_k\}$ is an increasing sequence of positive integers and $E$ is
    the set of all $x \in (-\pi,\pi)$ at which $\{\sin n_k x\}$ converges. Prove
    that $m(E) = 0$.

Amit Awasthi · Accepted Answer

\begin{proof}
      For any $A \subset E$, we have
      \begin{equation*}
        \int_A \sin n_kx\,dx 	o 0
      \end{equation*}
      as $k 	o \infty$
      since this gives Fourier coefficients for $\chi_A$, and then by Theorem
      $8.12$; similarly
      \begin{equation*}
        2\int_A (\sin n_kx)^2\,dx = \int_A (1-2\cos 2n_kx) \,dx 	o m(A).
      \end{equation*}
      \par Now for the actual problem, let $f(x)$ be the limit of the $\sin
      n_kx$ on $E$. Then, by the first fact above,
      \begin{equation*}
        \int_A f(x)\,dx = \lim_{k 	o \infty} \int_A \sin n_kx\,dx = 0
      \end{equation*}
      using the dominated convergence theorem, hence $f(x) = 0$ almost
      everywhere on $E$ by Exericse ef{1.2}. Let $A$ be where $f(x) = 0$.
      We then have
      \begin{equation*}
        \int_A \left(f(x)^2 - \frac{1}{2}ight)dx
        = \lim_{k 	o \infty} \int_A \left(\sin^2 n_kx - \frac{1}{2}ight) = 0
      \end{equation*}
      using the dominated convergence theorem and the second fact above, and 
      this implies $f(x) = 1/\sqrt{2}$ almost everywhere on $A$. Combining these
      two facts, we have that $m(A) = 0$. Similarly, the set on which $f(x) =
      1/\sqrt{2}$ also has measure zero, so $m(E) = 0$.
    \end{proof}

Exercise 11.16

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Exercise 11.16

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