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If $A \subset [0,2\pi]$ and $A$ is measurable, prove that
$$\lim_{n\to\infty} \int_A \cos nx\, dx = \lim_{n\to\infty} \int_A \sin nx\, dx = 0$$

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1. First, we show the result for open intervals in $[0,2\pi]$. Let $A = (a,b)$.
$$\left|\int_a^b \cos nx\, dx\right| = \left|\frac{1}{n}(\sin bx - \sin ax)\right| \leqslant \frac{2}{n} \xrightarrow{n\to\infty} 0$$
(similarly for $\int_a^b \sin nx\,dx$).

2. If $A$ is a **finite** disjoint union of intervals, we have the conclusion from (1).

3. Now suppose $A = \bigcup_{i=1}^\infty A_i$ is an open set in $[0,2\pi]$. Since $m(A)$ is finite, we can find some $M\in \mathbb N$ such that $\sum_{i=M+1}^\infty m(A_i) < \epsilon/2$.

\begin{align*}
\int_A \cos nx\, dx &= \sum_{i=1}^\infty \int_{A_i} \cos nx\, dx\\
&= \sum_{i=1}^M \int_{A_i} \cos nx\, dx + \sum_{i=M+1}^\infty \int_{A_i} \cos nx\, dx\\
&\le \sum_{i=1}^M \int_{A_i} \cos nx\, dx + \sum_{i=M+1}^\infty \int_{A_i} dx\\
&= \sum_{i=1}^M \int_{A_i} \cos nx\, dx + \sum_{i=M+1}^\infty m(A_i)\\
&< \sum_{i=1}^M \int_{A_i} \cos nx\, dx + \epsilon/2
\end{align*}
Taking $n\to\infty$ and noting that $A_i$'s are open intervals (for which we have already shown the result) we have $$\lim_{n\to\infty}\int_A \cos nx\, dx \le \epsilon/2 < \epsilon $$
Since $\epsilon > 0$ is arbitrary, $$\lim_{n\to\infty}\int_A \cos nx\, dx = 0$$

What do I do next, to extend to **measurable sets?**

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**Hint:** Prove the result when $A$ is an interval, then prove it when $A$ is a disjoint union of intervals. Then, use that any measurable set can be approximated by open sets from outside.

Sat, 03 Jul 2021 10:53 GMT