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First, we show the result for open intervals in $[0,2\pi]$. Let $A = (a,b)$.
$$\left|\int_a^b \cos nx\, dx\right| = \left|\frac{1}{n}(\sin bx - \sin ax)\right| \le \frac{2}{n} \xrightarrow{n\to\infty} 0$$
Now, suppose $A$ is a measurable set in $[0,2\pi]$. Fix some $\epsilon > 0$. There exists an open set $V$ such that $A \subset V$ and $m(V\setminus A) < \epsilon/2$. Since $V$ is open, it is a countable union of disjoint open intervals in $[0,2\pi]$, say $V = \bigcup_{i=1}^\infty A_i$. Clearly, $m(V) = \sum_{i=1}^\infty m(A_i)$. Since $m(V)$ 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 &\le \int_V |\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, $$\lim_{n\to\infty}\int_A |\cos nx|\, dx = 0$$
Now, $$\left| \int_A \cos nx\, dx  \right| \le \int_A |\cos nx|\, dx $$
gives $$\lim_{n\to\infty} \left| \int_A \cos nx\, dx  \right| = 0 \implies \lim_{n\to\infty} \int_A \cos nx\, dx = 0$$

Sat, 03 Jul 2021 07:57 GMT