Can/cannot one integrate over a measurable set $A \subset [-\pi, \pi]$?

Question

So this has caused some confusion. In one exercise one was asked to prove that

$$\lim_{k \rightarrow \infty} \int_A \cos(kt)dt=0$$

where $A \subset [-\pi, \pi]$ is a measurable set.

My initial idea was to take any $a,b \in A$ and then show that:

$$\lim_{k \rightarrow \infty} \int_a^b cos(kt)dt= \lim_{k \rightarrow \infty} \frac{\sin(bk)-\sin(ak)}{k} =0$$

But my instructor said that this wouldn't work out because the lengths of the partitions on $A$ would make my computation impossible. I'm not sure about this.

Answer 1

Trying to use an interval as a replacement for $A$ is not a good approach. $A$ could be a very complicated set, after all; it could be a countable union of intervals, or a set like the irrationals, or some totally disconnected Cantor set with positive measure. There is a general principle that measurable sets are like countable unions of intervals for many purposes, but this approach goes too far.

So there is very little relationship between the interval $[a, b]$ and the set $A$, even with $a, b \in A$.

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For an approach that you might find more successful: The integral you've written is the real part of the $k$-th Fourier coefficient of $\chi_A$ (the indicator function of $A$). What does Plancherel tell you?

Answer 2

Let $f=1_A$, and note that $f \in L^2[-\pi,\pi]$.

Let $\hat{f}_n = {1 \over 2 \pi} \int_{-\pi}^\pi f(t) e^{-int} dt$.

Bessel's inequality gives $\sum_n |\hat{f}_n|^2 \le \|f\|_2^2$, and so $\hat{f}_n \to 0$.

Another approach would be the Riemann Lebesgue lemma.

Answer 3

I think that your approach is a good start, but you have to extend your result to general measurable sets. You can do that by approximation.

Let $\mathbb T=[-\pi, \pi)$.

Step 1. Show that for any open set $O\subset \mathbb T$, it holds that $\lim_{k\to \infty} \int_O \cos(kt)\, dt =0$. To do this, use the fact that every open set in $\mathbb T$ is a countable disjoint union of intervals. Since $\mathbb T$ has a finite total measure, most of $O$ must be concentrated on a finite disjoint union of intervals. More precisely, for all $N\in\mathbb N$, $O$ can be decomposed as $$O=\bigcup_{j=1}^N (a_j, b_j) + R_N,\ \text{where }|R_N|\le\tfrac 1 N.$$ Now use the computation you have already done.

Step 2. Fix $\epsilon>0$. We know from measure theory that every measurable set $M\subset \mathbb T$ is contained in an open set $O_\epsilon$ in such a way that $|O_\epsilon \setminus M|\le \epsilon$. (Absolute value of a set denotes Lebesgue measure). So $$ \left| \int_M \cos(kt)\, dt\right|\le \left| \int_{O\setminus M} \cos(kt)\, dt \right| + \left| \int_O \cos(kt)\, dt\right| \le \epsilon + \eta_k, $$ where $\eta_k\to 0$ as $k\to \infty$.