Use squeeze theorem to integrate over a measurable set $A \subset [-\pi, \pi]$?

Question

In trying to prove that

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

where $A \subset [-\pi, \pi]$ is a measurable set, what is wrong with the following?

1. $$0 = \int_B 0 \times \cos(kt)dt \le \int_B 1_A \times \cos(kt)dt \le \int_B 1 \times \cos(kt)dt = \frac{\sin(\pi k)-\sin(-\pi k)}{k}$$

2. There is some $B$ for which $(1)$ is true.

3. Then squeeze theorem as $k \to \infty$

Related:

Limit of integral over set measurable

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

Answer 1

What is wrong with your reasoning is that $\cos(kt)$ may be negative, so it is not true that multiplying by it preserves inequalities, as you implicitly assumed. The fact that in the end you found $0 \le ... \le 2 \sin(k\pi)/k$ and that this last value may be negative should have phased you a little...