Interchange of limit and integral

Question

Let $f_m(x) =\lim_{n\rightarrow \infty}\cos(\pi m!x))^{2n}$; we know that it is a Lebesgue and Riemann integrable function, and $g(x)=\lim_{m\rightarrow \infty} f_m(x)$, which is not Riemann integrable but Lebesgue integrable. I want to know if $$\lim_{m\rightarrow \infty}\int_0^1f_m(x)dx = \int_0^1\lim_{m\rightarrow \infty}f_m(x)dx$$ I know a way to prove this is using the monotone convergence theorem but I don't think I can use it here since $f_m(x)$ is not monotone. Can someone help me?

Answer 1

Considering cosine to be a bounded function, it's trivial to apply the Dominated Convergence Theorem with a constant function, which is Lebesgue-integrable in a finite interval. The statement then follows