Mathematics Asked by fdez on October 8, 2020

Let $f_m(x) =lim_{nrightarrow infty}cos(pi m!x))^{2n}$; we know that it is a Lebesgue and Riemann integrable function, and $g(x)=lim_{mrightarrow infty} f_m(x)$, which is not Riemann integrable but Lebesgue integrable. I want to know if

$$lim_{mrightarrow infty}int_0^1f_m(x)dx = int_0^1lim_{mrightarrow infty}f_m(x)dx$$ 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?

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

Correct answer by Evaristo on October 8, 2020

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