Proving that $\lim_{m\to \infty }\lim_{n \to \infty }(\cos(m!\pi x))^{2n}$ is Lebesgue measurable.

Question

Proving that $$\lim_{m\to \infty }\lim_{n \to \infty }(\cos(m!\pi x))^{2n}$$ is Lebesgue measurable.

I was thinking of proving it this way: (three steps)

$1.$

Prove for fixed $m,n$ that $\cos(m!\pi x))^{2n}$ is measureable.

So I have $A=\{x\in \mathbb{R}| \cos(m!\pi x))^{2n}<c\}$

for $c>1: A \equiv \mathbb{R}\in \mathcal{B}(\mathbb{R})$

for $c<1: A \equiv \emptyset\in \mathcal{B}(\mathbb{R})$

for $c\in [-1,1]: A = \bigcup(a_i,b_i) \text{ because $\cos$ is a periodic function and to some power over 1 would still be within [-1,1]}\in \mathcal{B}(\mathbb{R})$

$2.$

We take now for $f_n(x)=(\cos(m!\pi x))^{2n}$. We already prove in class that if $\exists\lim_{n\to \infty}f_n(x)$ then $f(x)=\lim_{n\to \infty}f_n(x).$ is measurable I cannot prove that $\exists\lim_{n\to \infty}f_n(x)$ because if $\cos(m!\pi x)=-1$ i just dont know what $\lim_{n\to \infty}f_n(x)$ is in that case.

$3.$ I guess do something similar with step two...

Answer 1

Hints: For fixed $m$ you may take the limit in $n$ and obtain a function which is zero except for some point where it equals +1 (find out where and why it is measurable). The limit in $m$ is in fact a monotone limit (why?) and you may conclude. You may even describe the limit quite precisely.