Measurability of a set and calculating a special integral

Question

Let $f$ be a measurable function on $[0, 1]$ and let: A = {x $\in$ $[0,1]$| $f(x)$ is an integer}.

Prove that A is measurable and that

$\lim_{n \to \infty} \int_{0}^1 ((cos(\pi f(x)))^{2n} dx$ = $m(A)$.

Answer 1

Here is a sketch but you should be able to fill easily if you have tried,

$A$ is clearly measurable since $A=\bigcup _{j\in\mathbb{Z}} f^{-1}(j)$

Finally, the required integral $I$ can be broken into two parts, $I=\int_{A}+\int_{A^{c}}$

It is quite easy to see that, $\int_{A}$ part is $m(A)$. (Why??)

Lastly, on $A^{c}$ , we must have $|\cos f(x)|<1$, thus $(|\cos f(x)|)^{2n}\to 0$. Invoke dominated convergence theorem and conclude $\ldots$ $\int_{A^{c}}\to0$ as $n\to\infty$