Getting the limit of integral of $cos^{2n}(\pi f(x))$ with n→∞ [closed]

Let f be a measurable function on [0, ∞) and such that f(x) ∈ Z iff x ∈ [0, 1]

Evaluate

$$\lim_{n \to \infty} \int_{0}^\infty cos^{2n}(\pi f(x))dx$$

Does anyone know how shall I deal with f(x) part to solve its limit? (And where to begin)

Note that $$\begin{align} \lim_{n \to \infty} \int_{0}^\infty \cos^{2n}(\pi f(x))dx &= \lim_{n\to\infty} \left(\int_0^1 \cos^{2n}(\pi f(x))dx + \int_{1}^\infty \cos^{2n}(\pi f(x))dx \right) \\ &= \lim_{n\to\infty}\left(\int_0^1 1dx + \int_1^\infty\cos^{2n}(\pi f(x))dx\right) \\ &= 1 + \lim_{n\to\infty}\int_1^\infty\cos^{2n}(\pi f(x))dx \\ &\overset*= 1 + \int_1^\infty \lim_{n\to\infty}\cos^{2n}(\pi f(x))dx \\ &= 1\end{align}$$ where $(*)$ is by Beppo Levi's lemma.