Let $f_n =\big[\cos\left(1+\sin(x^5)\right)\big]^n$ and consider $L_1\big([0,1], \mathcal{L}, \lambda\big)$ ($\mathcal{L}$ denotes the Lebesgue $-\,\sigma$ algebra and $\lambda$ the lebesgue measure) compute $$\lim_{n \to \infty} \int_0^1 f_n\,\mathrm d\lambda$$
First I note that $|f_n| \leq 1$, $\forall n \in \mathbb{N}$ and since $1 \in L_1([0,1], \mathcal{L}, \lambda)$ then $f_n \in L_1([0,1], \mathcal{L}, \lambda)$ then I try to show that
$$\lim_{n \to \infty} f_n=0$$
and next used the Lebesgue dominated convergence theorem, but I don’t sure to show that $\lim_\limits{n \to \infty} f_n=0\,.$
Any hint or help I will be very grateful.
Best Answer
Note that $-1 \leqslant\sin(x^5) \leq 1$, and so $|\cos(1 + \sin(x^5))| < 1$ except when $\sin(x^5) = -1 \iff x^5 = 3\pi / 2 \pm 2k \pi$. Furthermore, we have that pointwise, whenever $|\cos(1 + \sin(x^5))| < 1$, $f_n \to 0$; so the only thing we need is that $x^5 = 3\pi / 2 \pm 2k \pi$ on a set of measure $0$. But this is easy, since there are countably many solutions, and countable sets have measure $0$.
And even then, there are no solutions on $[0,1]$, as Angelo pointed out. So you're good without even invoking anything about null sets!