Show that $\lim_{x \to \infty} x^p\cdot \mu(\{\omega: |f(\omega)| > x\}) = 0.$

Question

Assume $f \in \mathcal{L}^p(\Omega)$ with $p \in [1,\infty)$. Show that $$\lim_{x \to \infty} x^p\cdot \mu(\{\omega: |f(\omega)| > x\}) = 0.$$

Answer 1

$\int |f(\omega)|^p\{ \omega>x\}d\mu\ge x^p\mu(\{\omega: |f(\omega)| > x\})$ and $\int |f(\omega)|^p\{ \omega>x\}d\mu \to 0$ as $x\to\infty$ by dominated convergence.