Suppose that $E \subseteq \mathbb{R}$ is measurable and that the measurable functions $f: E \to \mathbb{R}$ satisfies $\int_E |f| ^p < \infty$. If $\omega$ is the distribution function of $f$, prove that $\lim_{\lambda \to 0} \lambda ^p \omega(\lambda) = 0$.
I'm completely lost. How do I solve this problem?
We define $\omega(\lambda) = \mu(\{x \in E, |f(x)| \geq \lambda\})$, then $\lambda^p \omega(\lambda) \to 0$ when $\lambda \to +\infty$ is well known, since $$\lambda^p \omega(\lambda) \leq \int_E |f|^p1_{|f|\geq\lambda}$$
It's interesting to see the convergence still holds when $\lambda \to 0$:
Take $\lambda_n \downarrow 0$, then we remark that
$$\int_E |f|^p \geq \sum_{n} |\lambda_{n+1}|^p\left(\omega(\lambda_{n+1}) - \omega(\lambda_n))\right) = \sum_{n} (\lambda_{n}^p - \lambda_{n+1}^p)\omega(\lambda_n)$$
Then for any $\epsilon >0$, $\exists N$ such that for all $k > N$ $$\sum_{n=k}^\infty (\lambda_{n}^p - \lambda_{n+1}^p)\omega(\lambda_n) < \epsilon$$
Then remark that $\omega(\lambda_n) \geq \omega(\lambda_k), \forall n \geq k$, so
$$\sum_{n=k}^\infty (\lambda_{n}^p - \lambda_{n+1}^p)\omega(\lambda_k) < \epsilon$$
i.e. $$\lambda_k^p\omega(\lambda_k)< \epsilon, \forall k > N$$
It's esay to complete the proof from here(suppose the conclusion is not true, then exists $\lambda_n$ such that blabla, and $\lambda_n$ has a decreasing subsequence, then apply what's above to get a contradiction.)