If $f$ is in $L^p$, prove that $\lim_{\lambda \to 0} \lambda ^p \omega(\lambda) = 0$ 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?
 A: 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.)
A: Here is another proof, using the identity $$p\int_{0}^{\infty} \lambda^{p-1} \omega(\lambda) \, d \lambda = \int_{E} |f|^{p} \, dm$$ which one can prove with Fubini-Tonelli. If $f\in L^{p}$, then the integral on the left converges; making the change of variables $\lambda = 1/t$, we have $$\int_{0}^{\infty} \lambda^{p-1} \omega(\lambda) \, d \lambda = \int_{\infty}^{0} \frac{1}{t^{p-1}} \omega\left(\frac{1}{t} \right) \frac{-1}{t^2} \, dt = \int_{0}^{\infty} \frac{1}{t^{p+1}} \omega\left(\frac{1}{t} \right) \, dt$$ Since this integral converges, and the function is nonnegative, we have $\lim_{c \to \infty} \int_{c}^{\infty} t^{-(p+1)} \omega(t^{-1}) \, dt =0$.
On the other hand, we can bound $$\int_{c}^{\infty} t^{-(p+1)} \omega(t^{-1}) > \int_{c}^{\infty} t^{-(p+1)}\, dt \cdot \omega(1/c) = \frac{1}{p}c^{-p} \omega(1/c)$$ Taking $c\to\infty$ gives the desired result by squeezing.
