$\lim_{n\to 0} n^p\mu(\{x\in X:|f(x)|>n\})=0$

Question

Let $f\in L^p(X)$, how to show $\lim_{n\to 0} n^p\mu(\{x\in X:|f(x)|>n\})=0$?

My attempt: Let $E$ be $\{x\in X:|f(x)>n|\}$, then by $\int_En^p<\int_E|f|^p<\int_X|f|^p<\infty$, we have $\mu(E)\cdot n^p<\infty$. But I can't show it's going to $0$.

Answer 1

For $\alpha >0$, define $f_\alpha(x) = f(x) \chi_{\{|f| \leq \alpha\}}(x)$, we have $f_\alpha(x) \to 0$ for all $x\in X$. Moreover, $|f_\alpha|^p \leq |f|^p$ then by dominated convergence theorem, we have $$\lim_{\alpha \to 0} \int_X |f_\alpha|^p d\mu = 0.$$ For any $\epsilon >0$, we can choose $\alpha_0$ such that $$\int_X |f_\alpha|^p d\mu \leq \epsilon,\qquad \forall\, \alpha \leq \alpha_0.$$ For $n < \alpha_0$, we have $\mu(\{|f|> n\}) = \mu(\{|f| > \alpha_0\}) + \mu(\{n < |f|\leq \alpha_0\})$, hence $$n^p\mu(\{|f|> n\}) = n^p \mu(\{|f| > \alpha_0\}) + n^p\mu(\{n < |f|\leq \alpha_0\})\leq \int_X |f_{\alpha_0}|^p d\mu +n^p\mu(\{|f|> \alpha_0\}) \leq \epsilon + n^p\mu(\{|f|> \alpha_0\}).$$ Therefore $$\limsup_{n\to 0}n^p\mu(\{|f|> n\}) \leq \epsilon,$$ for any $\epsilon >0$. This finishes the proof.

Answer 2

Let's take $p=1$ just to simplify things.

It's easy to see the result is true if $f$ is a simple function in $L^1.$ For the general $f\in L^1,$ let $\epsilon>0.$ Then there is a simple function $s\in L^1$ such that $\int_X |f-s| <\epsilon.$ Note that $\{|f|>t\}\subset \{|s|>t/2\} \cup \{|f-s|>t/2\}.$ We thus have

$$\tag 1 t\mu(\{|f|>t\}) \le t\mu(\{|s|>t/2\}) + t\mu(\{|f-s|>t/2\}).$$

But notice the last term is no more than $2\int_X|f-s| < 2\epsilon$ for any $t>0.$ Thus as $t\to 0^+,$ the $\limsup$ of the left side of $(1)$ is no more than $0+2\epsilon.$ Because $\epsilon$ is arbitrarily small, we have shown the desired limit is $0.$