Let $E_{n}= \lbrace x \in X \: | \: |f(x)| \geq n \rbrace$, then $\mu(E_{n}) \to 0$ when $n \to \infty$.

Question

Let $f \in L^{p}$ for $p \in [1, \infty)$ and $E= \lbrace x \in X \: | \: |f(x)| \neq 0 \rbrace$. If $$E_{n}= \lbrace x \in X \: | \: |f(x)| \geq n \rbrace$$, prove that $\mu(E_{n}) \to 0$ when $n \to \infty$.

Already prove that $E$ is $\sigma$ finite and I feel this is somehow related to prove what I stated before but for being honest I dont know how to attack this problem. Any help will be aprecciated. Thanks

Answer 1

The following weak type inequality is useful: \begin{align*} \alpha^{p}|(|f|\geq\alpha)|\leq\int|f|^{p}. \end{align*}

Answer 2

Note that $n \cdot 1_{E_n} \le |f|$ and so $n^p \mu E_n \le \int |f|^p$.

Answer 3

Note that

$$\mu(E_n) = \mu\{|f|^p/n^p \geq 1\} = \int I_{\{|f|^p /n^p \geq 1\}}d \mu \leq \int |f|^p/n^p d \mu = 1/n^p \int|f|^p d \mu \stackrel{n \to \infty}{\to} 0$$

since $$\int|f|^p d \mu < \infty$$

and where we used the trivial inequality

$$I_{\{|f|^p /n^p \geq 1\}} \leq|f|^p/n^p$$