For a continuous function $f :\mathbb{R}\rightarrow \mathbb{R}$ satisfying $\int_{\mathbb{R}}|f(x)|dx<\infty$, which of the following is true? Problem: For a continuous function $f :\mathbb{R}\rightarrow \mathbb{R}$ satisfying $$\int_{\mathbb{R}}|f(x)|dx<\infty$$ and for some $\alpha >0$ let $d_f(\alpha)$ be the Lebesgue measure of the set $$\{x\in \mathbb{R} : |f(x)|>\alpha\}$$
Then, for all $\alpha \geq 0$ which of the following is true?

*

*$\alpha d_f(\alpha)\leq \int_{\mathbb{R}}|f(x)|dx$

*$\alpha^2 d_f(\alpha)\leq \int_{\mathbb{R}}|f(x)|dx$

*$d_f(\alpha)\leq \alpha \int_{\mathbb{R}}|f(x)|dx$

*$d_f(\alpha)\leq \alpha^2 \int_{\mathbb{R}}|f(x)|dx$
My solution: As $\alpha<|f(x)|$ on $\{x\in \mathbb{R} : |f(x)|>\alpha\}$ we have
$$\int_{\{x: \alpha<|f(x)|\}}\alpha<\int_{\{x: \alpha<|f(x)|\}}|f(x)|<\int_{\mathbb{R}}|f(x)|dx$$
i.e., $$\alpha.d_f(\alpha)<\int_{\mathbb{R}}|f(x)|dx$$
So, I can see that first option is true.
I belive all the other three options are not necessarily true but i could not come up with an example.
Please help me to see if my argument for first bullet is sufficient/clear and help me to see other in detail.
Thank you.
 A: Question: I believe all the other three options are not necessarily true but i could not come up with an example.
Answer: Yes, you are right. Examples are as follows:
For option $\bf{(2)}$, Take $$f(x)=\begin{cases} 
      x &\text{if}~~~~ x\in [0,20] \\
      40-x &\text{if}~~~~ x\in [20,40]\\
      0 &\text{elsewhere}
         \end{cases}$$
Take $~\alpha=10~$ and $~d_f(\alpha)=20~$ as $~d_f(\alpha)=m\left(\{x\in\mathbb R:|f(x)|>10\}\right)=m(10,30)~$$\therefore~\alpha^2d_f(\alpha)=100\times20=2000~$ and $~\int_{\mathbb R}|f(x)|dx=\frac{x^2}{2}\big|_0^{20}+\left(40x-\frac{x^2}{2}\right)\big|_{20}^{40}=400~$Thus $~\alpha^2d_f(\alpha)\not\le\int_{\mathbb R}|f(x)|dx~.$Hence option $(2)$ is incorrect.
For option $\bf{(3)}$, Take $$f(x)=\begin{cases} 
      x &\text{if}~~~~ x\in [0,3] \\
      6-x &\text{if}~~~~ x\in [3,6]\\
      0 &\text{elsewhere}
         \end{cases}$$
Take $~\alpha=0.1~$ and $~d_f(\alpha)=5.8~$ as $~d_f(\alpha)=m\left(\{x\in\mathbb R:|f(x)|>0.1\}\right)=m(0.1,5.9)~$ and $~\int_{\mathbb R}|f(x)|dx=9\implies \alpha\int_{\mathbb R}|f(x)|dx=0.1\times 9=0.9~$ Clearly $~d_f(\alpha)\not\le\alpha\int_{\mathbb R}|f(x)|dx~$ and hence option $(3)$ is incorrect.
For option $\bf{(4)}$, From the previous example as on option $(3)$, $~\alpha^2=0.01~$ and $~\alpha^2\int_{\mathbb R}|f(x)|dx=0.01\times 9=0.09~$Thus $~d_f(\alpha)\not\le\alpha^2\int_{\mathbb R}|f(x)|dx~.$Hence option $(4)$ is incorrect.

Note: I know that the answer given here is an accepted answer. But I think my answer will also useful for solving the problem to the future reader. So if you downvote my answer, please tell me, where I am wrong. Thank you.
A: your proof is correct but none of the others are correct. For the last 2 pick a small alpha. For 2nd one, take a big alpha and take a function which decays slower than than alpha squared. Something like 1/sqrt{x} seems a good candidate
