How to bound this integration when we know preimage is bounded?

Question

Problem : Let $f:\mathbb{R} \to \mathbb{R}$ be measurable function. Suppose $\exists C > 0$ such that $\forall y \in [0,\infty),$ $$m\{x \in \mathbb{R}: |f(x)| \geq y \} \leq \frac{C}{y^2}.$$ Show that $\exists C' >0$ such that $$ \int_{E}|f(x)|dx \leq C' \sqrt{m(E)}. $$ for any measurable set $E$.

Usually, I questioned in this stackexchange with partial result of what I had. However, in this time, I tried many things, applying Egorov theorem with $1_{\{ x \in \mathbb{R}: |f(x)| \geq n\}}f,$ consider Dominated convergence theorem or Monotone Convergence theorem, etc. However, I didn't get any partial result.

If you give me a hint or clue to approach this problem, it will be greatly helpful for me.

Answer 1

\begin{align*} \int_{E}|f(x)|dx&=\int_{0}^{\infty}m(\{x\in E: |f(x)|\geq y\})dy\\ &=\int_{0}^{C^{1/2}/m(E)^{1/2}}m(\{x\in E: |f(x)|\geq y\})dy\\ &~~~~+\int_{C^{1/2}/m(E)^{1/2}}^{\infty}m(\{x\in E: |f(x)|\geq y\})dy\\ &\leq\int_{0}^{C^{1/2}/m(E)^{1/2}}m(E)dy+\int_{C^{1/2}/m(E)^{1/2}}^{\infty}\dfrac{C}{y^{2}}dy\\ &=C^{1/2}m(E)^{1/2}+C^{1/2}m(E)^{1/2}\\ &=C'm(E)^{1/2}, \end{align*} note that $m(\{x\in E: |f(x)|\geq y\})\leq\min\{m(E),C/y^{2}\}$.

Answer 2

Hint: Fubini shows that $$\int_E f(x)\,dx=\int_0^\infty m(\{t\in E:f(t)>y\})\,dy.$$

(Yes, $\int_0^\infty\frac1{y^2}\,dy=\infty$. But there are two things you know about $m(\{t\in E:f(t)>y\})$; one is that it's no larger than $C/y^2$, the other is...)

Answer 3

From @David's hint and @user284331's help, I just leave my own answer. Thank you very much.

If $m(E)=\infty$ then the inequality trivially holds. Suppose $m(E)< \infty.$ Then, as David hinted out, $m(\{ x\in E : |f(x)| \geq y \})$ is bounded by both $Cy^{-2}$ and $m(E)$. Thus, $$ m(\{ x\in E : |f(x)| \geq y \}) \leq \min(Cy^{-2}, m(E))$$ Also, $Cy^{-2}= m(E)$ when $y = \sqrt{Cm(E)}$, as @user284331 implicitly showed above. Also, as a function of $x$ and $y$, both $m(\{y \in \mathbb{R}: y< |f(x)| \}$ and $m(\{x \in E : |f(x)|\geq y \}$ is integrable since $f$ is bounded by above. Thus, we can apply Fubini's theorem, as @David hinted, that

\begin{align} \int_{E}|f(x)|dx &= \int_{E}m(\{y \in \mathbb{R}: y< |f(x)| \}dx \\ & =\int_{E}\int_{\{y < |f(x)| \}}dy dx \\ & = \int_{0}^{\infty}\int_{|f(x)|\geq y}dx dy &\textrm{ by Fubini's theorem} \\ & = \int_{0}^{\infty}m(\{x \in E : |f(x)|\geq y \} \\ \end{align}

And the rest part is the same as what @user284331 showed.

Thank you very much.