$|\{ |f| > t \}| \leq Ct^{-2} \implies \int_{E}|f| \leq D\sqrt{|E|}$

Question

Let $f \colon \mathbb{R}^{n} \to \mathbb{R}$ be a measurable function such that there is a constant $C > 0$ such

$$ |\{|f| > t \}| \leq Ct^{-2} ~~ \text{for all } t>0$$

Show that there exists a constant $D > 0$ such that for every measurable set of finite measure $E$:

$$ \int_{E} f(x) dx \leq D\sqrt{|E|}$$

Answer 1

This is related to the "duality characterization" of weak $L^2$- space. For a given $\lambda>0$ (which will be chosen later), let $\displaystyle E = \left(E\cap \{|f|>\lambda\}\right) \cup \left( E\cap \{|f| \le \lambda\}\right) = E' \cup E'' $. Then we have \begin{align*} \left|\int_E f\right| \le& \int_{E'} |f| + \int_{E''}|f| \\ \le& \int_\lambda^\infty \left|\{ |f|>t\}\right| dt + \int_{E}\lambda \\ \le& \frac{C}{\lambda} + |E|\lambda \end{align*} from the fact that $\displaystyle \int \varphi =\int_0^\infty \left|\{\varphi > t\}\right|dt$ for all non-negative measurable $\varphi$. Now we choose $\lambda = \left(\frac C {|E|}\right)^{\frac 1 2}$ to have that $$ \left|\int_E f\right| \le 2 C^{\frac 1 2} |E|^{\frac 1 2 }. $$