Given $\mu(|f(x)|>\lambda)\leq C'\lambda^{-2}$ for all $\lambda>0$ show there exists $C''$ such that $\int_E|f(x)|\leq C''\sqrt{\mu(E)}$

Question

Given $\mu(|f(x)|>\lambda)\leq C'\lambda^{-2}$ for all $\lambda>0$ show there exists $C''$ such that $\int_E|f(x)|\leq C''\sqrt{\mu(E)}$

I am not sure how to solve this problem. Here is my attempt that gets a bound but not the one asked. Given the bound, we can integrate both sides on $[\epsilon,\infty)$ as $\frac{1}{x^2}$ is integrable on that. Doing that and applying Fubini to change the order of integration we get $$\int_{|f|\geq \epsilon}|f|\leq M$$ Using this we can show that $\int_E |f|\leq \epsilon\mu(E)+M$ where $\epsilon>0$

Do I go from here or is this not close to a solution?

Answer 1

The claim is, quite obviously, true if $\mu(E)=\infty$, so we can assume $E$ (is measurable) and it's measure is finite.

To make notation a bit simpler let's assume that $f\ge 0$ and define $\mu_E(X) := \mu(E\cap X)$, the restriction of $\mu$ to $E$.

Note that, in general, $\int_X f(x) d\mu(x) = \int_0^\infty \mu_X\{f>\lambda\} \, d\lambda$ (see, e.g., Theorem 8.16 in Rudin's Real and Complex Analysis).

Then, for any $t>0$, \begin{eqnarray} \int_E f(x) \,d\mu &=& \int_0^\infty \mu_E(\{f>\lambda\}) \,d\lambda \\ &=& \int_0^t \mu_E(\{f>\lambda\})\, d\lambda + \int_t^\infty \mu_E(\{f>\lambda\}) \, d\lambda \\ &\le & t \mu(E) + C^\prime \int_t^\infty \frac{1}{\lambda^2}\, d\lambda \\ & = & t \mu(E) - \left. 2C^\prime\frac{1}{\lambda}\right|_t^\infty \\ &=& t\mu(E) + 2C^\prime \frac{1}{t} \end{eqnarray} Now choose $$t= \frac{1}{\sqrt{\mu(E)}}$$ The desired result then holds with $ C^{\prime\prime} := 1+ 2C^\prime$