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

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?

• I don't know why this is true. Using Fubini $$\int_E |f(x)|\mu(dx)=\int_0^\infty \mu\{|f(x)|>\lambda \}\,\mathrm d \lambda \leq C'\int_0^\infty \frac{1}{\lambda ^2}\,\mathrm d \lambda =\infty .$$
– Surb
Apr 15, 2020 at 20:07
• @Surb You only showed that in integral is perhaps infinite depending on $E$. We are only proving an upper bound after all. s Apr 15, 2020 at 20:14
• I didn't proved anything ! I just show that given the information you have, we a priori can't conclude anything...
– Surb
Apr 15, 2020 at 21:28
• you are quite close, but for some reason did not find the right place to use the hypothesis... Apr 18, 2020 at 14:48

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 $$C^{\prime\prime} := 1+ 2C^\prime$$