Show that if $\delta \in ]0,1[$ and $E_\delta = \{x \in E \mid f(x) \geq \alpha\delta\}$, then $m(E_\delta) \geq (1 - \delta^2)\alpha^2/\beta$. 
Let $f \colon E \to [0,+\infty]$ be a measurable function, with $0 < m(E) < +\infty$ (where $m(E)$ is the Lebesgue measure of $E$). Assume that there exists $\alpha, \beta$ such that $$\frac{1}{m(E)}\int_E f > \alpha \quad \textrm{and} \quad \frac{1}{m(E)}\int_E f^2 \leq \beta.$$
Show that if $\delta \in ]0,1[$ and $E_\delta = \{x \in E \mid f(x) \geq \alpha\delta\}$, then $m(E_\delta) \geq (1 - \delta)^2\alpha^2/\beta$.

Hi! I've been studying measure theory for the past months now and i stumbled upon this question. I nor know how to answer it neither have an intuition about what it means. Could you help me?
 A: I think that your inequality should be $$m(E_\delta )\geq \frac{(1-\delta )^{\color{red}2}\alpha ^2}{\beta },$$
instead of what you wrote.

Suppose WLOG that $m(E)=1$, i.e. $m$ is a probability measure on $E$. I'll denote $\mathbb E$ the expectation w.r.t. $m$. Your question is equivalent than proving that if there are $\alpha \in \mathbb R$ and $\beta>0$ s.t. $$\mathbb E[f]\geq \alpha \quad \text{and}\quad \mathbb E[f^2]\leq \beta ,$$
then $$m\{f\geq \alpha \delta \}\geq\frac{(1-\delta )^2\alpha ^2}{\beta }.$$
So,
\begin{align*}
\alpha &\leq \mathbb E[f]\\
&=\mathbb E[f\boldsymbol 1_{\{f\geq \alpha \delta \}}]+\mathbb E[f\boldsymbol 1_{\{f<\alpha \delta \}}]\\
&\leq \sqrt{\mathbb E[f^2]}\sqrt{m\{f\geq \alpha \delta \}}+\alpha \delta\\
&\leq \sqrt \beta \sqrt{m\{f\geq \alpha \delta \}}+\alpha \delta,  
\end{align*}
and thus $$\sqrt{m\{f\geq \alpha \delta \}}\geq \frac{\alpha (1-\delta )}{\sqrt\beta }\implies m\{f\geq \alpha \delta \}\geq \frac{(1-\delta )^2\alpha ^2}{\beta }.$$
The claim follows.

I let you adapt the proof whenever $m(E)\neq 1$.
