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Let $f\ge 0$ be a measurable real-valued function defined on $\mathbb R$. For each $c>0$, Let $E_c:=\{x \in\mathbb R :f(x)\ge c \}$. Prove that \

$$\int_{E_c}f\,dm\ge c \cdot m(E_c).$$

My idea is to construct a sequence of simple functions approximation of $f$\

$\Phi_n=\sum_{i=1}^N a_i\chi_{E_i}$. since we are integrating over $E_c:=\{x \in\mathbb R :f(x)\ge c \}$, I will let $a_i \ge c \quad \forall a_i $ \

By definition, $\int_{E_c}f\,dm= \sup\left\{\int_{E_C}\Phi \, dm, \Phi\le f \right\}$ \and

$\int_{E_C}\Phi \, dm =\sum_{i=1}^N a_i \cdot m(E_i)$. So In this case $E_c=\bigcup E_i$ \

But

$$\sum_{i=1}^N a_i \cdot m(E_i) \color{red}{\ge} \sum_{i=1}^N c\cdot m(E_i) = c \sum_{i=1}^N m(E_i) \text{ since } a_i \ge c \quad \forall a_i $$

by additivity of measure \

$c\sum_{i=1}^N m(E_i)=c\cdot m\left(\bigcup E_i\right)=c\cdot m(E_c)$ but I feel something is off.

Can you please help me out. thank you all in advance.

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    $\begingroup$ You should know that if $f \ge g,$ then $\int f \ge \int g.$ Do you see how this could help you out? $\endgroup$ – Kore-N Nov 6 '16 at 16:18
  • $\begingroup$ Yes,it helps a lot. Thank you $\endgroup$ – J. Kyei Nov 6 '16 at 19:58
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I think you're overcomplicating things: note that $f(x)\geq c\chi_{E_c}(x)$ for all $x\in E_c$ by the definition of $E_c$. So what do you get if you integrate both sides of this inequality?

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  • $\begingroup$ If $f(x)\ge c\chi_{E_c}(x)$ for all $ x \in R $ then $\int f(x)dm \ge \int c\chi_{E_c}(x)dm $ $\endgroup$ – J. Kyei Nov 6 '16 at 19:34
  • $\begingroup$ Exactly. And the right-hand side is just $cm(E_c)$. $\endgroup$ – carmichael561 Nov 6 '16 at 19:52
  • $\begingroup$ Oh okay , I see. It looks clearer now. thank you very much. $\endgroup$ – J. Kyei Nov 6 '16 at 19:57

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