# Let $E_c:=\{x \in \mathbb R: f(x) \ge c\}$ Prove that $\int_{E_c}f\,dm \ge c\cdot m(E_c)$

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.

• You should know that if $f \ge g,$ then $\int f \ge \int g.$ Do you see how this could help you out? – Kore-N Nov 6 '16 at 16:18
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?
• If $f(x)\ge c\chi_{E_c}(x)$ for all $x \in R$ then $\int f(x)dm \ge \int c\chi_{E_c}(x)dm$ – J. Kyei Nov 6 '16 at 19:34
• Exactly. And the right-hand side is just $cm(E_c)$. – carmichael561 Nov 6 '16 at 19:52