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.
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?