Proposition $4.5$ Royden's Real Analysis

Question

From Real Analysis Halsey Royden

$5$. Proposition:If $f$ and $g$ are bounded measurable functions defined on a set $E$ of finite measure,then:

$\int_E (af+bg)=a\int_Ef+b\int_Eg$

the solution is given as,

For the second case $a<0$ they wrote,$inf_{\phi \leq f} \int_E a\phi =a sup_{\phi \leq f}\int_E \phi$,can someone explain me how?I am not getting how they got this.

Answer 1

Observe for any bounded subset $A$ of $\mathbb R$ and $b>0$ $$-\sup\{x|x\in A\}=\inf\{-x|x\in A\} \text{ and }\sup\{bx|x\in A\}=b(\sup\{x|x\in A\})$$ Since $a<0$ we take $a=-b$ for some $b>0$ then we have $$inf_{\phi \leq f} \int_E a\cdot\phi =inf_{\phi \leq f} \int_E (-b)\cdot\phi= - sup_{\phi \leq f}\int_E b\cdot\phi=(-b)sup_{\phi \leq f}\int_E \phi$$$$ =a\cdot sup_{\phi \leq f}\int_E \phi$$

Answer 2

This is a property of "infimum" and "supremuum".

In fact, if $A \subseteq \mathbb{R}$ and $\lambda \in \mathcal{R}$ with $\lambda <0$ then you have $\inf \lambda A=\lambda\sup A$ where I mean $\lambda A=\{\lambda a : a \in A\}$.