properties of Lebesgue integration on the set of all measurable functions $\mathcal{L}^0_+([0,\infty])$

Question

I am going though the definition of Lebesgue integrals and I am having troubles with showing two properties.

Let $\mathcal{L}^0_+([0,\infty])$ be the set consisting of all measurable $[0,\infty]$-valued functions. How do I show that for $f_1,f_2\in \mathcal{L}^0_+([0,\infty])$ and $a\in [0,\infty]$ we have:

• If $f_1(x)\leq f_2(x)\forall x\in S$,where $S$ is nonempty, then $\int f_1d\mu\leq\int f_2d\mu$
• $\int \alpha fd\mu=\alpha \int fd\mu$

Answer 1

Remember $\int f d\mu = \sup\{h \in \mathscr{E}^+ : h \leq f\}$ where $\mathscr{E}^+$ is the set of positive simple functions. Define $A_f :=\{h \in \mathscr{E}^+ : h \leq f\}$, such that $\int f d\mu =\sup A_f$

(1) If $f_1 \leq f_2$, then $A_{f_1} \subset A_{f_2}$, therefore $\sup A_{f_1} \leq \sup A_{f_2}$, meaning $\int f_1 d\mu \leq \int f_2 d\mu$

(2) Let $\alpha \in \mathbb{R}^+$, one has $A_{\alpha f} = \alpha A_{f}$, therefore $\sup A_{\alpha f} = \alpha \sup A_{f}$, meaning $\int \alpha f d\mu = \alpha \int f d\mu$