Proof of A Property of Lebesgue Integral I am proving the following property of Lebesgue Integral: 

Let $(E,\mathcal A,\mu)$ be a measurable space. If $a\le f(x)\le b$ for $x\in E$, and $\mu(E)\lt \infty$, then $$a\mu(E)\le \int_E f\ d\mu\le b\mu(E).$$

The following was what I tried: 
First assume that $f(x)\le0$ for some $x\in E$ and $f(x)\ge0$ for some $x\in E$, and then $b\ge 0$ and $a\le 0$. Let $f=f^+-f^-$, and then $f^+ \le b$ and $f^- \le -a$. Let $s$ be the simple function such that $$0\le s \le f^+ \le b$$ and$$0\le s \le f^- \le -a.$$Then, we have: $$0\le \int_E f^+\ d\mu =\sup_{0\le s\le f}\int_Es d\mu \le b\cdot\mu(E)$$and $$0\le \int_E f^-\ d\mu =\sup_{0\le s\le f}\int_Es d\mu \le -a\cdot\mu(E).$$ Since by the definition, we have $$\int_E f\ d \mu=\int_E f^+\ d \mu-\int_E f^-\ d \mu. $$
Then $$\int_E f\ d \mu=\int_E f^+\ d \mu-\int_E f^-\ d \mu \le \int_E f^+\ d \mu \le b\cdot\mu(E)$$ and $$\int_E f\ d \mu=\int_E f^+\ d \mu-\int_E f^-\ d \mu \ge -\int_E f^-\ d \mu \ge a\cdot\mu(E). $$Then the desired result follows. 
The cases where $f$ is strictly negative or strictly positive can be proved separately. 
Does this proof make sense? Is there any simpler way to prove this? Thanks in advance. 
 A: It's important to first recall the definition of an integral on a subset of a measure space. Say $(X,\mathcal{M},\mu)$ is a measure space. If $E\subseteq X$ and $E\in\mathcal{M}$ is a measurable set, then we define:

$\int_{E}f\ d\mu=\int_{X} \chi_{E}(x)\cdot f(x)\ d\mu.$

Now, we need one basic property of the Lebesgue integral. Namely, if we have two functions $p(x),q(x):X\to \mathbb{R}$ and $p(x)\le q(x)$ a.e., then 

$\int p(x)\ d\mu\le \int q(x)\ d\mu$ (1)

It is good practice for you to prove this straight from the definitions (it follows from the fact that every simple function dominated by $p(x)$ is dominated from $q(x))$. Now, the statement that $f(x)\le a$ on $E$ is precisely the statement that $\chi_{E}(x)\cdot f(x)\le a\chi_{E}(x)$ a.e., and likewise the statment $b\le f(x)$ on $E$ is precisely the statement that $\chi_{E}(x)\cdot f(x)\ge b\chi_{E}(x)$ a.e. (work this part out!). Consequently, by $(1)$, we have that 

$b\mu(E)=\int b\chi_{E}(x)\ d\mu\le\int \chi_{E}(x)\cdot f(x)\ d\mu\le \int \chi_{E}(x)\cdot a \ d\mu=a\mu(E)$

