If is measurable and $\leq ()\leq $ for $\in$, and if $\mu(E)<\infty$, then prove that $$a\mu(E)\leq \int_E fd\mu \leq b \mu(E).$$
Work done:
Define two constant functions $g$ and $h$ such that $g(x)=a$ and $h(x)=b$ for all $x$. Now,
Clearly, $$\int_E g d\mu=a\mu (E)$$ and $$\int_E h d\mu =b\mu(E)$$ Hence, $$a\mu(E)\leq \int_E f d\mu\leq b\mu(E).$$
I hope this is fine.
But if I wanted to prove this without using the $$f\leq g\implies \int f\leq \int g$$ then how to prove this?
Since according to rudin, this result comes after the result whatever i have used in my proof.
Using the defn:
Let $0\leq s\leq f$ where $s$ is any simple function.
$$\int f d\mu= sup \int s d\mu = sup \sum c_i\mu(E_i)\leq b sup \sum \mu (E_i)=b\mu(E)$$
For the lower bound, consider the function $a$. Note that $a$ is a simple function.
Now, $$a\mu(E)=\sum_{i=1}^n a \mu(E_i)=\int a d\mu \leq sup \int s d\mu =\int f d\mu$$
Hence the proof.