For (Lebesgue) measurable functions $f$ and $g$, if $f=g$ a.e., then $ \int_{E} f=\int_{E} g. $

Question

Problem

Let $f$ and $g$ be bounded (Lebesgue) measurable functions defined on a set $E$ of finite measure. If $f=g$ a.e., then
$$
\int_{E} f=\int_{E} g.
$$

Attempt

Let $f,g:E\to \mathbb{R}$ be a (Lebesgue) measurable function with $m(E)<\infty$. Assume $f=g$ a.e. on $E$. It is enough to show that
$$
\int_{E} f-g=0.
$$
Let $\varphi$, $\psi$ be simple functions with $\varphi \leq f-g \leq \psi$. Since $f-g=0$ a.e., we have $\psi \geq 0$ a.e. and $\varphi\leq 0$ a.e.. Then,
$$
0 \leq \int_{E} \psi {\rm ~and~} 0\geq \int_{E} \varphi,
$$
${\it whence} $
$$
\int_{E} f-g \geq 0 {\rm ~and~} \int_{E} f-g \leq 0,
$$
as required.

Question

1. The hypothesis that there is a simple function $\varphi$ is vaild?
2. What is the exact meaning of 'whence' which is italics?
3. For the end of sentence, what did the right expression of the two?: a.e. or a.e..

Thanks!

Answer 1

The assertion about the simple functions is not clear neither justified. A simpler approach: let $A:=\{x\in \mathbb{R}:f(x)-g(x)=0\}$, then $A$ is measurable and

$$ \int_{E}(f-g) \mathop{}\!d \lambda =\int_{E \cap A}(f-g) \mathop{}\!d \lambda +\int_{E \cap A^\complement }(f-g) \mathop{}\!d \lambda =\int_{E \cap A}0 \mathop{}\!d \lambda+0=0 $$

where we used the fact that if a set $B$ have measure zero then $\int_B f \mathop{}\!d \lambda =0$ for any measurable function $f$.