Showing that if $f=g$ a.e. $\int_E f = \int_E g$. Given the following: $f$ is a bounded measurable function on a set of finite measure $E$.  Assume $g$ is also bounded and $f=g$ a.e. on $E$.  Show that $\int_E f = \int_E g$.
The most "powerful" tool right now is the bounded convergence theorem.  So I'm proceeding from there.
WLOG suppose $f,g \geq 0$.  Since $f \geq 0$, there exists a sequence of simple functions $\phi_n$ such that $\phi_n \leq f$ and $\phi_n \rightarrow f$.  So by the Bounded Convergence Theorem, $\lim\limits_{n \rightarrow \infty} \int_E \phi_n = \int_E f$.
Because $f = g $ a.e. on $E$, then $\lim\limits_{n \rightarrow \infty} \int_E \phi_n = \int_E g$.
Is this a proper application of the BCT?
 A: As Harald Hanche-Olsen said, your proof is perfectly correct and I like his second solution.
Here's yet another approach, using the inequality $\lvert \int f\rvert \leq \int \lvert f\rvert$ which follows directly from the definition of the integral as a supremum over simple functions and $|f| = f^+ + f^-$ where $f = f^+ - f^-$ is the decomposition of $f$ into positive and negative parts.


*

*If $h$ is bounded, measurable and $h = 0$ a.e. on $E$ then $\int_E h = 0$: there exists a null set $N \subset E$ such that $h = 0$ on $E \setminus N$. Then
$$
0 \leq \left\lvert \int_E h \right\rvert \leq 
\int_N \underbrace{\lvert h\rvert}_{\leq C} + \int_{E \setminus N}\underbrace{\lvert h\rvert}_{=0} \leq C \mu(N) = 0
$$ 

*Apply 1. to $h = f-g$.

A: Let $A\subset E$ be the set of measure zero on which $f\ne g$. Then
    $A\cup E\setminus A=E$ yields that $E\setminus A$ is measurable, as it is the
    complement in a measurable set of a set of measure zero.
    Note also that $g$ is measurable on $E$. Where $f=g$ this is clear. Let
    $M\in \mathbb{R}$ then
    $$
\{ x: g(x)>M\}\cap A\subseteq A
$$
    a set of measure zero. But subsets of measure zero sets have measure zero, and
    thus are measurable.
    Finally, since both $A$ and $E\setminus A$ are measurable,
    so is $f$ on the sets, since
    \begin{align*}
  &\{ x:\;x\in A, \;f(x)>m\}=\{ x:f(x)>m\}\cap A\\
  &\{ x:\;x\in E\setminus A, \;f(x)>m\}=\{ x:f(x)>m\}\cap E\setminus A
\end{align*}
    the intersection of measurable sets.
    Thus,
    \begin{align*}
  &\int_{E\setminus A}f=\int_{E\setminus A}g\\
  &\stackrel{m(A)=0\;\text{and} \; g\;\text{bounded}}{\implies}
  \int_{E\setminus A}f+\int_{A}f=\int_{E\setminus A}g+\int_{A}g\\
  &\implies
  \int_Ef=\int_Eg
\end{align*}
    With the final implication following since 
$$
A\cap B=\emptyset\implies \int_{A\cup B}f=\int_Af+\int_Bf
$$
for $A,B$ measurable subsets of a set of finite measure and $f$ bounded.
