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?