I am looking for feedback/corrections on the following solution attempt. I have only typed up one direction of the inequality; if this direction is correct, then I am certain that the other direction is as well. Thank you so much!
$\textbf{Problem:}$ Let $f$ be a bounded measurable function on a set of finite measure $E$. Show that if $A \subset E$ is measurable, then $\int_{A} f \> dm = \int_{E} f\cdot \chi_{A} \> dm$, where $m$ is the Lebesgue measure.
$\textbf{ Solution:}$ By definition, $$\int_{E} f \cdot \chi_{A} \> dm = \inf\{\int_{E} \psi \> dm: \psi \> \text{ is simple and} \> f\cdot \chi_{A} \leq \psi \}.$$ Take any simple function $\psi=\sum^{n}_{i=1} c_i \chi_{C_i}$ defined on $E$ such that $f \cdot \chi_{A} \leq \psi$, where $\sum^{n}_{i=1} c_i \chi_{C_i}$ is the canonical representation of $\psi$. Then, we know $f \leq \psi$ on $A$. Now, observe that $\psi'=\sum^{n}_{i=1} c_{i} \chi_{(C_{i} \cap A)}$ is a simple function on $A$ with $f \leq \psi'$ and, by defintion of $\int_{A} f \> dm$, we get that $$ \int_{A} f \> dm \leq \int_{A} \psi' \> dm = \sum^{n}_{i=1} c_{i} m(C_{i} \cap A) \leq \sum^{n}_{i=1} c_i m(C_i)= \int_{E} \psi \> dm.$$ Since $\psi$ was arbitary, we conclude $\int_{A} f \> dm$ is a lower bound for the set $\{\int_{E} \psi \> dm: \psi \> \text{ is simple and} \> f\cdot \chi_{A} \leq \psi \}.$ Thus, $\int_{A} f \> dm \leq \int_{E} f\cdot \chi_{A} \> dm.$