here is a solution solved by my good friend Zeming Bi. I think it is more accurate.
p.s. any theorems refered are from Royden 4th edition chapter 4.
Since$f$ is a bounded measurable function on a set of finite measure $E$, by Theorem 4, $f$ is integrable over $E$. Since Lebesgue intergal is equal to the upper Lebesgue integral, $\int_E f\cdot\chi_A=\inf\{\int_E \psi\ |\psi \text{ simple and}\ \psi \geq f\cdot\chi_A \text{ on}\ E\}.$
Let $\psi$ be any simple function for which $\psi \geq f\cdot \chi_{A}$ on $E$. Then, $\psi \geq f $ on $A$ and $\psi \geq 0$ on $E\setminus A$. \
We show that $\int_E \psi \geq \int_A \psi$. \
Since $\psi$ is a simple function, we may choose a finite disjoint collection $\{E_i\}_{i=1}^n$ of measurable subsets of $E$. For each $i$, $1\leq i\leq n$, let $a_i$ be the values taken by $\psi$. Thus,
\begin{align*}
\psi&=\mathop{\sum}\limits_{i=1}^{n}a_i\cdot\chi_{E_i}\\
&=\mathop{\sum}\limits_{i=1}^{n}a_i\cdot\chi_{E_i\cap A}+\mathop{\sum}\limits_{i=1}^{n}a_i\cdot\chi_{E_i\cap A^c}\\
&=\mathop{\sum}\limits_{i=1}^{n}a_i\cdot\chi_{E_i\cap A}+\mathop{\sum}\limits_{a_i\geq 0}a_i\cdot\chi_{E_i\cap A^c}+\mathop{\sum}\limits_{a_i<0}a_i\cdot\chi_{E_i\cap A^c}
\end{align*}
Since $\psi \geq 0$ on $E\setminus A$, then for each $i$ such that $a_i<0$, $E_i\cap A^c=\emptyset$. Thus,
\begin{align*}
\psi &=\mathop{\sum}\limits_{i=1}^{n}a_i\cdot\chi_{E_i}\\
&=\mathop{\sum}\limits_{i=1}^{n}a_i\cdot\chi_{E_i\cap A}+\mathop{\sum}\limits_{a_i\geq 0}a_i\cdot\chi_{E_i\cap A^c}\\
\end{align*}
Therefore,
\begin{align*}
\int_E \psi&=\mathop{\sum}\limits_{i=1}^{n}a_im(E_i\cap A) +\mathop{\sum}\limits_{a_i\geq 0}a_im(E_i\cap A^c)\\
&\geq \mathop{\sum}\limits_{i=1}^{n}a_im(E_i\cap A)\\
&=\int_A \psi
\end{align*}
Then we have
\begin{align*}
\int_E f\cdot \chi_A &=\inf\{\int_E \psi\ |\psi \text{ simple and}\ \psi \geq f\cdot\chi_A \text{ on}\ E\}\\
&\geq \inf\{\int_A \psi\ |\psi \text{ simple and}\ \psi \geq f\cdot\chi_A \text{ on}\ E\}\\
&\geq \inf\{\int_A \psi\ |\psi \text{ simple and}\ \psi \geq f \text{ on}\ A\}\\
&=\int_A f.
\end{align*}
Thus, $\int_E f\cdot\chi_A\geq \int_A f$.\
Since Lebesgue intergal is equal to the lower Lebesgue integral, $\int_E f\cdot\chi_A=\sup \{\int_E \varphi\ |\varphi \text{ simple and}\ \varphi \leq f\cdot\chi_A \text{ on}\ E\}.$
Let $\varphi$ be any simple function for which $\varphi \leq f\cdot \chi_{A}$ on $E$. Then, $\varphi \leq f $ on $A$ and $\varphi \leq 0$ on $E\setminus A$. Similarly, we can prove that $\int_E \varphi\leq \int_A \varphi$. Then we have
\begin{align*}
\int_E f\cdot \chi_A &=\sup\{\int_E \varphi\ |\varphi \text{ simple and}\ \varphi \leq f\cdot\chi_A \text{ on}\ E\}\\
&\leq \sup \{\int_A \varphi\ |\varphi \text{ simple and}\ \varphi \leq f\cdot\chi_A \text{ on}\ E\}\\
&\leq \sup\{\int_A \varphi\ |\varphi \text{ simple and}\ \varphi \leq f \text{ on}\ A\}\\
&=\int_A f.
\end{align*}
Thus, $\int_E f\cdot\chi_A\leq \int_A f$.\
Therefore, $\int_E f\cdot\chi_A= \int_A f$.\