Restriction of a Lebesgue integral to a subset of a measurable set. Let $f$ be a bounded measurable function on a set of finite measure $E$.  For a measurable subset $A$ of $E$, show that $\int_A f = \int_E f \cdot \chi_A$

Lemma: Let $f$ be a bounded measurable function on a set of finite measure $E$.  Suppose $A$ and $B$ are disjoint measurable subsets of $E$.  Then $\int_{A \cup B} f = \int_A f + \int_B f$.

My first approach was to view the problem like this utilizing the above lemma:
$$\int_E  f \cdot \chi_A = \int_{E\setminus A} f \cdot X_A + \int_A f \cdot \chi_A $$
I'm not really sure how the result follows from the above.
 A: $f\chi_A =f$ on $A$, and $f\chi_A = 0$ on $E\setminus A$.
A: 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, $\int_E f\cdot\chi_A= \int_A f$.
A: Denote,
A = $ \{\int_E \phi$ | $\phi \text{ is simple and } \phi \leq f.\chi_{A}\}$
B = $ \{\int_E \psi$ | $\psi \text{ is simple and }  f.\chi_{A} \leq \psi\}$
THEN
sup A = $\int_E f.\chi_A$ = inf B.
Hint-: For every pair of simple functions $\phi $ and $\psi$ on A such that, $\phi \leq f \leq \psi$ on A, show that $\int_A \phi$ and $\int_A \psi$ belong to A and B respectively.
