The question was:(From Royden "Real Analysis" fourth edition)
Let $f$ be a bounded measurable function on a set of finite measure $E$. For a measurable subset $A$ of $E$, show that $\int_{A} f = \int_{E} f\cdot \chi_{A}.$
My proof was:
Let $f$ be a bounded measurable function on a set of finite measure $E.$ Then by Theorem 4 on page 74, $f$ is integrable over $E.$
Now, by The definition on page 73, $f$ is Lebesgue integrable over $A$ (because $A$ is a measurable subset of $E$ by the assumption of the problem and hence has finite measure) if and only if the following holds: $$\begin{align*}\int_A f &= \sup\{\int_A\varphi : \varphi \text{ is simple and } \varphi \leq f\} \\ &= \inf\{\int_A\psi: \psi \text{ is simple and } f \leq \psi\}.\end{align*}$$
Also, $f\cdot \chi_A$ is integrable if and only if the the following holds: $$\begin{align*}\int_E f \cdot \chi_A &= \sup\{\int_E\varphi : \varphi \text{ is simple and } \varphi \leq f\cdot \chi_A\} \\ &= \inf\{\int_E\psi: \psi \text{ is simple and } f \cdot \chi_A \leq \psi\}.\end{align*}$$
Now, since $\int_A f=\inf\{ \int_A \psi: \psi \text{ is simple and } \psi\geq f \text{ on }A\}$ and $\int_E f\cdot\chi_A = \inf \{ \int_E \phi: \phi \text{ is simple and }\phi\geq f\cdot\chi_A \text{ on }E \}.$
For any given simple function $\psi$ such that $\psi\geq f$ on $A,$ we can extend it so that $\psi=0$ on $E\setminus A$ and this extension is still a simple function.
Therefore, for any $x\in E,$ $$(f \cdot \chi_A)(x) = \begin{cases} f(x) & \text{ if } x\in A \\ 0 & \text{ if } x\in E\setminus A \end{cases} \leq \begin{cases} \psi(x) & \text{ if }x\in A \\ 0 & \text{ if }x\in E\setminus A \end{cases} = \hat{\psi}(x).$$
Now, if $\psi \geq f$ on $A$, then $\psi \cdot \chi_A \geq f \cdot \chi_A$ on $E$ by monotonicity of integration proposition 2 or Theorem 5 and because for simple functions we have $\int_A \psi = \int_E \psi \cdot \chi_A$.
Thus,
$$\int_A \psi = \int_E \psi \cdot \chi_A \geq \inf_{\hat{\psi} \geq f \cdot \chi_A} \int_E\hat{\psi} = \int_E f \cdot \chi_A.$$
Taking the infimum of the LHS, we obtain
$$\int_A f = \inf_{\psi \geq f} \int_A \psi \geq \int_E f \cdot \chi_A.$$
Hence, $\int_A f \geq \int_E f\cdot\chi_A$.
Now,to show that $\int_A f \leq \int_E f \cdot \chi_A$, let $\phi$ be a simple function such that $\phi \leq f$ on $A$. It follows that $\phi \cdot \chi_A \leq f \cdot \chi_A$ on $E$ and
$$\int_A \phi = \int_E \phi \cdot \chi_A \leq \sup_{\hat{\phi} \leq f \cdot \chi_A}\int_E \hat{\phi} = \int_E f \cdot \chi_A.$$
Taking the supremum of the LHS, we obtain
$$\int_A f = \sup_{\phi \leq f} \int_A \phi \leq \int_E f \cdot \chi_A.$$
But there were some comments I received on my solution :
1-Why is $f$ measurable on $A$?
2-Why is $f\cdot \chi_{A}$ measurable?
3- Prove that for simple functions we have $\int_{A} \psi = \int_{E} \psi \cdot \chi_{A}$?
Could anyone help me in answering those comments please?
Note: we are not allowed to use any material from the book after page 79.