Prove that $\int_A f = \int_E f\chi_A$ using definition of Lebesgue integral The following question is taken from Royden's Real Analysis $4$th edition, Chapter $4,$ exercise $10,$ page $79.$

Question: 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 attempt: 
Since $f\cdot \chi_A$ is bounded measurable on a set of finite measure $E,$ therefore it is integrable.
    Recall that 
    $$\int_A f=\inf\left\{ \int_A \psi: \psi \text{ is simple and }\psi\geq f \text{ on }A \right\}$$
    and 
    $$\int_E f\cdot\chi_A = \inf\left\{ \int_E \phi: \phi\text{ is simple and }\phi\geq f\cdot\chi_A \text{ on }E \right\}.$$
    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 $x\in E,$
    $$(f\cdot\chi_A)(x) = \begin{cases} 
      f(x) & \text{ if }x\in A \\
      0 & \text{ if }x\in E\setminus A 
   \end{cases} \leq \begin{cases} 
      \psi(x) & \text{ if }x\in A \\
      0 & \text{ if }x\in E\setminus A
   \end{cases} = \psi(x).$$
   Hence, $\int_A f \geq \int_E f\cdot\chi_A$.
   Similarly, let $\phi$ be a simple function such that $\phi\geq f\cdot\chi_A$ on $E.$
   Then for any $x\in A,$
   $$f(x) = (f\cdot\chi_A)(x)\leq \phi(x).$$
   Therefore, $\int_A  f \leq \int_E f\cdot\chi_A$ and hence $\int_A  f = \int_E f\cdot\chi_A.$
Is my proof correct?
 A: Your argument that $\int_A f \geqslant \int_E f \cdot \chi_A $ is essentially correct.  
To fill in some necessary detail, if $\psi \geqslant f$ on $A$, then $\psi \cdot \chi_A \geqslant f \cdot \chi_A$ on $E$ and for simple functions we have $\int_A \psi = \int_E \psi \cdot \chi_A$.
Thus,  
$$\int_A \psi = \int_E \psi \cdot \chi_A \geqslant \inf_{\hat{\psi} \geqslant f \cdot \chi_A} \int_E\hat{\psi} = \int_E f \cdot \chi_A.$$
Taking the infimum of the LHS, we obtain
$$\int_A f = \inf_{\psi \geqslant f} \int_A \psi \geqslant \int_E f \cdot \chi_A.$$
To show that $\int_A f \leqslant \int_E f \cdot \chi_A$, let $\phi$ be a simple function such that $\phi \leqslant f$ on $A$. It follows that $\phi \cdot \chi_A \leqslant f \cdot \chi_A$ on $E$ and
$$\int_A \phi = \int_E \phi \cdot \chi_A \leqslant \sup_{\hat{\phi} \leqslant f \cdot \chi_A}\int_E \hat{\phi} = \int_E f \cdot \chi_A.$$
Taking the supremum of the LHS, we obtain
$$\int_A f = \sup_{\phi \leqslant f} \int_A \phi \leqslant \int_E f \cdot \chi_A.$$
