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.

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$f\chi_A =f$ on $A$, and $f\chi_A = 0$ on $E\setminus A$.

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Does the first part make sense $\int_E f \chi_A$? At first, I wanted to say $\int_E f \chi_E$, but I didn't think that made much sense. – emka Nov 12 '12 at 6:13
The idea is good. $\int_E f\chi_A$ makes sense. – aid Nov 12 '12 at 6:14
I guess, to put my question succinctly, why? My group and I arrived to that by consensus and comfort, but we don't really understand it. – emka Nov 12 '12 at 6:18
(1) The hypotheses of the lemma apply. (2) Applying the lemma how you did and then simplifying gives the desired result. ------- What part is unclear? – aid Nov 12 '12 at 6:22

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, $\int_E f\cdot\chi_A= \int_A f$.\

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