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I am looking for feedback/corrections on the following solution attempt. I have only typed up one direction of the inequality; if this direction is correct, then I am certain that the other direction is as well. Thank you so much!

$\textbf{Problem:}$ Let $f$ be a bounded measurable function on a set of finite measure $E$. Show that if $A \subset E$ is measurable, then $\int_{A} f \> dm = \int_{E} f\cdot \chi_{A} \> dm$, where $m$ is the Lebesgue measure.

$\textbf{ Solution:}$ By definition, $$\int_{E} f \cdot \chi_{A} \> dm = \inf\{\int_{E} \psi \> dm: \psi \> \text{ is simple and} \> f\cdot \chi_{A} \leq \psi \}.$$ Take any simple function $\psi=\sum^{n}_{i=1} c_i \chi_{C_i}$ defined on $E$ such that $f \cdot \chi_{A} \leq \psi$, where $\sum^{n}_{i=1} c_i \chi_{C_i}$ is the canonical representation of $\psi$. Then, we know $f \leq \psi$ on $A$. Now, observe that $\psi'=\sum^{n}_{i=1} c_{i} \chi_{(C_{i} \cap A)}$ is a simple function on $A$ with $f \leq \psi'$ and, by defintion of $\int_{A} f \> dm$, we get that $$ \int_{A} f \> dm \leq \int_{A} \psi' \> dm = \sum^{n}_{i=1} c_{i} m(C_{i} \cap A) \leq \sum^{n}_{i=1} c_i m(C_i)= \int_{E} \psi \> dm.$$ Since $\psi$ was arbitary, we conclude $\int_{A} f \> dm$ is a lower bound for the set $\{\int_{E} \psi \> dm: \psi \> \text{ is simple and} \> f\cdot \chi_{A} \leq \psi \}.$ Thus, $\int_{A} f \> dm \leq \int_{E} f\cdot \chi_{A} \> dm.$

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  • $\begingroup$ In those inequalities in the third from last line, I originally noted that $\psi \geq 0$ on $E \backslash A$ since $f \cdot \chi_{A} =0$ on $E \backslash A$ and $f \cdot \chi_{A} \leq \psi.$ $\endgroup$ – dgc1240 Dec 7 '13 at 22:18
  • $\begingroup$ Hmmm... Then, I am quite puzzled. I got this problem and definition out of Royden 4th edition (section $4.2$). It makes no mention of $f$ being positive. Another definition, for nonnegative measurable functions, is introduced in the next section. $\endgroup$ – dgc1240 Dec 7 '13 at 22:27
  • $\begingroup$ No, sorry, for bounded functions with finite-measure support, I think your definition works as well (although, as you understand, I haven't seen it being used as such before). $\endgroup$ – Jonathan Y. Dec 7 '13 at 22:34
  • $\begingroup$ Indeed, the only definition I have at my disposal, at this stage of development, is that a bounded measurable function $f$ is Lebesgue integrable if its upper and lower Lebesgue integrals agree. For the direction of my solution above, I used the upper Lebesgue integral to prove my desired result; I use the lower Lebesgue integral to prove the reverse inequality. $\endgroup$ – dgc1240 Dec 7 '13 at 22:41
  • $\begingroup$ I'm lost; upper and lower Lebesgue integrals? Used for bounded measurable functions with finite-measure support? Every such function should be $L^1$ $\endgroup$ – Jonathan Y. Dec 7 '13 at 22:46

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