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Problem 28: Let $f$ be integrable over $E$ and $C$ a measurable subset of $E$. Show that $\int_C f = \int_E f \cdot \chi_C$.

This problem comes from section 4.4 in Royden and Fitzpatrick's Real Analysis. Seeking verification of my own proof, I came across this hint:

HINT: For $h$ be a bounded measurable function of finite support $E_0$ and $0 \le h \le f^+ \cdot \chi_C$ on $E$, prove that $\int_E h = \int_C h$ and use this to show $\int_E f^+ \cdot \chi_C \le \int_C f^+$. For $h$ be a bounded measurable function of finite support $E_0$ and $0 \le h \le f^+$ on $C$, extend $f$ from $C$ to $h'$ on $E$ and use this extension to show that $\int_E f^+ \cdot \chi_C \ge \int_C f^+$.

This hint suggests a solution that is relatively more complicated than the one I came up with, which causes me to wonder whether I've made a mistake somewhere.

In section 4.3., we developed all the tools needed to compute integrals of nonnegative functions over arbitrary measurable subsets of $\Bbb{R}$. In section 4.4., the integral of $f$ over $E$ is defined as $\int_E f = \int_E f_+ - \int_E f_-$, where $f_+$ and $f_-$ are nonnegative functions over $E$ and therefore their integrals are computed using the tools of 4.3. Hence, it seems the above problem is solved like so:

$$\int_C f = \int_C f_+ - \int_C f_- = \int_E f_+ \cdot \chi_C - \int_E f_- \cdot \chi_C = \int_E f \cdot \chi_C$$

So, what mistake am I making?

EDIT (there's still a gap!):

Okay. Here are some results and definitions I have available to me:

Problem 10: Let $f$ be a bounded measurable function on a set of finite measure $E$. For a measurable subset $A$ of $E$, show that $\int_A f = \int_E f \cdot \chi_A$.

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Definition: For $f$ a nonnegative measurable function on $E$, we define the integral of $f$ over $E$ by $$\int_E f = \sup \{ \int_E h \mid h \mbox{ bounded, measurable, of finite support and } 0 \le h \le f \mbox{ on } E \}$$

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Theorem 11 (Additivity Over Domains of Integration) Let $f$ be a nonnegative measurable function on $E$. If $A$ and $B$ are disjoint measurable subsets of $E$, then $$\int_{A \cup B} = \int_A f + \int_B f$$

Obviously theorem 11 can extended to countable collections of disjoint measurable sets (I've already done this)

To prove the problem 28, I am going to show that it suffices to consider $f$ nonnegative $E$ with $m(E) < \infty$. Clearly it suffices to consider $f$ nonnegative, given what I wrote about before I edited this post. Let $n \in \Bbb{Z}$ and define $E_n = E \cap [n,n+1)$. Then $m(E_n) < \infty$ and $\{E_n\}$ forms a pairwise disjoint collection of measurable sets such that $E = \bigcup_{n \in \Bbb{Z}} E_n$. Then, using problem 10, using theorem 11, and assuming 28 holds for $f$ nonnegative over sets of finite measure, we get

\begin{align} \int_C f &= \int_{\bigcup_{n \in \Bbb{Z}} (E_n \cap C)} f \\ &= \sum_{n \in \Bbb{Z}} \int_{E_n \cap C} f \\ &= \sum_{n \in \Bbb{Z}} \int_{E_n} f \cdot \chi_C \\ &= \sum_{n \in \Bbb{Z}} \int_{E_n} f \cdot \chi_C \\ &= \int_E f \cdot \chi_C \\ \end{align}

...Nope...Doesn't work...In the above calculation, I am presupposing that $f$ is bounded over $E_n$...I don't know how to solve this problem.

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  • $\begingroup$ Try to prove the result when $f$ is a characteristic function, and work your way up from there: $$\text{characteristic} \rightarrow \text{simple} \rightarrow \text{bounded nonnegative} \rightarrow \text{nonnegative} \rightarrow \text{general}$$ This procedure is useful to prove lots of things. $\endgroup$ – Umberto P. Mar 24 '18 at 11:59
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You seem to be begging the question. In your solution you write (in part) $$ \int_C f_+ = \int_E f_+ \cdot \chi_C$$ which is what you are trying to prove in the first place.

The difference is that $f_+$ is nonnegative. Did you already prove the result for nonnegative functions?

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  • $\begingroup$ I don't see how I'm begging the question. The first equality is just the definition of the integral over $C$ found in 4.4; the second equality is the result of using the tools in 4.3.; and the final equality is just using the definition in 4.4. again. $\endgroup$ – user193319 Mar 22 '18 at 23:05
  • $\begingroup$ If that is the definition, then you are fine. There is a slight issue that if $(X,\mu)$ is a measure space and $C$ is a measurable set, then $(C,\mu)$ is also a measure space (with the appropriate $\sigma$-algebra) and then $\int_C f$ has its usual definition. The point of the hint is to show these integrals are the same. $\endgroup$ – Umberto P. Mar 22 '18 at 23:08
  • $\begingroup$ Hmm...Now that I read section 4.3. more carefully, I doesn't seem that $\int_C f = \int_E f \cdot \chi_C$ has been proven for nonnegative functions. But given how the integral is defined in section 4.4., it obviously suffices to prove it for nonnegative functions. But then this is precisely what the hint in my OP is hinting at, right? $\endgroup$ – user193319 Mar 22 '18 at 23:09
  • $\begingroup$ I could use some help...See my edit. $\endgroup$ – user193319 Mar 23 '18 at 20:51
  • $\begingroup$ kudos for correct usage of "begging the question" $\endgroup$ – zhw. Mar 23 '18 at 22:34

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