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.
Best Answer
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?