I want to prove:

If $A \subset E$, where $E$ is measurable, then $\int_A f = \int_E f \chi_{A}$, where $f$ is a bounded measurable function and $m(E) < \infty$

Solution:

Let $f: E \to \mathbb{R}$ be a bounded measurable function on $E$, where $E$ has finite measure. Let $A \subset E$ to be measurable as well. Notice that $f\chi_A$ is integrable if the the following holds. $$\int_E f\chi_A = \sup\{\int_E\varphi: \varphi\text{ is simple and } \varphi \leq f\chi_A\} = \inf\{\int_E\psi: \psi\text{ is simple and } f\chi_A \leq \psi\},$$ and $f$ is integrable over $A$ if $$\int_A f = \sup\{\int_A\varphi: \varphi\text{ is simple and } \varphi \leq f\} = \inf\{\int_A\psi: \psi\text{ is simple and } f \leq \psi\}.$$ Now, $$f\chi_A = \begin{cases} f & \text{ if } x\in A, \\ 0 & \text{ if } x \in E\sim A, \end{cases}$$ and notice that for each $x \in E\sim A$, we have $f = 0$, but if $f$ is integrable we must have that the supremum and infinimum of simple functions coincide, thus the condition $$ x \in E\sim A \text{ implies that } \varphi \leq 0 \text{ and } 0\leq \psi,$$ and $\varphi = 0 = \psi$ if we want $f\chi_A$ to be integrable over $E$. But this means that we only need to consider the restriction $f|_A$, that is, $$f:A \rightarrow \mathbb{R},$$ and in this case we have $$\int_E f\chi_A = \int_Af\chi_A = \int_A f,$$ as required. To see this, notice that for $f$ restricted to $A$, we have, $$\int_E f\chi_A = \sup\{\int_A\varphi: \varphi\text{ is simple and } \varphi \leq f\chi_A\} = \inf\{\int_A\psi: \psi\text{ is simple and } f\chi_A \leq \psi\},$$ but for every $x \in A$, we have that $\chi_A = 1$, therefore $$\int_E f\chi_A = \sup\{\int_A\varphi: \varphi\text{ is simple and } \varphi \leq f\} = \inf\{\int_A\psi: \psi\text{ is simple and } f \leq \psi\} = \int_A f.$$ Is this solution correct?

  • You have to have that $A$ is measurable as well. You need $f$ to be integrable too. – skyking Nov 16 '15 at 14:47

By definition $\int_E f=\int f\chi_E$. So by definition $$\int_E f\chi_A=\int f\chi_A\chi_E=\int f\chi_A,$$because $\chi_A\chi_E=\chi_A$.

  • I am not understanding your argument. I tried to modify my aproach but not sure if it is correct. – richitesenpai Dec 8 '15 at 23:06
  • What part don't you understand? The definitions or the fact that $\chi_A\chi_E=\chi_A$? – David C. Ullrich Dec 8 '15 at 23:56
  • Yesterday I didn't see what aspect of the argument you could be failing to follow. Today I have a conjecture: Perhaps when I said that $\int_Ef=\int\chi_E f$ by definition you thought I meant that that followed from some definition and you didn't see how? If so: No, what I meant by that is that the equation $\int_Ef=\int\chi_Ef$ is the definition of $\int_E f$. – David C. Ullrich Dec 9 '15 at 14:42

Notice that $$\int_{E} f \chi_{A} \, d\mu = \int_{E \setminus A} f \chi_{A} \, d\mu + \int_{A} f \, d\mu,$$ so it suffices to show that $$\int_{E \setminus A} f \chi_{A} \, d\mu = 0.$$

This can be done rather easily by the definitions.

  • I modified the title, because I am only allowed to use the definition of the integral. I need to use the result in the question to prove the relation solution you give, sorry for the misunderstanding of the title. – richitesenpai Dec 6 '15 at 0:31

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