0
$\begingroup$

The proof I do not understand comes from the text The Elements of Integration by Bartle on page 28. It goes as follows:

Observe that $$\varphi\,\chi_E=\sum_{j=1}^{n}a_j\,\chi_{E_j\cap E}.$$ Hence $$\lambda(E)=\int \varphi\,\chi_E\, d\mu=\sum_{j=1}^{n}a_j\int \chi_{E_j\cap E}=\sum_{j=1}^{n}a_j\, \mu(E_j\cap E)$$ where $\varphi$ is a simple function and $\mu$ is the measure defined in the measure space.


How does the author go from $\int \varphi\,\chi_E\, d\mu$ to $\sum_{j=1}^{n}a_j\int \chi_{E_j\cap E}$? I initially thought one will be able to construct the representative form of the simple function $\varphi\,\chi_E$ and go directly from $\int \varphi\,\chi_E\, d\mu$ to $\sum_{j=1}^{n}a_j\, \mu(E_j\cap E)$, what is the extra step intended for?

$\endgroup$
  • $\begingroup$ As $\varphi$ is a simple function, its general form is nothing but $\sum_{j=1}^na_j\chi_{E_j}$, and then you just use $\chi_A\chi_B=\chi_{A\cap B}$. $\endgroup$ – b00n heT May 31 '16 at 7:32
  • $\begingroup$ @b00nheT Yes, I know that, I just don't get what $\int \chi_{E_j\cap E}$ means. $\endgroup$ – Kun May 31 '16 at 7:37
  • $\begingroup$ By the very definition of Lebesgue integral: $\int\chi_{A}d\mu=\mu(A)$. Is this sufficient? $\endgroup$ – b00n heT May 31 '16 at 7:38
  • $\begingroup$ I don't think it's that important, either you do as you said and leave out the extra step or you use linearity of the integral what he is doing here. $\endgroup$ – YannickSSE May 31 '16 at 7:38
  • 1
    $\begingroup$ @b00nheT Yes, that's it! I missed that point.... Thanks $\endgroup$ – Kun May 31 '16 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.