5-1

Question: If f is a simple measurable function (not necessarily nonnegative) taking values a_{j} on E_{j}, j= 1,2,3...N. Show that \int _{E}f = \sum_{j=1}^{N} a_{j} |E_{j}|.(Use Theorem 5.24. )
Proof:
Theorem 5.24: If \int _{E}f exists and E =\bigcup_{k}^{} E_{k} is the countable union of disjoint measurable sets E_{k}, then \int _{E}f = \sum_{k}^{} \int _{E_{k}}f .
Since f=\sum_{k=1}^{N} a_{k} \chi_{E_{k}} is a simple function, f is measurable and its Lebesgue integral exists. From theorem 5.24, we could write the Lebesgue integral in this way:
\begin{aligned}
\int _{E}f &= \sum_{k}^{} \int _{E_{k}}f\\
&=\sum_{k}^{} \int _{E_{k}}\sum_{i=1}^{N} a_{i} \chi_{E_{i}}\\
&=\sum_{k}^{} \sum_{i=1}^{N} a_{i} \int _{E_{k}}\chi_{E_{i}}
\end{aligned}
When k \ne i, \int _{E_{k}}\chi_{E_{i}} = \int _{E_{k}} 0 =0; When k=i, \int _{E_{k}}\chi_{E_{i}} = \int _{E_{k}} 1 = |E_{k}|. 
Hence, \int _{E}f=\sum_{k}^{} \sum_{i=1}^{N} a_{i} \int _{E_{k}}\chi_{E_{i}}= \sum_{j=1}^{N} a_{j} |E_{j}|.