Proposition

If $C$ is a bounded set of measure 0 and $\int_A \chi_C$ exists, show that $\int_A \chi_C = 0$.

Solution

Let $P$ be a partition. Suppose $L(f, P) \ne 0$. Since $\chi_C(x)$ is either 0 or 1 for any $x \in A$, $m_S$ is 0 or 1 for any subrectangle $S \in P$. If $m_S = 0$ for all subrectangles, $L(f, P) = 0$. Thus there must exist a subrectangle $S_0 \in P$ such that $m_{S_0} = 1$. In other words, $\forall x \in S_0, \chi_C(x) = 1$, so $S_0 \subset C$.

As shown here, $S_0$ does not have content 0. Since $S_0$ is compact and does not have content 0, it does not have measure 0 by Theorem 3-6. Then $C$, which contains $S_0$, does not have measure 0. This is a contradiction, so $L(f, P) = 0$.

We showed that $L(f, P) = 0$ for all partitions. Thus $\sup\{ L(f, P) \} = 0$. We know that $\int_A \chi_C$ exists, and it must be equal to $\sup\{ L(f, P) \}$. Therefore, $\int_A \chi_C = 0$.