Theorem 4.3.1 (Criteron for integrability). A function $f: \R^n \to \R$ is integrable if and only if it is bounded with bounded support, and for all $\epsilon > 0$, there exists $N$ such that
$$ O_{N, \epsilon} = \sum_{\{C\in \mathcal{D}_N \mid \text{osc}_C(f) > \epsilon\}} \text{vol}_nC < \epsilon. $$
Proof. Say $f$ has tame oscillation when the following is true.
Notice that
$$ U_N - L_N = \sum_{\{C\in \mathcal{D}_N \}} (M_C(f) - m_C(f))\text{vol}_nC \\
\epsilon \cdot \sum_{\{C\in \mathcal{D}_N \mid \text{osc}_C(f) > \epsilon\}} \text{vol}_nC . $$
If $f$ does not have tame oscillation, then for some $\epsilon_0 > 0$, $O_{N, \epsilon_0} \geq \epsilon_0$ for all $N$.Therefore, $U_N - L_N \geq \epsilon^2$ for all $N$ and $f$ is not integrable.
Now suppose $f$ has tame oscillation. We must show that $U(f) = L(f)$. Let $\delta > 0$. We want to show that for large $N$, $U_N(f) - L_N(f) < \delta$. For any $\epsilon > 0$,
$$ U_N(f) - L_N(f) = \sum_{C\in \mathcal{D}N} osc_C(f)vol(c)\\ \leq \sum{\{C\in \mathcal{D}N \mid \text{osc}C(f) > \epsilon\}}2\sup|f| \cdot vol(C) + \sum{\{C \mid osc_C(f) \leq \epsilon\}\text{and} C \text{ is bounded by support}}\epsilon \cdot vol(C)\\ \leq 2\sup|f|O{N, \epsilon} + \epsilon R^n . $$
Setting $\epsilon = \frac{\delta}{2\sup|f| + R^n}$, then
$$ U_N(f) - L_N(f) \le 2 \sup |f| \epsilon + \epsilon R^n = \epsilon(2\sup|f| + R^n) = \delta. $$
Example 4.3.3 (Nonintegrable function). The function that is 1 on rationals and 0 elsewhere is not integrable. This is because no matter how small of a cube you take, it will contain both rationals and irrationals. Therefore, $osc = 1$ and enver decreases.
Proposition 4.3.4 (Bounded support of graph of integrable function has volume 0). Let $f: \R^n \to \R$ be an integrable function with graph $\Gamma(f)$, and let $C_0 \subset \R^n$ be any dyadic cube*.* Then
$$ \text{vol}_{n+1}(\Gamma(f) \cap (C_0\times \R)) = 0. $$
This intuitively makes sense because a curve in the plane has area 0. A surface in 3-space has volume 0.