Lemma 0.9
A function is Darboux integrable if and only if it is Reimann integrable. Moreover, \(I(f) = \int_a^b f(x)dx\).
Proof
Suppose that \(f\) is Darboux integrable. Let \(\epsilon > 0\) be given. By definition, there exists a partition
$$
\mathcal{P}_{\epsilon} = \{x_0,x_1,\ldots,x_n\}
$$
such that
$$
U(f,\mathcal{P}_{\epsilon}) - L(f,\mathcal{P}_{\epsilon}) < \epsilon \quad \quad \quad (1)
$$
Furthermore, by this Lemma, then
$$
\begin{align*}
(U) \int_a^b f(x)dx = (L) \int_a^b f(x)dx
\end{align*}
$$
where by definition,
$$
\begin{align}
(U) \int_a^b f(x)dx &= \inf\{U(f, \mathcal{P}): \mathcal{P} \text{ is any partition of } [a,b]\} \\
(L) \int_a^b f(x)dx &= \sup\{L(f,\mathcal{P}): \mathcal{P} \text{ is any partition of } [a,b]\}
\end{align}
$$
Now, Define \(I\) to be
$$
\begin{align}
I(f) = \int_a^b f(x)dx := (U) \int_a^b f(x)dx = (L) \int_a^b f(x)dx.
\end{align}
$$
Hence, \(I(f)\) is the infimum of all upper sums for every partition \(\mathcal{P}\). That is
$$
\begin{align}
I(f) \leq U(f,\mathcal{P})
\end{align}
$$
At the same time, \(I\) is the supremum of all lower sums for every partition \(\mathcal{P}\). That is
$$
\begin{align}
I(f) \geq L(f,\mathcal{P})
\end{align}
$$
Therefore,
$$
\begin{align}
L(f,\mathcal{P}) \leq I(f) \leq U(f,\mathcal{P})
\end{align}
$$
for any partition and in particular, it is true for \(\mathcal{P}_{\epsilon}\)
$$
L(f,\mathcal{P}_{\epsilon}) \leq I(f) \leq U(f,\mathcal{P}_{\epsilon})
$$
But recall by(1) that \(U(f,\mathcal{P}_\epsilon) - L(f,\mathcal{P}_\epsilon) < \epsilon\) since \(f\) is Darboux integrable. Hence we get
Hence,
$$
U(f,\mathcal{P}_{\epsilon}) < I(f) + \epsilon
$$
and
$$
L(f,\mathcal{P}_{\epsilon}) > I(f) - \epsilon
$$
So now given partition \(\mathcal{P} \supset \mathcal{P}_{\epsilon}\), we know that the Riemann sum with respect to some tagged partition must satisfy
$$
S(f,\mathcal{P}, t_j) \leq U(f,\mathcal{P}) \quad\quad\quad (2)
$$
since by defintion, \(U(f,\mathcal{P})\) is the upper Riemann sum for \(\mathcal{P}\). Moreover, since \(\mathcal{P}\) is a refinement over \(\mathcal{P}\), then
$$
U(f,\mathcal{P}) \leq U(f,\mathcal{P}_{\epsilon}) \quad\quad\quad (3)
$$
Combining (2) and (3)
$$
S(f,\mathcal{P},t_j) \leq U(f,\mathcal{P}) \leq U(f,\mathcal{P}_{\epsilon}) \leq I(f) + \epsilon
$$
Similarly
$$
I(f) - \epsilon \leq L(f,\mathcal{P}_{\epsilon}) \leq L(f,\mathcal{P}) \leq S(f,\mathcal{P},t_j)
$$
Hence
$$
|S(f,\mathcal{P},t_j) - I(f) | \leq \epsilon
$$
Now suppose that \(f\) is Riemann integrable. Then by definition, for all \(\epsilon > 0\), we can find a partition \(P_{\epsilon}\) such that if \(\mathcal{P} \supset \mathcal{P}_{\epsilon}\), then for any choice of tags
$$
|S(f,\mathcal{P},t_j) - I(f) | \leq \epsilon
$$
Fix one such partition. We showed in this lemma, that if \(f\) is Riemann integrable, then \(f\) must be bounded. Therefore, we can find tags \(t_j\) and \(u_j\) such that \(t_j, u_j \in [x_{j-1}, x_j]\) and
$$
f(t_j) \geq M_j(f) - \frac{\epsilon}{2} \quad \text{ and } \quad f(u_j) \leq m_j(f) + \frac{\epsilon}{2}
$$
Hence
$$
\begin{align*}
U(f,\mathcal{p}) - L(f,\mathcal{p}) &= \sum_{j=1}^n (M_j(f) - m_j(f))(x_j - x_{j-1}) \\
&\leq \sum_{j=1}^n (f(u_j) + f(t_j) + \epsilon)(x_j - x_{j-1}) \\
&\leq \left| \sum_{j=1}^n f(u_j)(x_j - x_{j-1}) - I(f) \right| + \left| \sum_{j=1}^n f(t_j)(x_j - x_{j-1}) - I(f) \right| \\
&< \epsilon + \epsilon + (b - a)\epsilon = (2 + b-a)\epsilon
\end{align*}
$$
References