Theorem 0.1
(a) (Linearity) Let \(f,g\) be Riemann integrable functions on \([a,b]\). Then
$$
\begin{align*}
\int_a^b (f(x) + g(x))\ dx &= \int_a^b f(x)dx + \int_a^b g(x) \ dx, \quad \text{ and } \\
\int_a^b \alpha f(x) \ dx &= \alpha \int_a^b f(x) \ dx
\end{align*}
$$
(b) (Disjoint Additivity)
$$
\int_a^b f(x) \ dx = \int_a^c f(x)dx + \int_c^b f(x) \ dx \\
$$
Proof
Let \(\epsilon > 0\). Since \(f\) and \(g\) are Riemann integrable, then by definition, there exists a partition \(\mathcal{P}_{\epsilon}\) such that for any partition \(\mathcal{P} \supset \mathcal{P}_{\epsilon}\) and for any stags \(t_j\),
$$
\left| S(f,\mathcal{P},t_j) - \int_a^b f(x)dx \right| < \frac{\epsilon}{2} \quad \text{ and } \quad \left| S(g,\mathcal{P},t_j) - \int_a^b g(x)dx \right| < \frac{\epsilon}{2}
$$
By the triangle inequality,
$$
\begin{align*}
\left| S(f,\mathcal{P},t_j) - \int_a^b f(x)dx + S(g,\mathcal{P},t_j) - \int_a^b g(x)dx \right| &\leq \left| S(f,\mathcal{P},t_j) - \int_a^b f(x)dx \right| + \left| S(g,\mathcal{P},t_j) - \int_a^b g(x)dx \right| \\
\left| S(f+g,\mathcal{P},t_j) - \left( \int_a^b f(x)dx + \int_a^b g(x)dx\right) \right| &\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}
$$
Hence the Riemann sum \(S(f+g,\mathcal{P},t_j)\) can be made arbitrary close to \(\int_a^b f(x)dx + \int_a^b g(x)dx\). Hence by the definition of Riemann integral, \(\int_a^b f(x) + g(x)dx = \int_a^b f(x)dx + \int_a^b g(x)dx\)
(b) We first need to show that \(f\) is Reimann integrable on \([a,c]\). Since \(f\) is Riemann integrable on \([a,b]\). Then, let \(\epsilon > 0\). There exists partition \(\mathcal{P}\) of \([a,b]\) where
$$
\begin{align*}
\mathcal{P} = \{a, x_0, x_1, \ldots, x_k, b\}
\end{align*}
$$
such that
$$
\begin{align*}
U(f,\mathcal{P}) - L(f, \mathcal{P}) < \epsilon \quad\quad\quad (1)
\end{align*}
$$
Let \(\mathcal{P}' = \mathcal{P} \cup \{c\}\) where
$$
\begin{align*}
\mathcal{P}' = \{a, x_0, x_1, \ldots, c, \ldots, x_k, b\}
\end{align*}
$$
Hence, \(\mathcal{P}'\) is a refinement of \(\mathcal{P}\) by adding the single point \(c\). Since it’s a refinement, then
$$
\begin{align}
U(f,\mathcal{P}') \leq U(f,\mathcal{P}) \quad \text{ and } \quad L(f,\mathcal{P}') \geq L(f, \mathcal{P})
\end{align}
$$
Hence
$$
\begin{align}
U(f,\mathcal{P}') - L(f,\mathcal{P}') \leq U(f,\mathcal{P}) - L(f, \mathcal{P}) \quad\quad\quad (2)
\end{align}
$$
Now, let \(\mathcal{P}_1 = \mathcal{P}' \cap [a,c]\). Note that \(\mathcal{P}_1\) contains all the partition points in \(\mathcal{P}'\) that lie exactly in \([a,c]\). Then
$$
\begin{align*}
\mathcal{P}_1 = \{a, x_0, x_1, \ldots, c\}
\end{align*}
$$
Since we \(\mathcal{P}_1\) only holds a subset of the points in the bigger partition, then
$$
\begin{align}
U(f,\mathcal{P}_1) - L(f,\mathcal{P}_1) \leq U(f,\mathcal{P}') - L(f, \mathcal{P}') \quad\quad\quad (3)
\end{align}
$$
If we combine (1),(2),(3) we get
$$
\begin{align}
U(f,\mathcal{P}_1) - L(f,\mathcal{P}_1) \leq \epsilon
\end{align}
$$
Hence, \(f\) is Riemann integrable on \([a,c]\). With a similar argument, we can also show that \(f\) is integrable on \([b,c]\).
Next, we prove the identity in the claim. Hence let \(\mathcal{P}\) be any partition of \([a,b]\). Then define
$$
\begin{align*}
\mathcal{P}_0 = \mathcal{P} \cup \{c\}, \mathcal{P}_1 = \mathcal{P}_0 \cap [a,c], \mathcal{P}_2 = \mathcal{P}_0 \cap [c,b]
\end{align*}
$$
Then
Then \(\mathcal{P}_0\) is a refinement of \(\mathcal{P}\), so
$$
\begin{align}
U(f,\mathcal{P}) \;\ge\; U(f,\mathcal{P}_0)
= U(f,\mathcal{P}_1) + U(f,\mathcal{P}_2).
\end{align}
$$
For any partition \(\mathcal{Q}\) of \([a,c]\) we have \(U(f,\mathcal{Q}) \ge \int_a^c f(x)\,dx\), and
for any partition \(\mathcal{R}\) of \([c,b]\) we have \(U(f,\mathcal{R}) \ge \int_c^b f(x)\,dx\). In
particular,
$$
\begin{align}
U(f,\mathcal{P}_1) + U(f,\mathcal{P}_2)
\ge \int_a^c f(x) \ dx + \int_c^b f(x) \ dx,
\end{align}
$$
and hence
$$
\begin{align}
U(f,\mathcal{P}) \ge \int_a^c f(x) \ dx + \int_c^b f(x) \ dx.
\end{align}
$$
So the upper sum of any partition is greater than the two integrals. Now, recall by definition that the upper Darboux integral is defined to be the infimum over all upper sums of all partitions
$$
\begin{align}
(U) \int_a^b f(x)dx &:= \inf\{U(f,\mathcal{P}): \mathcal{P} \text{ is any partition of } [a,b]\} \\
\end{align}
$$
hence, this upper sum is also greater than the two integrals
$$
\begin{align}
(U)\int_a^b f(x)\,dx \ge \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.
\end{align}
$$
But \(f\) is Riemann integrable, then \((U)\int_a^b f = \int_a^b f\), so
$$
\begin{align}
\int_a^b f(x)\,dx \ge \int_a^c f(x)\,dx + \int_c^b f(x)\,dx. \tag{4}
\end{align}
$$
Now consider the lower Darboux sum,
$$
\begin{align}
L(f,\mathcal{P}) \leq L(f,\mathcal{P}_0) = L(f,\mathcal{P}_1) + L(f,\mathcal{P}_2) \leq \int_a^c f(x)dx + \int_c^b f(x)dx
\end{align}
$$
By definition, the lower Darboux integral is defined by taking the supremum over all partitions and using
$$
\begin{align}
(L) \int_a^b f(x)dx &:= \sup\{L(f,\mathcal{P}): \mathcal{P} \text{ is any partition of } [a,b]\}
\end{align}
$$
This lower sum is must still be lower than both integrals
$$
\begin{align}
(L)\int_a^b f(x)\,dx \le \int_a^c f(x)\ dx + \int_c^b f(x) \ dx.
\end{align}
$$
And again since f is Riemann integrable, $$(L)\int_a^b f = \int_a^b f$$. Hence
$$
\begin{align}
\int_a^b f(x)\,dx \le \int_a^c f(x)\,dx + \int_c^b f(x) \ dx \tag{5}
\end{align}
$$
Combining (4) and (5)
$$
\begin{align}
\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx
\end{align}
$$
References