How to prove that ##f## is integrable given that ##g## is integrable?

[pasmith]

There is a more elegant proof.

Notation-wise, you will confuse the reader and yourself if you have $x_i$ being both the points at which $f(x) \neq g(x)$ and the points of an arbitrary partition. So I will call the former $\{\xi_k : k = 1, \dots, n\}$.

We can write $$f(x) = g(x) + \sum_{k=1}^n (f(\xi_k) - g(\xi_k)) h_k(x)$$ where $$h_k : [a,b] \to \mathbb{R} : x \mapsto \begin{cases} 0 & x \neq \xi_k \\ 1 & x = \xi_k.\end{cases}$$ It suffices to show that each $h_k$ is (Darboux) integrable, since a linear combination of finitely many (Darboux) integrable functions is (Darboux) integrable. (The one with the upper and lower sums being arbitrarily close to each other is Darboux integrability; it is however completely equivalent to Riemann integrability.)

Let $\epsilon > 0$ and $D = \{x_0 = a, \dots, x_m = b\}$ a partition of $[a.b]$ with $\|D\| = \max\{\delta_i = x_i - x_{i-1}, i = 1, \dots, m\} < \frac12 \epsilon$. Then there exists a minimal $1 \leq i_1\leq m$ such that $x_{i_1-1} \leq \xi_k \leq x_{i_1}.$ We then have that $M_i - m_i = 0$ for every $i$ except for $i_1$ and, if $x_{i_1} = \xi_k$ and $i_1 < m$, also $i_1+1$. For these intervals $M_i - m_i = 1$. Thus $(M_i - m_i)\delta_i = 0$ except for at most two intervals for which $(M_i - m_i)\delta_i = \delta_i \leq \|D\|$. Thus $$U(h_k,D) - L(h_k,D) \leq 2\|D\| < \epsilon.$$