Prove that any function $f:[a,b]\rightarrow\mathbb{R}$ is Riemann integrable if it is bounded and continuous except a finite number of points.
I will use the following criterion of Riemann integrability:
A function $g$ is Riemann integrable on an interval $[a,b]$ if and only if for any $\epsilon>0$ there exist step functions $g_1$, $g_2$ such that
- $g_1(x)\le g(x)\le g_2(x)$
- $ \int_a^b (g_2(x)-g_1(x))dx <\epsilon $.
Let $A=\{d_1,\ldots, d_k\}$ be a set of discontinuities of $f$. Choose one of those points, $d$, and an interval $[\alpha,\beta]$ such that $d\in [\alpha,\beta]\subset [a,b]$, $d\ne \alpha, \beta $ and $A\cap [\alpha, \beta]=d$. If we prove that $f$ is integrable on such $[\alpha,\beta]$, then from additivity it will be integrable on the whole $[a,b]$.
We know that $|f(x)|<M$. Fix $\epsilon>0$. Since $f$ is continuous (and hence integrable) on an interval $[\alpha, d-\frac{\epsilon}{8M}]$, there are some step functions $f_1^{\alpha}, f_2^{\alpha} : [\alpha, d-\frac{\epsilon}{8M}]\rightarrow \mathbb{R}$ such that $f_1^{\alpha}(x)\le f(x) \le f_2^{\alpha} (x)$ for all $x\in[\alpha, d-\frac{\epsilon}{8M}]$. Similarly, since $f$ is continuous on $[d+\frac{\epsilon}{8M}, \beta]$, we choose step functions $f_1^{\beta}, f_2^{\beta} : [d+\frac{\epsilon}{8M}, \beta]\rightarrow \mathbb{R}$ such that $f_1^{\beta}(x)\le f(x) \le f_2^{\beta} (x)$ for all $x\in[d+\frac{\epsilon}{8M}, \beta]$.
Now define
$$f_1(x) = \begin{cases} f_1^{\alpha}(x) &\mbox{if } x\in [\alpha, d-\frac{\epsilon}{8M}] \\ -M &\mbox{if } x\in (d-\frac{\epsilon}{8M},d+\frac{\epsilon}{8M}) \\ f_1^{\beta}(x) &\mbox{if } x\in [d+\frac{\epsilon}{8M}, \beta] \end{cases} $$ and $$f_2(x) = \begin{cases} f_2^{\alpha}(x) &\mbox{if } x\in [\alpha, d-\frac{\epsilon}{8M}] \\ M &\mbox{if } x\in (d-\frac{\epsilon}{8M},d+\frac{\epsilon}{8M}) \\ f_2^{\beta}(x) &\mbox{if } x\in [d+\frac{\epsilon}{8M}, \beta] \end{cases} $$
It is clear that $f_1(x)\le f(x) \le f_2(x)$ for al $x\in[\alpha, \beta]$.
We have
$$\Big| \int_{\alpha}^{\beta} (f_2(x)-f_1(x)) dx\Big| = \Big| \int_{\alpha}^{d-\frac{\epsilon}{8M}} (f_2(x)-f_1(x)) dx + \int_{d-\frac{\epsilon}{8M}}^{d+\frac{\epsilon}{8M}} (f_2(x)-f_1(x)) dx + \int_{d+\frac{\epsilon}{8M}}^{\beta} (f_2(x)-f_1(x)) dx\Big|< \frac{\epsilon}{4}+\frac{\epsilon}{4M}\cdot 2M+\frac{\epsilon}{4}=\epsilon \mbox{.}$$
I would be very happy if somebody verified my "proof".
(Absolute value signs are not necessary here, but I used them for clarity)