I've found some proofs (on StackExchange and elsewhere) for the statement: "A function with finitely many points of discontinuity is Riemann integrable" but I've found all too much complicated as compared to the one I've come up with. Then I want to verify if what I've done is wrong or right. Can anyone verify?
Proof:
Let $f \in \mathbb{R}^{[a,b]}$ be a bounded function, i.e. |f(x)| < M.
Suppose $A=\{a_1, ..., a_k\}$ is the set of $k$ points of discontinuity of $f$. Let $\textbf{b}$ be a dissection (partition) such that every discontinuity point is singlely contained in an interval $I_i=[b^*_i, b^*_{i+1}]$. Let $I_p=[b_p, b_{p+1}]$ denote the $n$ intervals that doesn't contain a discontinuity point. We have that the length of each interval must be higher than 0. Let $b^*_i - b^*_{i+1}=: \epsilon>0$. We can take the $\textbf{b}$-upper Riemann sum as (the lower is defined anagolously):
$\overline{R}_b =\sum^k_{i=1} [sup f ([b^*_{i-1}, b^*_i])]*(b^*_i - b^*_{i-1})$ + $\sum^n_{p=1} [sup f ([b_{i-1},b_i])]*(b_i - b_{i-1})$
Since we choose $\epsilon>0$ to be as small as we want, we have that as $\epsilon \rightarrow 0$, the first term also tends to zero, since
$\sum^k_{i=1} [sup f ([b^*_{i-1},b^*_i])]*(b^*_i - b^*_{i-1}) \leq Mk\epsilon$ (and $kM$ is finite)
Since $f$ restricted to $[a,b]-A$ is uniformly continuous, it is Riemann integrable (I can use this result since it has already been proved).
Does this count as a valid proof?
Thanks in advance
It doesn't count as a valid proof since you use the fact that $f|_{[a,b]\setminus A}$ is uniformly continuous and therefore Riemann-integrable. This doesn't make sense, because the concept of Riemann-integrable function is defined only for functions whose domain is an interval which is closed and bounded.
There are other problems, but this one is enough.