Let $f:[a,b]\to \mathbb{R}$ be a bounded function and $A$ be the set of its discontinuities. I am asking for a (direct) proof that if $A$ is countable then $f$ is Riemann integrable in $[a,b]$ that doesn't explicitely, or implicitly, require the notion of sets of measure $0$ ( and of course without the use of the Lebesgue Criterion).
One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that $A$ is countable. A proof that can't be trivially altered so that it holds even if $\lambda(A)=0$
EDIT: Here is the proof of WimC with all the details: Let $\epsilon>0$ and $D=\left\{d_1,d_2,...\right\}\subseteq A$ be the countable set of discontinuities of $f$. Define: $I=\left\{x\in [a,b]:\exists \delta>0: \omega f((x-\delta,x+\delta)\cap [a,b])<\epsilon\right\}$ Now $x\in I\iff \exists \delta>0: \omega f((x-\delta,x+\delta)\cap [a,b])<\epsilon\iff [\left|y-x\right|<\delta\implies \omega f(y)<\epsilon]$ and because $\epsilon$ is in fact arbitrary, $x\in I\iff \text{ $f$ is continuous at $x$}$ In addition, if $x\in I$, $\exists \delta>0$ so that $\omega f((x-\delta,x+\delta)\cap [a,b])<\epsilon$ If $y\in B(x,\delta)\cap [a,b]$. Then $y\in I$ and so $I$ is open relative to $[a,b]$.
Because $I=[a,b]\setminus D$, $[a,b]=I\cup D$. For $k\in \mathbb{N}$ define $D_k=\left(d_k-\frac{\epsilon}{M2^{k+1}},d_k+\frac{\epsilon}{M2^{k+1}}\right)\cap [a,b]$ Obviously $D\subset \bigcup_{k=1}^{\infty}D_k$ and $[a,b]=I\cup \bigcup_{k=1}^{\infty}D_k$ (since $D_k\subseteq [a,b]$). The compactness of $[a,b]$ implies $[a,b]=I\cup \bigcup_{k=1}^{N}D_k$. Now $[a,b]\setminus \bigcup_{k=1}^{N}D_k$ is compact (closed and bounded) and included in $I$. As such it can be covered by $F_x=(x-\delta_x,x+\delta_x)$ where $ \delta_x>0:$ is chosen so that $\omega f((x-\delta_x,x+\delta_x)\cap [a,b])<\epsilon$. Compactness implies the existence of a finite subcover, $[a,b]\setminus \bigcup_{k=1}^{N}D_k\subseteq \bigcup_{i=1}^{M}(x_i-\delta_i,x+\delta_i)$ As we can replace the intervals that intersect we can suppose $ \bigcap_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\emptyset$ Therefore, $[a,b]= \bigcup_{k=1}^{N}\overline{D}_k\cup \bigcup_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\bigcup_{i=0}^{n}[t_{i-1},t_i]$ where for $i\le n$, $[t_{i-1},t_i]= [x_k-\delta_k,x+\delta_k]$ or $[t_{i-1},t_i]= \overline{D}_k$ because $\bigcup_{k=1}^{N}\overline{D}_k\cap \bigcup_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\emptyset$ Considering all the endpoints of the above (that are pairwise different) we can create a partition $\mathcal{P}=\left\{a=t_0<...
My questions are: Is this proof correct? ( I doubt the point: "where for $i\le n$, $[t_{i-1},t_i]= [x_k-\delta_k,x+\delta_k]$ or $[t_{i-1},t_i]= \overline{D}_k$ because $\bigcup_{k=1}^{N}\overline{D}_k\cap \bigcup_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\emptyset$") Second, can it be simplified?