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 $\mathcal{P}=\left\{a=t_0<…<t_n=b\right\}$ of $[a,b]$. We separate the indices: $A=\left\{i:[t_{i-1},t_i]= [x_k-\delta_k,x+\delta_k]\right\}$ and $B=\left\{i:[t_{i-1},t_i]= \overline{D}_k \right\}$. Therefore,
\begin{gather}U_{f,\mathcal{P}}-L_{f,\mathcal{P}}=\sum_{i=1}^n\omega f([t_{i-1},t_i])(t_i-t_{i-1})=\sum_{i\in A}\omega f([t_{i-1},t_i])\ell([t_{i-1},t_i])+\sum_{i\in B}\omega f([t_{i-1},t_i])\ell([t_{i-1},t_i])\\
\le \sum_{i\in A}2\left\|f\right\|\ell([t_{i-1},t_i])+\sum_{i\in B}\epsilon\ell(\overline{D}_k)\le 2\left\|f\right\|\frac{\epsilon}{M}+\epsilon(b-a)=2\epsilon+\epsilon(b-a)\end{gather}
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?
Best Answer
Let $M = \sup(f) - \inf(f)$ on $[a,b]$ and $D = \{d_0, d_1, \dotsc \} \subset [a,b]$ be the countable set of discontinuities of $f$. Let $\delta > 0$ and define $I \subseteq [a,b]$ as $$I = \{ x \in [a,b] \mid \sup(f) - \inf(f) < \frac{\delta}{b-a} \textrm{ on some neighbourhood of }x \}.$$ Then $I$ is open and contains all points where $f$ is continuous. In particular $[a,b] = I \cup D$. For $k \geq 0$ define the interval $D_k$ by $$D_k = \left(d_k-\frac{\delta}{M2^{k+2}}, \, d_k+\frac{\delta}{M2^{k+2}} \right) \cap [a,b].$$ Then $I$ together with all these intervals cover $[a,b]$ and because $[a,b]$ is compact it is already covered by a finite union $$[a,b] = I \cup D_0 \cup D_1 \cup \dotsc \cup D_n$$ for some $n \geq 0$. The complement $$[a,b] \setminus (D_0 \cup \dotsc \cup D_n)$$ is compact and contained in $I$ and can therefore be covered by open intervals such that $\sup(f)-\inf(f) < \delta/(b-a)$ on the closure of each. (Every point in $I$ has such a neighbourhood by definition.) By compactness a finite subcover of such intervals exists. Considering all end points in this subcover as possible cut points one can partition the complement into finitely many closed intervals such that $\sup(f) - \inf(f) \leq \delta/(b-a)$ on each. Finally the closure of $D_0 \cup \dotsc \cup D_n$ is itself a union of closed intervals with a total length less than $\delta/M$.
The entire interval $[a,b]$ is now partitioned in a finite number of closed intervals that admit upper and lower Riemann sums of $f$ that differ by at most $2\delta$. Since $\delta$ can be chosen arbitrarily small, $f$ is Riemann integrable.
In fact this argument works just as well for any $D$ that can be covered by a countable union of intervals of arbitrarily small total length, i.e. for $D$ of measure $0$.