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?

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    $\begingroup$ Do you mean Riemann integrable? $\endgroup$ – Quinn Culver Dec 22 '12 at 21:58
  • $\begingroup$ @QuinnCulver Yes. $\endgroup$ – Nameless Dec 22 '12 at 21:59
  • $\begingroup$ @Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $\epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well. $\endgroup$ – jdods Apr 20 '18 at 8:05
  • $\begingroup$ You asked for a proof that wasn't just the typical Lebesgue Criterion proof. Then you accepted a proof that is just the typical Lebesgue Criterion proof. $\endgroup$ – MichaelGaudreau Jan 31 '19 at 16:18

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 $0$.

  • $\begingroup$ First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct? $\endgroup$ – Nameless Dec 24 '12 at 7:52
  • $\begingroup$ 2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition? $\endgroup$ – Nameless Dec 24 '12 at 8:13
  • $\begingroup$ @Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $\sup - \inf$ is sufficiently small. Since $I \setminus D_0 \cup \dotsc \cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition. $\endgroup$ – WimC Dec 24 '12 at 8:54
  • $\begingroup$ I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas. $\endgroup$ – Nameless Dec 24 '12 at 8:57
  • $\begingroup$ Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on? $\endgroup$ – Nameless Dec 25 '12 at 18:39

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