Claim: If $f \in {\mathscr R[a,b]}$, then $f$ has infinitely many points of continuity.

1.) I read that it is a corollary of the Lebesgue integrability criterion. Is it possible to prove the claim without invoking the concept of measure(or using less abstraction) ?

2.) Here is attempt: Given $\epsilon > 0, \exists$ Partition, $P = \left\{ x_0 =a,...,x_n =b \right\} $ of $[a,b]$ such that $\sum^n_{i=1} (M_i -m_i)\Delta x_i < (b-a)\epsilon$, where $M_i= \sup \left\{ f(x) : x \in \Delta x_i \right\}$ and $ m_i= \inf \left\{ f(x) : x \in \Delta x_i \right\} $

Let $(M_j -m_j)=\min \left\{ M_i -m_i : i=0,...,n \right\} $. $ \implies (M_j-m_j)(b-a) \leq \sum^n_{i=1} (M_i -m_i)\Delta x_i < (b-a)\epsilon$ $ \implies (M_j-m_j)<\epsilon. $

Let $ c\in (x_{j-1},x_j)$ and $\delta$ be any positive number such that $(c-\delta, c+\delta) \subseteq (x_{j-1}, x_j)$.

It follows that $ \left| f(x) - f(c) \right| < \epsilon, $ whenever $ \left| x - c \right| < \delta $.

Since $c$ is arbitrary and $[x_{j-1},x_j]$ is an interval, there are infinitely many points of continuity.

There is something wrong with my proof since Thomae's function is a counterexample. Could anyone point out the mistakes in my proof? Thank you.

  • 4
    $\begingroup$ $R[a,b]$ seems to be the space of Riemann-integrable functions on $[a,b]$? $\endgroup$ Oct 5, 2013 at 21:06
  • $\begingroup$ Certainly, it can be done comprehensively based only on the concepts of sets of 'zero measure' and 'zero content', which is to say you don't need to develop any part of measure theory to get there. But I suspect that this can be achieved more cheaply, as this corollary is considerably weaker than Lebesgue's integration criterion. $\endgroup$ Oct 5, 2013 at 21:17
  • $\begingroup$ I know this is an old question, however, for anyone wondering where is the error in the, otherwise elegant, proof of the OP: the sub-interval $[x_{j-1}, x_j]$ depends on $\epsilon$. $\endgroup$ Jul 20, 2022 at 10:36

3 Answers 3


You find the statement in J. Pierpont - Lectures on the Theory of Real Variables Vol. 1 (1905) $\S$ 508 p. 348.
The theorem is announced saying ...There is, however, a limit to the discontinuity of a function beyond which it ceases to be integrable ...
I rewrite the proof as to the modern standard of rigor, omitting some detail.

Because of the Riemann criterion if $f$ is integrable on a (compact) interval, then, for every $\varepsilon>0$, there exists a subinterval on which the oscillation of $f$ is less than $\varepsilon$. In fact starting from $$\sum \omega_j\,(x_j-x_{j-1})<\varepsilon(b-a)$$ it is enough to consider the smallest of the $\omega_j$'s.
Remember also that, if $f$ is integrable on an interval, then it is integrable on every subinterval.

Then, if $I$ is a subinterval of $[a,b]$, you can construct a nested sequence $\{I\}_n$ of subintervals of $I$ such that $I_{n+1}$ has no extreme in common with $I_n$ and the oscillation of $f$ on $I_n$ is less than $1/n$.
By a well known property of the real number system, the intersection of the intervals of $\{I\}_n$ is not empty.
It is easy to prove that, if $c$ is a point of the intersection, then $f$ is continuous in $c\;$(note that $c$ is a point interior of every $I_n\,$).
So any subinterval of $[a,b]$, however small, contains a point of continuity and the theorem is proved.

Note that, if you don't suppose that $I_{n+1}$ has no extreme in common with $I_n$, only one-sided continuity can be assured.

  • $\begingroup$ thanks for the lucid explanation! May I know how you prove $f$ is continuous at $c$ ? I tried using sequential criterion to no avail. $\endgroup$ Oct 13, 2013 at 10:19
  • $\begingroup$ If $\varepsilon>0$, there exists $n$ such that $1/n<\varepsilon$. Then on $I_n$ the oscillation is less $\varepsilon$: the risk is that $c$ is an extreme of $I_n$ but this doesn't happen. In fact also $c \in I_{n+1}$ and $I_{n+1}$ has no extreme in common with $I_n$. Then $I_n$ is a right neighbourhood of $c$ to prove the (two-sided) continuity of $f$ at $c$. $\endgroup$ Oct 13, 2013 at 11:07
  • $\begingroup$ thanks for great explanation! Suppose $f$ is bounded on $[a,b]$(but not necessarily integrable on $[a,b]$), does it still imply that $f$ has infinitely points of continuities? $\endgroup$ Oct 13, 2013 at 11:42
  • $\begingroup$ See the Dirichlet's function: it is nowhere continuous. $\endgroup$ Oct 13, 2013 at 15:18
  • 1
    $\begingroup$ You start with an $\varepsilon$ and find $c$. Then take another $\varepsilon$, say $\varepsilon_1$: you use the Riemann criterion again. Are you sure you find a partial interval in wich the oscillation is less than $\varepsilon_1$ (we know it exists) and such that it contains that $c\,$ ? $\endgroup$ Oct 14, 2013 at 17:17

Without loss of generality, we'll work with $[0,1]$.

LEMMA 1 Suppose that $$U(P,f)-L(P,f)<\frac{1}n$$ for some partition $P$ and natural number $n$. Then there exists $i$ such that $M_i-m_i<\frac 1 n$.

P If not, we would have $$U(P,f)-L(f,P)=\sum_{i=1}^m (M_i-m_i)\Delta x\geqslant\frac 1 n\sum_{i=1}^m \Delta x=\frac 1 n$$

contrary to our hypothesis.

THEOREM If $f\in{\mathscr R}[0,1]$, then $f$ is continuous at infinitely many points of $[0,1]$, moreover, this set is everywhere dense in $[0,1]$.

P Pick a partition $P_1=\{t_1,t_2,\ldots,t_n\}$ such that $U-L<1$. Then there exists an $i$ for which $M_i-m_i<1$. If $i\neq 1,n$; choose $a_1=x_{i-1},b_1=x_i$. Else, choose $a_1\in(0,t_1)$ or $b_1\in (t_{n-1},b)$ (if $i=1$ or $i=n$, resp.). Then $$\sup_{x\in [a_1,b_1]} f(x)<1$$

Since $f\in{\mathscr R}[a_1,b_1]$, we may repeat the procesto obtain $a_1<a_2<b_2<b_1$ such that $\sup_{x\in [a_2,b_2]} f(x)<\frac 1 2$. Continuing the process, we obtain a sequence of nested closed intervals $I_n=[a_n,b_n]$ such that $$(\sup-\inf)\{f:I_n\}<\frac 1 n$$

By Cantor's theorem there exists $\xi\in \bigcap_{n\geqslant 1}I_n$. In particular, since $a_n<a_{n+1}<b_{n+1}<b_n$, we have $\xi\neq a_i,b_i$. (Ex. Show $f$ is continuous at $x=\xi$.) Now, $f$ is integrable on $[0,\xi]$, $[\xi,1]$, so we may repeat the process to obtain $\xi_1,\xi_2$ two different points of continuity. By construction, $\xi_1,\xi_2\neq \xi,0,1$. We may continue this process to obtain, for each $j$, $2^{j+1}-1$ points of continuity. Thus this set is infinite.

Pick now $x\in[0,1]$, and consider $\left[x,x+\dfrac 1{2^n}\right]$ (the argument should be slightly modified for endpoints. Then for each $n$ we can find by the above construction a point of continuity $\mu_n \neq x$, and it is evident $\mu_n\to x$.


@Tony Piccolo:

May I know if I have interpreted correctly?

Given $\epsilon_1 > 0$, $\exists$ closed interval $ I_{1} \subset [a,b], $ such that oscillation of $f$ on $ I_{1} < \epsilon_1$ .

Given $\epsilon_2 < \epsilon_1, \exists I_2 \subset I_1$ such that oscillation of $f$ on $I_2< \epsilon_2$.... Proceeding in a similar fashion, we can construct a nested sequence $\{I\}_n$ of $I_1$ such that $I_n$ has no common extreme points with $I_{n+1}$.

If we only require $f$ to be bounded on $[a,b]$, then we might not be able to construct such $\{I\}_n$ as described above. For eg, define $f:[0,1] \to \mathbb{R}$ by $f(x) = 1,$ if $x \in \mathbb{Q}$ and $f(x) =0$, otherwise, i.e. $f \notin R[0,1]$. Then $\exists \epsilon_o = \frac{1}{2}$ such that oscillation of $f$ on $I > \epsilon_o,\forall I\subseteq [0,1]?$

  • $\begingroup$ Perfect ! It is enough that $\varepsilon_n \rightarrow 0$ as you suppose ($\{\varepsilon\}_n$ is monotone). The Dirichlet function has oscillation $1$ on every subinterval of $[0,1]$, so the construction of those nested subintervals is impossible. $\endgroup$ Oct 14, 2013 at 22:52

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