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

Question

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.

Answer 1

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.

Answer 2

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 $\mu_n\to x$.

Answer 3

@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 $I > \epsilon_o,\forall I\subseteq [0,1]?$