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.
Best Answer
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.