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Let $f$ be bounded and Riemann-integrable on $[a,b]$. Prove that $f$ is continuous at some point $x \in [a,b]$ and is also continuous on a dense subset $D$ of $[a,b].$

Let $\epsilon >0$. Since $f$ is Riemann-integrable, there exists a partition $P$ such that $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$, where $\alpha$ is an increasing function. Are we able to conclude the first part of the proof that wants us to show that $f$ is continuous at some point $x \in [a,b]$ by taking the partition $P$ to be $[a,b]$? If not, how should we proceed/conclude?

Some definitions:

A set $E$ is dense in $X$ if every point of $X$ is a limit point of $E$, or is a point of $E$ (or both).

Let $A \subset \mathbb{R}$, let $f:A \to \mathbb{R}$, and let $c \in A$. $f$ is continuous at $c$ if for any $\epsilon >0$ there exists $\delta >0$ such that $x \in A$ implies that if $|x-c| < \delta$ then $|f(x)-f(c)| < \epsilon$.

A function is Riemann-integrable if $\inf U(P,f,\alpha) = \sup L(P,f,\alpha)$, where the inf and sup are taken over all partitions.

EDIT:

For any real function $f$ bounded on $[a,b]$ denote

$\displaystyle U(P,f,\alpha ) = \sum_{i=1}^n M_i \Delta \alpha _i$ and

$\displaystyle L(P,f,\alpha ) = \sum_{i=1}^n m_i \Delta \alpha _i$, where

$M_i = \sup \{ f(t): x_{i-1} \leq t \leq x_i \}$ and

$m_i = \inf \{ f(t): x_{i-1} \leq t \leq x_i \}$.

A function is Riemann-integrable if $\inf U(P,f,\alpha) = \sup L(P,f,\alpha)$, where the inf and sup are taken over all partitions. (This is Rudin's definition of Riemann-integrable. $\alpha$ is an increasing function so that $\Delta \alpha _ i = \alpha (x_i) - \alpha (x_{i-1})$. For example, if $\alpha = x$, we get the definition of $\Delta x$.)

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    $\begingroup$ Sorry, what's your definition of Riemann-integrable? Where does $\alpha$ come in? $\endgroup$ – jgon Feb 3 '18 at 2:17
  • $\begingroup$ @jgon I just made an edit with clarifications. $\endgroup$ – Dominated Convergence Theorem Feb 3 '18 at 2:38
  • $\begingroup$ See math.stackexchange.com/a/519921/72031 and your second claim follows from the first. $\endgroup$ – Paramanand Singh Feb 3 '18 at 4:26
  • $\begingroup$ That sounds like the definition of Stieltjes integral, not just Riemann. $\endgroup$ – Martin Argerami Feb 4 '18 at 1:05
  • $\begingroup$ R. integrable implies continuous a.e. implies the continuity set is dense $\endgroup$ – MathematicsStudent1122 Feb 4 '18 at 3:55
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Define the oscillation, $\omega_f(U)$ for an open set $U$ as $\omega_f(U)=\sup_{x\in U} f(x)-\inf_{x\in U}f(x)$. Then if $x\in \newcommand{\RR}{\mathbb{R}}\RR$, define $$\omega_f(x) = \lim_{\epsilon\to 0} \omega_f(B_\epsilon(x)).$$

It's easy to see that $f$ is continuous at $x$ if and only if $\omega_f(x)=0$. Now consider $A_n:=\omega_f\newcommand{\inv}{^{-1}}\inv([0,1/n))$ for $n > 0$. I claim that $A_n$ is open and dense for all $n$, then we'll have that the points of continuity, $\bigcap_{n\in \Bbb{N}_+} A_n$ will be dense by the Baire category theorem.

Thus we just need to show that $A_n$ is open and dense. Suppose $x\in A_n$. Then $x$ has a neighborhood $V$ such that $\omega_f(V)< 1/n$. Hence for all $y\in V$, $\omega_f(y)\le \omega_f(V)<1/n$. Thus $V$ is an open nhood of $x$ contained in $A_n$. Thus $A_n$ is open. As for density, here we need to use integrability of $f$. Suppose $(r,s)$ is a nonempty interval contained in the complement of $A_n$. Then for all $x\in (r,s)$, $\omega_f(r,s) \ge 1/n$. Now take any partition $P$ containing $r$ and $s$. Then $U(P,f,\alpha)- L(P,f,\alpha) \ge 1/n(\alpha(s)-\alpha(r))$ for any such $P$, so the Riemann integral doesn't converge. (Assuming $\alpha$ is strictly increasing. If $\alpha$ weren't strict, bad stuff could happen when it remained constant. Therefore $A_n$ is open and dense, so by the Baire category theorem the points of continuity are dense.

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You can prove this by contradiction.

Suppose there is no dense subset of $[a,b]$ such that f is continuous.

There exists a point $x_1\in[a,b]$ that in a finite interval, $c>0$, and $[x_1-c,x_1+c]$ f is not continuous.

Since f is not continuous in $[x_1-c,x_1+c]$, there exist $p > 0 $ such that $|f(x)-f(t)|\geq p$ for $x,t\in[x_1-c,x_1+c]$

Then $U(P,f,\alpha)-L(P,f,\alpha) > p[\alpha(x_1+c)-\alpha(x_1-c)]$. This contradicts to f is integrable.

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  • $\begingroup$ But Thomae's function is discontinuous on the rationals (which are dense), yet it is Riemann integrable. $\endgroup$ – Martin Argerami Feb 3 '18 at 20:46
  • $\begingroup$ You can see my edit. The statement is slightly different from what's in Thomae's function. $\endgroup$ – Math The Novice Feb 4 '18 at 3:49

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