Boundedness and Riemann-integrability implies Continuity at some point (Analysis) 
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 $\Delta x$.)
 A: 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.
