I'm trying to prove that if $f:[a,b] \rightarrow \mathbb{R}$ is a bounded function, and assuming that $f$ is continuous on $(a,b)$ then $f$ is Riemann integrable on $[a,b]$. Here's my attempt at the proof:
Since $f$ is continuous, by definition $\exists \delta$ such that $\forall \epsilon > 0$:
$|x - y| < \delta \rightarrow |f(x) - f(y)| < \epsilon$
Now I'm going to guess that I'm supposed to use either the theorem:
$f$ is Riemann integrable iff $\forall \epsilon > 0, \exists \text{ partition, } P \text{ s.t } U(f,p) - L(f,p) < \epsilon$
or the theorem:
$f$ is Riemann integrable with Riemann integral I iff $\forall \epsilon > 0, \exists \text{ a partition P s.t. } U(f,p)-\epsilon < I < L(f,p) + \epsilon$
to create some sort of epsilon argument. I'm going to try using the first one because if I'm being honest, I wouldn't know how to go about with the second.
Let $P$ be a partition of $[a,b]$ such that $P=\{a = x_0, x_1, \ldots, x_n = b\}$ where the distance between each $x_i$ and $x_{i+1}$ is less than $\delta$ (because of continuity?). And then our sup and inf will exist somewhere within our partition:
$M_i = \sup(f(x))$ where $x\in (x_i, x_{i+1})$
$m_i = \inf(f(x))$ where $x \in (x_i, x_{i+1})$
And here's where I'm stuck. Help or advice would be much appreciated!