I need to prove that if a function $f: [a,b] \to \mathbb{R}$ has bounded variation, than $f$ is integrable on $[a,b]$.
This is what I have tried:
Let $S(P)$ and $s(P)$ denote the upper and lower Riemann sums of $f$ over a partition $P$ of $[a,b]$. Then $S(P) - s(P) = \sum \limits_{i = 1}^{n} (M_i - m_i) \Delta x_i $, where $M_i = \sup_{x_{i-1} \leq x \leq x_i} f(x) $ and $m_i = \inf_{x_{i-1} \leq x \leq x_i} f(x) $. Now as $f$ must be bounded, $M_i - m_i \leq 2|f(a)$|. Then for all partitions $P$ with $\||P|| < \frac{\epsilon}{2|f(a)|} $, we have that $S(P) - s(P) \leq 2|f(a)| \cdot ||P|| < \epsilon \implies f$ is integrable.
However this argument doesn't work, as if it did this would imply bounded functions are integrable, which isn't true. So I'm not sure where it goes wrong, and how to fix it. I most likely need to incorporate the definition of bounded variation somewhere, but I am not sure where. If anyone could help me fix this I would appreciate it.
You are almost correct. Let $\epsilon >0$ and $P$ be a partition of $[a, b]$ such that $(M+1)||P||< \epsilon$ (see definition of $M$ below). Then
$$S(P) - s(P) = \sum_i^n (M_i - m_i) \Delta x_i\ .$$ By definition of $M_i$ and $m_i$, there exists $a_i$ and $b_i$ in $[x_i, x_{i+1}]$ such that
$$|f(a_i) - f(b_i)| > M_i - m_i - 1/n$$
Then
$$S(P) - s(P) < \sum_{i=1}^n (|f(a_i) - f(b_i)| + 1/n) \Delta x_i \leq ||P||\big(\sum_{i=1}^n (|f(a_i) - f(b_i)|\big) +||P||$$.
Let all these $a_i$ and $b_i$ be a subset of some partition of $[a, b]$, as $f$ is of bounded variation, $\sum_i (|f(a_i) - f(b_i)|<M$ for some $M$ (which depends only on $f$). Thus
$$S(P) - s(P) < (M+1)||P|| < \epsilon\ .$$
Thus $f$ is Riemann integrable. Indeed, there is a result showing that every function of bounded variation can be written as the difference of two increasing functions. As increasing functions are Riemann integrable, so are functions of bounded variations.
See this Bounded variation, difference of two increasing functions for the last fact mentioned.