I'm trying to complete the proof in this answer that if $f: [a, b] \to \mathbb{R}$ is a Riemann integrable function, then $|f|$ is an integrable function also.
I understand the proof that $$ \sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x) + \epsilon $$
but I need to translate this to $$ \sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x) $$ because otherwise the summations break down (see below). Can I just say that because $\epsilon > 0$ can be made as small as possible, then $$ \sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x) $$ in the limit? Once I have that, it's easy, because then I just sum the inequality across every interval to get $$ \sum_{i=0}^{n-1} (\sup |f| - \inf |f|)(x_{i+1} - x_i) \leq \sum_{i=0}^{n-1} (\sup f - \inf f + \epsilon)(x_{i+1} - x_i) $$ which gives me the correct upper and lower sums. If I can't just remove the $\epsilon$, then I get \begin{align} U(|f|, P) - L(|f|, P) &= \sum_{i=0}^{n-1} (\sup |f| - \inf |f|)(x_{i+1} - x_i) \\ &< \sum_{i=0}^{n-1} (\sup f - \inf f + \epsilon)(x_{i+1} - x_i) \\ \end{align}
where the supremum and infimum are taken over the interval $[x_i, x_{i+1}]$. I somehow need to prove that $U(|f|, P) - L(|f|, P) < \epsilon$, and I can't see how to manipulate that summation to do so.
EDIT: To clarify, I know that once I have $$ U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P) $$ the proof is complete because for any $\epsilon > 0$, I can pick a partition $P$ such that $U(f,P) - L(f,P) < \epsilon$. My question is about using the supremum and infimum properties above to show that $$ U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P) $$ That's where I'm stuck.
You get $U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$, and since $f$ is integrable, you can choose a partition so the right hand side is as small as you want. (Also, we have $0 \le U(|f|,P) -L(|f|,P)$.)
Addendum:
You have $\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x)$ for all intervals in a partition $P$. Hence $\sup_{x\in I}|f(x)|l(I)-\inf_{x\in I}|f(x)|l(I)\le \sup_{x\in I}f(x)l(I)-\inf_{x\in I}f(x)l(I)$.
Since (abusing notation a little, as in $I \in P$), and similarly for $L(g,P)$, we have $U(g,P) = \sum_{I \in P} \sup_{x\in I}g(x)l(I)$.
Hence $U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$.