Rudin's Riemann-Stieljes Integral Proof

Rudin provides the following Theorem (modified to isolate my question):

$f$ is Riemann Integrable on $[a,b]$ if and only if for all $\epsilon>0$ there exists a partition $P$ such that $$U(P,f)-L(P,f) < \epsilon$$

Proof (<= direction):

For each partition $P$ we have: $$L(P,f) < \sup_{P} L(P,f) < \inf_{P} U(P,f) < U(P,f)$$ Implies that: $$0 \leq \sup_{P} L(P,f) - \inf_{P} U(P,f) < \epsilon$$ Hence for all $\epsilon > 0$, we have $$\sup_{P} L(P,f) = \inf_{P} U(P,f)$$ Which, by definition, means $f$ is Riemann integrable.

I'm confused by the part in the bold. It's certainly true that $$0 \leq \sup_{P} L(P,f) - \inf_{P} U(P,f) < \epsilon$$ for any $\epsilon$. But for any $\epsilon$, no matter how small, $\epsilon$ is still greater than 0. How do we get from $\sup_{P} L(P,f)$ and $\inf_{P} U(P,f)$ being arbitrarily close to actually equal? I think there's a skipped step, but I can't say it precisely.

Actually it should be $0\leq b-a<\epsilon$, where $b=\inf_{P}U(P,f)$, $a=\sup_{P}L(P,f)$. Assume that $b-a>0$, then take $\epsilon=b-a$, one gets $b-a<b-a$, contradiction, so $b=a$.
• Thanks it should be $b-a$, indeed! Commented Jan 23, 2018 at 14:13
The numbers $\sup_{P} L(P, f), \inf_{P} U(P, f)$ are fixed and they can't be arbitrarily close to each other. Rather they maintain a fixed distance (namely $\inf_{P}U(P, f) - \sup_{P} L(P, f)$) to each other. If this fixed distance is smaller than every positive number $\epsilon$ then it must be zero. Always remember the trivial/obvious fact from high school: