Cauchy - Criterion for integrability: A bounded function $f$ on $[a,b]$ is Riemann-ntegrable iff for any $\varepsilon > 0$ there is a partition $P_{\varepsilon}$ such that $U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon$
Consider $f(x) = \frac{1}{x+1}$ on $I = [0, c]$, $c>0$. note that $f$ is decreasing on $I$. We want to show that $f$ is Riemann-integrable using the Cauchy criterion, i.e. for $\varepsilon > 0$, we must find a partition $P_{\varepsilon}$ such that $U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon$.
If $0 < \varepsilon < c$, I tried $P_{\varepsilon} = \{0, \varepsilon, c\}$ but
$$U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) = (\varepsilon + \frac{c - \varepsilon}{1 + \varepsilon}) - (\frac{\varepsilon}{1 + \varepsilon} + \frac{c - \varepsilon}{1 + c}),$$
which is messy and complicated.
As for $0 < c < \varepsilon$, I'm not even sure where to begin.
Any hints on how to partition $I$? Thanks!
For $n \in \mathbb N$ let $P:=\{\frac{j}{n}c: j=0,....,n\}.$
Show that
$$U(f, P)- L(f, P)= \frac{c}{n}(1- \frac{1}{c+1}).$$
If $ \varepsilon>0$ is given, then choose $n$ such that
$$\frac{c}{n}(1- \frac{1}{c+1} ) < \varepsilon.$$