Let $f : [a, b] → \mathbb{R}$ be monotonic decreasing. Prove that $f$ is Riemann integrable on $[a, b]$.
I have the proof for the question when $f$ is monotonic increasing, how can I change it to get the answer for when $f$ is monotonic decreasing?
Proof for increasing $f$:
Note that $f$ is automatically bounded (an upper bound is $f(b)$ and a lower bound is $f(a)$). Fix $\epsilon > 0$ and let $P_n$ be the partition ${a+ \frac{i(b−a)}{n}: i = 0, 1, . . . , n}$ of $[a, b]$ into $n$ equally sized pieces. Then, as $f$ is increasing we have
$$U(f, P_n) − L(f, P_n) = \sum_{i=1}^n{\frac{b-a}{n}\left(f\left(a+\frac{i(b-a)}{n}\right)-f\left(a+\frac{(i-1)(b-a)}{n}\right)\right)}=\frac{b-a}{n}\left(f(b)-f(a)\right).$$
By choosing $n$ with $n > \frac{(b−a)(f(b)−f(a))}{\epsilon}$, we have $U(f, P_n) − L(f, P_n) < \epsilon$. Therefore $f$ is Riemann integrable by the $\epsilon$-test.