So I was having some difficulty coming up with a conceptual reason for why an increasing function on a closed interval would be Riemann Integrable, when without effort a proof seemingly fell out of some computations. Could anyone verify if it is correct? Thank you very much. Here's what I have.
$\textbf{Proof:}$
Suppose that $f:[a,b]\to\mathbb{R}$ is increasing. WLOG we can assume $f(b)>f(a)$, for else $f$ is constant, and trivially integrable. Let $\epsilon>0$. Choose a partition $P=\{x_i\}_0^n$ of $[a,b]$ such that for all $1\leq i,j\leq n$ it follows that $x_i-x_{i-1}=x_j-x_{j-1}<\epsilon/(f(b)-f(a))$. We compute
$$U(f,P)-L(f,P)=\sum_{i=1}^n\bigg(\sup_{[x_{i-1},x_i]}f(x)-\inf_{[x_{i-1},x_i]}f(x)\bigg)(x_i-x_{i-1})$$
$$=\sum_{i=1}^n\bigg(\sup_{[x_{i-1},x_i]}f(x)-\inf_{[x_{i-1},x_i]}f(x)\bigg)(x_1-x_{0})$$
$$=\sum_{i=1}^n(f(x_i)-f(x_{i-1}))(x_1-x_{0})=(f(b)-f(a))(x_1-x_0)$$
$$<(f(b)-f(a))(\epsilon/(f(b)-f(a)))=\epsilon.$$
This completes the proof.$\square$
Best Answer
Yes, the proof is correct.
Every step is clearly explained.
The increasing property does the trick.