# Verifying an increasing function on a closed interval is Riemann Integrable

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.$$\square$$

• The proof is correct. The intuition for integrability is that monotone functions on closed intervals are relatively well behaved, in the sense that they have at most countably many discontinuities. – rubikscube09 Nov 7 '18 at 23:27