# 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. Nov 7, 2018 at 23:27

Theorem: A function $$f$$ is Riemann integrable on $$[a,b]$$ if and only if $$f$$ is bounded on $$[a,b]$$ and the set of discontinuities of $$f$$ in $$[a,b]$$ has measure 0.
Since an increasing function on $$[a,b]$$ is clearly bounded, and the set of discontinuities of an increasing function is countable and hence is of measure 0. Then we directly prove your question.