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.


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.$\square$

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    $\begingroup$ 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. $\endgroup$ Nov 7, 2018 at 23:27

2 Answers 2


Yes, the proof is correct.

Every step is clearly explained.

The increasing property does the trick.


Another much easier proof can be provided if you know this theorem.

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.


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