# Integrability of increasing function

I am trying to prove the following (from Axler's Measure, Integration and Real Analysis):

"Suppose $$f:[a,b]\to\mathbb{R}$$ is an increasing function (i.e. $$c,d\in [a,b], c). Prove that $$f$$ is Riemann integrable on $$[a,b]$$."

Now, I have tried to use the fact that $$f$$ is integrable if $$\forall\varepsilon>0$$ there exists a partition $$P$$ of $$[a,b]$$ such that $$U(f,P,[a,b])-L(f,P,[a,b])<\varepsilon$$.

EDITED ACCORDING TO ANSWER BY Fred BELOW:

if for example I fix $$\varepsilon>0$$ and use the equally spaced partition of $$[a,b], x_j-x_{j-1}=\frac{b-a}{n}<\frac{\varepsilon}{f(b)-f(a)+1}$$ (such an $$n$$ exists since $$\frac{const}{n}\overset{n\to\infty}{\to}0$$) I get
$$U(f,P,[a,b])-L(f,P,[a,b])=\sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}f-\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f= \sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j) -f(x_{j-1}))=\frac{(b-a)}{n}\sum_{j=1}^{n}(f(x_j)-f(x_{j-1}))=\frac{(b-a)}{n}(f(b)-f(a))<\varepsilon\frac{f(b)-f(a)}{f(b)-f(a)+1}<\varepsilon$$

• Do you know that a bounded function is RI iff it is continuous almost everywhere? – Kavi Rama Murthy Feb 22 at 11:42
• @KaviRamaMurthy thank you for your comment, but thanks to the answer below by Fred I think I have been able to amend my original proof – lorenzo Feb 22 at 12:17

You wrote $$\sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j-f(x_{j-1}))=(b-a)(f(b)-f(a)).$$ But this is not correct. Correct is:
$$\sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j-f(x_{j-1}))=\frac{b-a}{n}(f(b)-f(a)).$$