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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. $c,d\in [a,b], c<d\Rightarrow f(c)\leq f(d)$). 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
$$U(f,P,[a,b])-L(f,P,[a,b])=\sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}f-\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f= \sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j) -f(x_{j-1}))=\frac{(b-a)}{n}\sum_{j=1}^{n}(f(x_j)-f(x_{j-1}))=\frac{(b-a)}{n}(f(b)-f(a))<\varepsilon\frac{f(b)-f(a)}{f(b)-f(a)+1}<\varepsilon$$

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    $\begingroup$ Do you know that a bounded function is RI iff it is continuous almost everywhere? $\endgroup$ – Kavi Rama Murthy Feb 22 at 11:42
  • $\begingroup$ @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 $\endgroup$ – lorenzo Feb 22 at 12:17
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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:

$$\sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j-f(x_{j-1}))=\frac{b-a}{n}(f(b)-f(a)).$$

Can you procced ?

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