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Let $I=[a,b]$ be closed and bounded and $f:I\rightarrow\mathbb{R}$ be a monotone function show or contradict that $f$ is Riemann integrable.

Any hint are welcome since I have no idea where to start.

Dave L. Renfro
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R.vW
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  • Do you know the Lebesgue criterion? If so there is a way to proceed based on understanding the set of discontinuities of a monotone function, once you notice that $f$ has to be bounded. – Ian Jun 19 '17 at 17:41
  • I have never heard of it – R.vW Jun 19 '17 at 17:42
  • OK. What about the equivalence of Riemann and Darboux integration? The idea of looking into how badly discontinuous a monotone function can be is still useful in that framework. – Ian Jun 19 '17 at 17:43

2 Answers2

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For a monotone function and any partition of $[a,b]$ finer than $\delta$ the difference between lower and upper sum is at most $\delta\cdot|f(b)-f(a)|$. (Can you see why?)

Ian
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Hagen von Eitzen
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We may assume that $f$ is increasing. Let $\epsilon>0$, and let $a=x_0<x_1<\dots<x_n=b$ be a subdivision of $[a,b]$ with $\max_{1\leq k\leq n}(x_k-x_{k-1})<\delta$ where $\delta>0$ has to be determined. Then the upper sum and the lower sum are $$U=\sum_{k=1}^n f(x_k)(x_{k}-x_{k-1}),\quad\mbox{and}\quad L=\sum_{k=1}^n f(x_{k-1})(x_{k}-x_{k-1}).$$ Now we evaluate the difference: $$U-L=\sum_{k=1}^n (f(x_k)-f(x_{k-1}))(x_{k}-x_{k-1})<\delta\sum_{k=1}^n (f(x_k)-f(x_{k-1}))=\delta(f(b)-f(a))=\epsilon$$ where $\delta:=\epsilon/(f(b)-f(a))$. So $f$ is Riemann integrable over $[a,b]$.

Robert Z
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