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Let $f\colon[a,b]\to\mathbb{R}$ be a bounded function, and let $\mathcal{P}_{n}$ be a sequence of partitions (not necessarily evenly spaced) with mesh $m(\mathcal{P}_{n}) \to 0$ as $n\to \infty$. Then $$|S_{L}(\mathcal{P}_{n})-S_{R}(\mathcal{P}_{n})|\to 0 \text{ for } n\to\infty$$

Note:

$$S_{L}(\mathcal{P}_{n}) = \sum^{n}_{k=1}f(c_{k-1})[x_{k}-x_{k-1}] \text{ and }S_{R}(\mathcal{P}_{n})=\sum^{n}_{k=1}f(c_{k})[x_{k}-x_{k-1}]$$ where $c_{k}$ is an internal point, ie. $c_{k}\in[x_{k-1},x_{k}]$

Ok so I thought of the function $f\colon[0,1]\to\mathbb{R}$: $$f(x) = \begin{cases} 1, & \text{if } x\in\mathbb{Q}, \\ 0, & \text{if } x \notin \mathbb{Q}.\end{cases}$$

But I'm not sure how to progress with showing the difference between the Riemann sums doesn't converge to zero, i.e. showing the function is non-integrable.

How would I do this using only the information above?

user2850514
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1 Answers1

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For every partition, produce a finer partition an take all the $c_i$ to be rational, then take the same refinement partition and take all $c_i$ irrational.

One sum will be $0$ the other $1$.

  • I was thinking of doing this but surely we must use the same internal points for $S_{L}(\mathcal{P_{n}})$ and $S_{R}(\mathcal{P_{n}})$? If not can you explain why we can use different internal points for each one? To me it seems that taking different internal points results in two different regions being compared. – user2850514 May 09 '14 at 12:49
  • @user2850514 Nah! The definition (according to where you read) allows you to either use any points or to take the same points of the partition, or the maximum and/or minimum of each partition. Besides, when you change the partition you take new points. Notice I said refine the any partition given. So, we are not working with the given one, but a new one. –  May 09 '14 at 12:52
  • @user2850514 The definition says that for every $\epsilon>0$ there is a partition $P$ so that any finer partition has Riemann sum $\epsilon$-close to the (to be) value of the integral. So, above, for any given a partition we build two finer ones that remain apart from each other. –  May 09 '14 at 12:54