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I need to show that the following function is not Riemann Integrable on $[0,1]$,

$f(x) = \begin{cases} x & x\in \mathbb{Q} \\ \ 0 & x\in\mathbb{R}\backslash \mathbb{Q} \end{cases}$

My solution is:

Assume $f$ is Riemmann integrrable. Thus by Darboux's Criterion, given an $\epsilon \gt 0$ there exists a partition $P = \{x_{0},...,x_{n}\}$ of $[0,1]$ such that

$U(f,P) - L(f,P) \lt \epsilon \\ \iff \sum_{i=0}^{n}M_{i}(x_{i}-x_{i-1}) - 0 \lt \epsilon \\$

Now,

$M_{n}(x_{n} - x_{n-1}) \lt \sum_{i=0}^{n}M_{i}(x_{i}-x_{i-1})$ , since the interval has only positive values.

Thus,

$M_{n}(x_{n} - x_{n-1}) \lt \epsilon$

but, $M_{n} = x_{n} = 1$,

So,

$1 \lt \epsilon + x_{n-1}$

Am I on the right track, because I'm not sure where to got from here?

Botond
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TrendyParker
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  • No, you are not on the right track, since it's easy to find a partition that satisfies $1 \lt \epsilon + x_{n-1}.$ Is $M_i$ supposed to be the sup of $f$ on the $i$th interval? If so, it's equal to the right-hand endpoint. – saulspatz Aug 21 '18 at 12:52
  • I see. Yes, sorry, $M_{i}$ is the sup on the $i_{th}$ interval. I'm just not sure how else to approach it? Also, the right hand endpoint is not guarenteed to be rational, so wouldn't the sup in the interval be the largest rational number? – TrendyParker Aug 21 '18 at 14:04
  • The sup would be the supremum of all the rational numbers in the interval, which is the right-hand endpoint. What is the supremum of all rational number less than $\sqrt2?$ – saulspatz Aug 21 '18 at 15:09
  • I'm embarrassed to admit I've forgotten how to do this. It's easy to see that if you partition the unit interval into $n$ intervals of length $1/n$, then $U(P,f)>\frac12.$ I think the trick must be to show that you can approximate any upper sum by one of these "uniform" sums, but I can't quite see how to do that. – saulspatz Aug 21 '18 at 15:54
  • I mean, I could also use the regular Riemann definition of integrability by stating that $L(f,P)=0$ no matter the partition, while $U(f,P)$ is never zero no matter the partition. Thus the upper and lower integrals are not equal? – TrendyParker Aug 21 '18 at 18:16
  • Yes that works. –  Aug 21 '18 at 19:01

1 Answers1

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For a general partition $(x_0,x_1, \ldots,x_n)$ we have $M_j = \sup_{x \in [x_{j-1},x_j]}f(x) = x_j,$ even if $x_j$ is not rational. This follows because the rationals are dense and for any $\delta > 0$ there exists a rational $r$ such that $x_j - \delta < r = f(r) \leqslant x_j$. Unless $M_j = x_j$ we get a contradiction.

Consequently, the upper sum is

$$U(P,f) = \sum_{j=1}^n x_j(x_j - x_{j-1}) \\= \frac{1}{2}\sum_{j=1}^n x_j(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n x_j(x_j - x_{j-1}) \\ > \frac{1}{2}\sum_{j=1}^n x_j(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n x_{j-1}(x_j - x_{j-1}) \\ = \frac{1}{2} \sum_{j=1}^n (x_j^2 - x_{j-1}^2) = \frac{1}{2}(x_n^2 - x_0^2 ) = \frac{1}{2}.$$

Since the irrationals are dense, we have $L(P,f) = 0$. Thus, $U(P,f) - L(P,f) > \frac{1}{2}$ for every partition $P$ proving that $f$ is not Riemann integrable.

RRL
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