Let $f:[0,1]\to\mathbb{R}$ where $f(x)=x$ for $x\in[0,1]$ rational and $f(x)=0$ for $x\in[0,1]$ irrational. Prove that $f$ is not Riemann integrable on $[0,1]$.
I have a vague idea of what I'm supposed to do, but I don't know where to start.
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One way to do it is to show that the upper Riemann sum is at least $1/2$ and the lower Riemann sum is $0$ for any given partition on $[0,1]$. See http://en.wikipedia.org/wiki/Riemann_sum – Paul Dec 07 '13 at 01:07
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Are you wanting to show it from first principles? (Because we don't know what all theorems are at your disposal...) If so, one plan is to show that the difference between the upper and lower sums doesn't tend to zero as outlined here. – JohnD Dec 07 '13 at 02:30
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1Note a similar function, but where for rational arguments it has $f\left(\frac{p}{q}\right) = \frac{1}{q}$ where $p$ and $q$ are positive, relatively prime integers, is sometimes called the "Ruler Function", due to its roughly similar appearance to rulers. It is a special case of "Thomae's function", with more info at Wikipedia. – John Omielan Dec 30 '18 at 21:06
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Can you prove that $$\overline{\int_0^1}f=\int_0^1 xdx=\frac 1 2$$ and that $$\underline{\int_0^1}f=0\;?$$
Pedro
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First, show that $f$ is discontinuous everywhere on that interval. Therefore, the set of discontinuities of $f$ is $(0,1]$, and we know
$$ m((0,1]) = 1 - 0 = 1 > 0 $$.
Therefore, $f$ cannot be riemman integrable.
ILoveMath
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A little correction: $f$ is discontinuous except at the point $0$ for $x_n\to0$ in $[0,1]\implies$ for $\epsilon>0~\exists~k\in\mathbb Z^+$ such that $|x_n|<0~\forall~n\ge k\implies |f(x_n)|<\epsilon~\forall~n\ge k.$ – Sugata Adhya Dec 07 '13 at 02:40
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@AtahualpaInca What is the $m((0,1])$ mean? Specifically the $m$ part? – Desperate Fluffy Dec 09 '13 at 01:56
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