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One says a bounded $f$ is Riemann integrable on [a,b] if the Upper and lower Riemann integrals are equal. Another sufficient condition for Riemann integrablity is that the set of discontinuity of $f$ must be countable set. The following function is continuous only at $x=1/2$ and so the set of discontinuity of $f$ is uncountable set. This shows that $f$ is not Riemann integrable. But I want to justify this by computing the upper and lower Riemann integrals. Any one who can help me how to get the upper and lower Riemann integrals. Let $f\colon[0,1]\rightarrow\mathbb{R}$ such that $$f(x)=\left\{\begin{array}{c} x, x\in\mathbb{Q}\\ 1-x, x\notin\mathbb{Q} \end{array}\right.$$ Show that $f$ is not Riemann integrable on $[0,1].$

Przemysław Scherwentke
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  • Suppose I give you the partition : $0,\frac 1n,\frac 2n,...,\frac{n-1}n,1$ of $[0,1]$. With this partition, can you compute the upper and lower Riemann sum? Now let $n \to \infty$. – Sarvesh Ravichandran Iyer Dec 22 '17 at 05:12
  • @rtybase This $f(x)$ is continuous only at $x=1/2$. – Przemysław Scherwentke Dec 22 '17 at 05:19
  • You're right that this function isn't Riemann integrable, but you should know that the set of discontinuities being uncountable does not allow you to conclude that. There are functions with an uncountable number of discontinuities that are Riemann integrable, for example here. – JonathanZ Dec 22 '17 at 05:50
  • As you said, the set of discontinuity being countable is sufficient for Riemann integrability. It is not necessary. So the fact that your $f$ has an uncountable set of discontinuities does not show that $f$ is not Riemann integrable. – Robert Israel Dec 22 '17 at 05:50
  • Robert Israel and JonathanZ Yes you are right it is my mistake rather the set of discontinuity should have measure or content zero. Thanks – Math Dec 22 '17 at 06:52
  • I hate to keep nit-picking your statements, @Math, as they don't really affect the answers to this particular problem, but, y'know, we are doing math here. You should know that measure zero and content zero are not the same thing (example), and the criterion for being integrable uses measure zero. – JonathanZ Dec 22 '17 at 13:51

2 Answers2

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Let

$$g(x) = \left\{\begin{array}{lll} 1-x & if & 0\leq x\leq 1/2 \\ x & if & 1/2<x\leq 1 \\ \end{array}\right.$$

and

$$h(x) = \left\{\begin{array}{lll} x & if & 0\leq x\leq 1/2 \\ 1-x & if & 1/2<x\leq 1 \\ \end{array}\right.$$

Both $g$ and $h$ are continuous, and so Riemann integrable. Now show that the upper integral for $f$ is equal to the integral of $g$ and the lower integral of $f$ is equal to the integral of $h$. Here you need to use the following facts. 1. $h\leq f\leq g$\ 2. Both $\mathbb{Q}\cap[0,1]$ and $[0,1]\setminus \mathbb{Q}$ are dense in $[0,1]$. Therefore every interval $[a,b]\subseteq [0,1]$ with $a<b$ will intersect both $\mathbb{Q}\cap[0,1]$ and $[0,1]\setminus \mathbb{Q}$, so

$$ \sup_{x\in [a,b]} f(x) = \sup_{x\in [a,b]} g(x)~~{\rm and}~~ \inf_{x\in [a,b]} f(x) = \inf_{x\in [a,b]} h(x) $$

JH vd Walt
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HINT: The upper Riemann integral is the integral of $\max(x,1-x)$ on $[0,1]$ and the lower: $\min(x,1-x)$. (Why?)

Przemysław Scherwentke
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