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Construct a nontrivial example of a Riemann integrable function (on a bounded rectangle) that is discontinuous on a dense subset of the rectangle.

A (trivial) example would be to redefine a nice function like $f(x) = x^2$ for $0< x< 1$ on a null set.

Christmas Bunny
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    What notion of Riemann integrability are you using? – Cameron Williams Nov 14 '13 at 06:24
  • My book's definition: Suppose $f$ is a bounded function on a special rectangle I. Then $f$ is Riemann integrable if for every $\epsilon>0$ there exist step functions $\tau$ and $\sigma$ on I such that $\sigma\le f\le \tau$ and $\int_I(\tau-\sigma)d\lambda<\epsilon$ – Christmas Bunny Nov 14 '13 at 06:33
  • The general theorem is that the precise class of Riemann integrable bounded functions on a closed interval consists of those whose discontinuity sets have Lebesgue measure zero (i.e. are null sets, which concept seems to predate Lebesgue). – Ryan Reich Nov 14 '13 at 06:39

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The following is a favorite of mine (due to the fact that it nicely demonstrates the workings of Funbini's theorem); I can't currently remember where I fished it from, but it's likely it was one of Rudin's textbooks: $f:[0,1]^2\to[0,1]$ defined $$f(x,y) = \begin{cases}1 & x\not\in\mathbb{Q}\vee y\not\in\mathbb{Q}\\ 1-\frac{1}{q} & x,y\in\mathbb{Q}, x=\frac{p}{q}, \gcd(p,q)=1\end{cases}$$ Naturally, it's not different from your original suggestion (nor can it be, due to Lebesgue's integrability theorem).

Jonathan Y.
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  • So in this case what is the dense set? Thank you. – Christmas Bunny Nov 14 '13 at 06:47
  • @Yifeng Xu: This is a variant of the Thomae function. See here http://math.stackexchange.com/questions/530097/proof-of-continuity-of-thomae-function-at-irrationals –  Nov 14 '13 at 06:59
  • $\mathbb{Q}^2\cap[0,1]^2$, as $f$ has the limit $1$ in every point. – Jonathan Y. Nov 14 '13 at 07:01
  • @John, many thanks. I knew I was blundering the credit. – Jonathan Y. Nov 14 '13 at 07:03
  • So is the Thomae function Riemann integrable? Is the function $f(x,y)$ defined above also Riemann integrable? Why or why not? Thanks. – Christmas Bunny Nov 14 '13 at 22:16
  • Both are Riemann integrable. As mentioned above (an as @RyanReich also said), Lebesgue(?) showed that Riemann-integrable functions are exactly those whose set of discontinuities is null (and as RyanReich said, such sets were already studied as those that can be covered by countably many rectangles with arbitrary small overall volume). – Jonathan Y. Nov 14 '13 at 22:24