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For $\mathbf{I}$ a generalized rectangle in $\mathbb{R}^{n}$, define $f : \mathbf{I} \rightarrow \mathbb{R}$ to be the function with constant value $1$. Find a subset $D$ of $\mathbf{I}$ such that the restriction $f : D \rightarrow \mathbb{R}$ is not integrable.

I was thinking of taking $D$ to be the set of points in $\mathbf{I}$ with all $n$ components rational and coming up with a density argument to prove non-integrability, but I haven't been able to do so.

Also, I've learned about Jordan content, but I haven't learned about measure zero sets.

I haven't been able to make much progress, and I would appreciate some sort of help.

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  • $\begingroup$ Seems like your $D$ is in the right direction, because what would a Riemann integral over a discrete set be if not the sum of the values of the function at its points? $\endgroup$ Apr 26, 2019 at 7:00

1 Answer 1

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The Riemann integral $\int_D f$ is by definition $\int_I f \chi_D$ where $\chi_D$ is the indicator function taking the value $1$ for $x \in D$ and $0$ otherwise.

If $D$ is the subset of $I$ where for $x = (x_1,\ldots,x_n)$ the first component $x_1$ is rational, then $f \chi_D$ is not Riemann integrable. Any subrectangle of a partition of $I$ contains points in and not in $D$ where the integrand takes the values $1$ and $0$, respectively. Hence all lower sums equal $0$ and all upper sums equal the content of $I$ and the difference cannot be made arbitrarily small.

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