6

Probably a simple question, but I wonder about the following. I know that if a function $f : \mathbb{R} \rightarrow \mathbb{R} $ is (Riemann)integrable, then it is bounded. I wonder if I can generalize this to functions on $\mathbb{R}^3$ (now for an ingral over a volume).

It seemed logical to me that, because this theorem is true on $\mathbb{R}$, that it should be true on $\mathbb{R}^n$. But apparently it isn't, because the integral in theorem 10.1 of the document http://people.maths.ox.ac.uk/kirchhei/section_1008.pdf converges according to the theorem. The function under the integral is not bounded however. So can't I generalize the theorem that a integrable function should be bounded?

Rayman
  • 227

2 Answers2

10

It is not a theorem, it is part of the definition. The Riemann integral is defined for bounded functions on a bounded domain. If the function, the domain or both are unbounded, then the integral may exist as an improper integral.

Julián Aguirre
  • 76,354
6

For an unbounded smooth positive function whose improper Riemann integral exists, try the function $f$ defined by $$ f(x)=\sum_{n\geqslant1}n\mathrm e^{-n^5(x-n)^2}. $$ Note that $$ \int_\mathbb Rf(x)\mathrm dx=\sum_{n\geqslant1}n\int_\mathbb R\mathrm e^{-n^5(x-n)^2}\mathrm dx=\sum_{n\geqslant1}\frac1{n^{3/2}}\int_\mathbb R\mathrm e^{-x^2}\mathrm dx, $$ hence $f$ is integrable, while $f(n)\geqslant n$ for every $n$.

Did
  • 279,727