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Put simply, I would like to know if there is an unbounded open set (of $\mathbb{R}^n$) whose characteristic function is integrable in extended sense. I get the suspicion that something like the area between the graph of $f(x)=\frac{1}{x^2}$ for $1<x<\infty$ in $\mathbb{R}^2$ would do the trick, but I'm stuck extending this to higher dimensions.

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up vote 3 down vote accepted

Expanding on Dennis's answer:

Let $n \geq 2$ and let $f: \mathbb{R}^{n-1} \to [0,\infty)$ be a continuous integrable function whose support is all of $\mathbb{R}^{n-1}$. Its subgraph $U = \{(x,y) \in \mathbb{R}^{n-1}\times \mathbb{R}\,:\,0 \leq y \leq f(x)\}$ is an example of the kind of set you're looking for. Indeed, by Tonelli's theorem,

$$\mu(U) = \int_{\mathbb{R}^{n}} \chi_{U}\, d\mu^{n} = \int_{\mathbb{R}^{n-1}} \int_{\mathbb{R}} \chi_{U}\, d\mu^{1} \, d\mu^{n-1} = \int_{\mathbb{R}^{n-1}} f \, d\mu^{n-1} \lt \infty$$

since Lebesgue measure $\mu^{n}$ on $\mathbb{R}^{n}$ is the product measure $\mu^{n-1} \times \mu^{1}$.

To make this completely explicit, take e.g. $f(x) = e^{-\|x\|^2}$ whose integral is $\pi^{(n-1)/2}$ (after Poisson).

A way of constructing an example that works for all $n \geq 1$: let $x_{k} = (k,0,\ldots,0)$ for $k =1,2,\ldots$. Choose an open ball or cube $U_{k}$ of volume $1/{2^k}$ centered around $x_{k}$. Then the union $U = \bigcup U_{k}$ will have volume $1$ in $\mathbb{R}^{n}$ since the sets $U_{k}$ have pairwise empty intersection.

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Clearly, for any continuous function $f:[a,\infty)\to\mathbb{R}^+$ with unbounded support and $\int_a^\infty f(x)dx <\infty$ you can define $U=\{(x,y)|a<x, 0< y< f(x)\}$. $U$ will be open and unbounded and clearly $\chi_U(x)$ will be integrable.

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You should present it in a clearer way, "$f$ whose support is unbounded and $\int f dx<\infty$ ... " – Asaf Karagila May 16 '11 at 9:04
@Asaf: Remark accepted, answer updated :) – Dennis Gulko May 16 '11 at 9:06
This answer doesn't directly address the case of $\mathbb{R}^n$. You get an example by taking (the positive part of) the subgraph of any old non-negative integrable function on $\mathbb{R}^{n-1}$ with unbounded support. – t.b. May 16 '11 at 9:10
Theo- thanks, I was able to abstract to that from the given material. Explicitly stating that would be good, though. – Jadmrial May 16 '11 at 9:25

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