Unbounded open set whose characteristic function is integrable in extended sense 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.
 A: 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.
A: 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.
