Unbounded *open and connected* set with finite Lebesgue measure

Question

I need easy examples of unbounded open and connected subsets of $\mathbb{R}^n$ with finite Lebesgue measure.

The example that I have in mind belongs to the class of examples which are easy to draw but longer to write down (imagine a countable family of open balls in the plane whose radius decreases as $1/n$, the centers move towards infinity but with a velocity that preserves the overlap so that the connectedness is maintained).

Answer 1

There clearly is no such example if $n=1$.

For $n=2$ a simple example would be $\{(x,y):|y|<1/(x^2+1)\}$. You could give an analogous example for $n>2$.

Now it turns out that "I need easy examples of unbounded open and connected subsets of $\mathbb{R}^n$ with finite Lebesgue measure." is not what the OP actually wanted; he says "It is interesting to me to understand under which conditions on a metric measure spaces a connected open set is bounded if and only if it has finite measure."

Of course it's trivial to concoct a condition C such that C implies there is an unbounded connected open set of finite measure. The C I have in mind seems unreasonably strong to me, making the proof trivial, but I suspect it nonetheless holds in all the metric measure spaces the OP has in mind:

Triviality. Suppose $(X,d)$ is an unbounded measure space and $\mu$ is a Borel measure on $X$ such that for every $x,y\in X$ there exists a path from $x$ to y of measure zero. If every ball is connected and every ball has finite measure then then there is an unbounded connected open set.

Regarding my comment on whether that condition is unreasonably strong or not: Imagine a path in the plane. Of course that path can have positive two-dimensional Lebesgue measure. If you're the sort of person who thinks of a path in the plane with positive area as rare or pathological then the conditions in the triviality seem quite reasonable. Otoh if you're me you think that a "typical" path in the plane has positive measure, so the triviality seems quite weak, having unreasonably strong hypotheses. (Otoh assuming every ball is connected and has finite measure seems perfectly reasonable; certainly not necessary, but the sort of thing you'd expect to have to assume.)

Note to be clear, when I say there exists a path from $x$ to $y$ of measure $0$ I mean there exists a continuous $\gamma:[0,1]\to X$ with $\gamma(0)=x$, $\gamma(1)=y$, and $\mu(\gamma([0,1])=0$; in particular the measure of the path $\gamma$ is $\mu(\gamma([0,1])$.

Proof. Say $(x_n)$ is a sequence with $d(x_n,x_0)\to\infty$. Say $\gamma_n$ is a path from $x_n$ to $x_{n+1}$ of measure zero. Let $K_n=\gamma_n([0,1])$. For $r>0$ let $$V_n(r)=\bigcup_{p\in K_n}B(p,r).$$Now $K_n$ is compact, so $$\bigcap_{r>0}V_n(r)=K_n.$$And $V_n(1)$ is bounded, so $\mu(V_n(1))<\infty$, hence $$\lim_{r\to0}\mu(V_n(r))=\mu(K_n)=0.$$So you can choose $r_n>0$ such that $\mu(V_n(r_n))<1/n^2$, and now $$V=\bigcup_{n=0}^\infty V_n(r_n)$$is an unbounded connected open set with finite measure.

Note for anyone who detects a whiff of snarkiness here: It's irritating when someone posts a question, and then comments on an answer saying that's not what (s)he wanted! It wastes people's time. If "I need easy examples of unbounded open and connected subsets of $\mathbb{R}^n$ with finite Lebesgue measure." was really what the OP wanted then the first two sentences of this post would have made him happy - those two sentences give a much "easier" example of what he said he wanted.

Answer 2

For $n \ge 2$ take

$$U =\{(x_1, \dots x_n) \in \mathbb R^n \mid x_1 >0 \textrm{ and } x_2^2 + \dots x_n^2 < e^{-x_1}\}$$

$U$ is open as it is the intersection of two open subsets. Namely the half open plane $\{(x_1, \dots x_n) \in \mathbb R^n \mid x_1>0\}$ and the inverse image of the open subset of the reals $\{x \in \mathbb R \mid x>0\}$ under the continuous map $(x_1, \dots , x_n) \mapsto -(x_2^2 + \dots x_n^2) + e^{-x_1}$.

$U$ is connected as it is path connected.

And $\mu(U)$ is finite.