What is a tube in $\mathbb{R}^n$?

Question

Lebesgue's differentiation theorem states that if $x$ is a point in $\mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a Lebesgue integrable function, then the limit of $\frac{\int_B f d\lambda}{\lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.

But I just came across a journal paper which shows that if you take the collection of all "tubes" in $\mathbb{R}^n$ oriented in certain directions, then the Lebesgue differentiation holds true for this collection for $L^p$ functions with $p>1$. But my question is, what exactly is a tube in $\mathbb{R}^n$ as the term is used in this paper? The paper doesn't provide any definition as far as I can tell.

Is it like a cylinder, or what?

Answer 1

Based on reference 19 in the paper you linked, on page 224 of the corresponding article (or page 11 in the .pdf), it looks like a tube is the same as an "oriented" cylinder, and is determined by a direction $\gamma\in S^{n-1}$, and two parameters: a height, and a radius of the cylinder itself.

Answer 2

It should be a $n$ dimensional cylinder right? So pick a disk in $\mathbb{R}^{n-1}$ then map it's point set, $(x_1,\dots,x_{n-1},0)$ to $(x_1,\dots,x_{n-1},z)$ where $z \in \mathbb{R}$.

For the usual space, $n=3$, you get a regular cylinder which has it's base disk as a subset of the xy-plane.