1

I was reading about the different metric vs. abstract open set definitions of topologies and wondered, whether it is possible to define a measure on ($\mathbb{R}^n,\mathcal{B}^n)$, such that this measure is also a metric.

Of course, in $\mathbb{R}$ the Lebesgue measure $\lambda$ and the Euclidean distance $d$ coincide, in the sense that for any interval $[a,b)$, we have $\lambda[a,b)=d(a,b)$. But with $n\geq2$ this is obviously not the case.

I guess we would define a measure and then prove that it also satisfies the definition of a metric, but I don't know where to start in constructing such a measure.

mltm97
  • 53
  • What about the one-dimensional measure on $\mathbb{R}^n$; in that it would give the length of any curve in $\mathbb{R}^n$ and is defined on all Borel sets – almosteverywhere Jan 21 '22 at 21:38
  • 3
    I don't understand your question. How can a measure be a metric? What properties would you like such a measure to have on $\mathbb{R}^n$? – Michael Albanese Jan 21 '22 at 21:39
  • I assumed it to mean that @mltm97 wants the measure of lines between two points to be the distance between them? – almosteverywhere Jan 21 '22 at 21:41
  • Thanks @almosteverywhere, that was indeed what I was looking for! But I also see where my description of the problem was unclear: of course we would need two different functions, one defined on the Borel sets $\mathcal{B}^n$, one defined on $\mathbb{R}$, such that they take on the same value for the respective set or points. The one-dimensional measure definitely suits that. – mltm97 Jan 21 '22 at 21:49
  • I'm sorry I meant the one dimensional Hausdorff measure on $\mathbb{R}^n$. I forgot the word Hausdorff. – almosteverywhere Jan 21 '22 at 22:01
  • 1
    I agree that this question is too ambiguous. A measure would be a function $\Sigma \to [0,\infty]$ where $\Sigma$ is a $\sigma$-algebra over $\mathbb{R}^n$, while a metric would be a function $\mathbb{R}^n\times\mathbb{R}^n\to [0,\infty)$. It is unclear how you would want these two functions to coincide. – Jacob Jan 21 '22 at 22:08
  • It seems OP just wants the distance between any pair of points to equal the measure of the line segment joining them. But if line segments in ${\bf R}^n$ have positive measure (for $n>1$) then higher dimensional subsets will al have infinite measure, so doesn't seem like a very useful concept to me. – Gerry Myerson Jan 21 '22 at 22:23
  • See here – Jean Marie Jan 21 '22 at 22:54

1 Answers1

2

A measure is not a distance and the converse is not true as well, simply by definition. The best example I can think of is the following where you use a measure on $\mathbb{R}^n$ to actually define a distance between sets.

Denote with $|\cdot|$ the $n-$dimensional Lebesgue measure in $\mathbb{R}^n$. Fix a set $A \subset \mathbb{R}^n$, a point $x \in \partial A$, $H_x$ the upper hyperplane passing through $x$ and a radius $r$. Define then the following quantity:

$$D(x,r):= \frac{|(A \triangle H_x) \cap B_{r}(x)|}{\alpha_n r^n}$$

where $\alpha_n$ is the $n-$dimensional measure of the unit ball.

If you draw you see that the following quantity is exploiting a measure in $\mathbb{R}^n$ to define a distance (up to measure zero) between any set $A$ and the hyperplane passing through point $x$ (if we let them vary).

Son Gohan
  • 4,517