I was reading about the different metric vs. abstract open set definitions of topologies and wondere

Banguizb

Banguizb

Answered question

2022-07-08

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 ( R n , B n ), such that this measure is also a metric.
Of course, in R the Lebesgue measure λ and the Euclidean distance d coincide, in the sense that for any interval [a,b), we have λ [ a , b ) = d ( a , b ). But with n 2 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.

Answer & Explanation

Nirdaciw3

Nirdaciw3

Beginner2022-07-09Added 20 answers

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 R n to actually define a distance between sets.
Denote with | | the n−dimensional Lebesgue measure in R n . Fix a set A R n , a point x A, H x the upper hyperplane passing through x and a radius r. Define then the following quantity:
D ( x , r ) := | ( A H x ) B r ( x ) | α n r n
where α n is the n−dimensional measure of the unit ball.
If you draw you see that the following quantity is exploiting a measure in 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).

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