0

If I have a topological space $(E,t)$ and define a measure on $E$ will that imply that I have a metric space?

By that I mean defining a sigma-algebra $S$ in $E$ and then a measure function $m:S \to \overline{\Bbb R}$, then will there be an metric function $n$ such that $(E,n)$ is a metric space?

How can I properly verify this?

Any help will be appreciated.

Ben Grossmann
  • 234,171
  • 12
  • 184
  • 355
  • What do you intend to do with $t$? – Fabio Somenzi Dec 30 '16 at 17:47
  • a topology in $E$ –  Dec 30 '16 at 17:50
  • Of course. I'm asking what you plan to do with it in your construction. Do you ignore it? – Fabio Somenzi Dec 30 '16 at 17:51
  • yes. what do you mean by "my construction"? my attempts? –  Dec 30 '16 at 17:52
  • 1
    I mean the definition of the $\sigma$-algebra and the measure function. If you ignore $t$, you are essentially starting from a set. – Fabio Somenzi Dec 30 '16 at 17:54
  • @Crostul. Is the wikipedia definition wrong then? –  Dec 30 '16 at 17:55
  • 1
    A metric space is not a measure space. They are two different things. Why should you have a metric when you have a measure? – Crostul Dec 30 '16 at 17:57
  • @Fabio Somenzi. Oh, sorry, I misread your first comment with "what is $t$?". Actually, while reading the second comment of Crostul I recognized that maybe I'm not healthy enough to think at this right now. Thanks. But please (moderator) don't exclude the topic because I'll be back later. –  Dec 30 '16 at 17:59
  • this question seems relevant – Ben Grossmann Dec 30 '16 at 18:10
  • @Omnomnomnom, it was VERY relevant and informative. I don't need this question anymore. Thank you very much and to anyone in this thread. –  Dec 30 '16 at 19:45
  • 1
    You can turn the measurable $\sigma$-algebra module the null sets into a metric space (if the measure is finite), using $d(A,B) = \mu(A \Delta B)$, using the symmetric difference operator. There is no clear way to make the space itself a metric space that I heard of. – Henno Brandsma Dec 31 '16 at 15:26

1 Answers1

0

you won't have a metric space because in a metric space you can calculate DISTANCES between points and eventualy sets ( you can define that as the inf of the distances of points x in X and y in Y) . On the other hand , defining a measure allows you to calculate the SIZE of the sets which are in your sigma algebra.

Ariel Serranoni
  • 504
  • But what do I have in a metric space that makes me be able to calculate distances? –  Dec 30 '16 at 17:54
  • 2
    @Alnitak A metric. – Theoretical Economist Dec 30 '16 at 18:06
  • For a finite measure $\mu$, convergence in measure is pseudo-metrizable. – nullUser Dec 30 '16 at 18:15
  • Theoretical Economist, by calculating you meant then the possibility of assuming a value in a field for two points, but not a method on how to do it? I guess I don't know what calculate means. I can see why I can't take the infimum mentioned in the answer. Is that what calculate means? –  Dec 30 '16 at 18:18
  • Calculate can mean different things depending on the context you are working with ! That is , depending on the metric defined on your metric space . A metric can be any function satisfying : d(x,y)=0 if and only if x=y , d(x,y)=d(y,x) , d(x,z)<=d(x,y)+d(y,z), d(x,y)>=0.. sorry if the notation was a bit confusing , but d is a function defined on the space you are turning into a metric one and x,y and z are points in that space – Ariel Serranoni Dec 31 '16 at 00:16