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Measure of borel set minus open <e

  • Thread starter ArcanaNoir
  • Start date
  • #1
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Homework Statement


We have a metric space [tex] X=\cup X_k [/tex] where [itex] X_k\subset X_{k+1}[/itex] and each [itex] X_k[/itex] is open. Show that for any Borel set E, there is an open set U such that [itex] \mu (U-E)<\epsilon [/itex]. (Its supposed to be "U \ E".)


Homework Equations



[itex] \mu [/itex] is a measure, so probably the important thing is countable subadditivity.
A borel set is a set generated by countable union, countable intersection, and relative complement of open sets.

The Attempt at a Solution



I know that if I have an open set I can intersect it with an [itex] X_k[/itex] and still have an open set... In this way I believe I can chop up any open set to countable pieces. But how can I get the difference in measure to be less than [itex] \epsilon [/itex] ?
The only solutions to a problem like this that I have seen are in the context of Lebesgue measure and [itex]\mathbb{R}^n [/itex], but I cannot use this context. I must prove it in a general metric space with a general measure. Also, if anyone can recommend a book that has a good treatment of general measures instead of focusing on Lebesgue, I would appreciate the suggestion. So far I have Royden and Folland.
 

Answers and Replies

  • #2
92
8
Going back to set notation A/B= A[itex]\cap[/itex] B* (with * denoting the compliment).

This should make it easier to use like like countable subadditivity and finite intersections for measure.
 
  • #3
768
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Okay, good point, I like that. But... I still don't know how to get an open set (or countably many open sets) tightly wrapped around E.
 
  • #4
pasmith
Homework Helper
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Okay, good point, I like that. But... I still don't know how to get an open set (or countably many open sets) tightly wrapped around E.
You're working in a metric space so you should consider looking at a collection (or more than one collection) of open balls.
 
  • #5
92
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I'm just guessing because I haven't actually thought this problem out in detail but....

If we let E be generated by some of the Xk 's. I would think for a general borel set it doesn't have to be the X's but I'm not sue if the logic would flow nicely otherwise.


Then E= [itex]\cup[/itex]some k's Xk.
So E*= ([itex]\cup[/itex]k Xk)* = [itex]\cap[/itex] Xk*

and U/E is simply U[itex]\cap[/itex]([itex]\cap[/itex]k Xk) so this leaves you with a countable number of intersections.

And then you can try and work it from here. I think you may have to assume at some point that X has finite measure; not sure though.
 
  • #6
768
4
If we let E be generated by some of the Xk 's. I would think for a general borel set it doesn't have to be the X's but I'm not sue if the logic would flow nicely otherwise.


Then E= [itex]\cup[/itex]some k's Xk.
So E*= ([itex]\cup[/itex]k Xk)* = [itex]\cap[/itex] Xk*

and U/E is simply U[itex]\cap[/itex]([itex]\cap[/itex]k Xk) so this leaves you with a countable number of intersections.
But what if E is not a union of Xk's? What if E is only parts of the Xks?

And then you can try and work it from here. I think you may have to assume at some point that X has finite measure; not sure though.
Right, I forgot to mention [itex] \mu (X_k)<\infty [/itex]
 
  • #7
768
4
You're working in a metric space so you should consider looking at a collection (or more than one collection) of open balls.
Yeah, but how would I get only countably many open balls?
 
  • #8
92
8
Yeah, generating from the Xk 's is kind of nice but isn't really true.

But that is ok because if your generating sets are Ji then each Ji [itex]\subset[/itex] [itex]\cup[/itex] for some k's Xk.

And then you could use that the fact that if A[itex]\subset[/itex] B then μ(A)≤ μ(B).

I think that should be enough to get the ball rolling.

The goal, in my mind, is to find what the measure of set U has to be and so that you can try and construct it.

Also, a trivial answer is U = ∅. Cheap, but finding a solution sort of proves that a solution has to exist.
 
Last edited:
  • #9
768
4
Holy crap I solved it. :surprised
 

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