Measure of borel set minus open <e

[ArcanaNoir]

Homework Statement

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

Homework Equations

\mu 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 X_k 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 \epsilon ?
The only solutions to a problem like this that I have seen are in the context of Lebesgue measure and \mathbb{R}^n, 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.

 

[Jufro]

Going back to set notation A/B= A\cap B* (with * denoting the compliment).

This should make it easier to use like like countable subadditivity and finite intersections for measure.

 

[ArcanaNoir]

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.

 

[pasmith]

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.

 

[Jufro]

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 X_k '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= \cup_{some k's} X_k.
So E*= (\cup_k X_k)* = \cap X_k*

and U/E is simply U\cap(\cap_k X_k) 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.

 

[ArcanaNoir]

If we let E be generated by some of the X_k '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= \cup_{some k's} X_k.
So E*= (\cup_k X_k)* = \cap X_k*

and U/E is simply U\cap(\cap_k X_k) 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 \mu (X_k)&lt;\infty

 

[ArcanaNoir]

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?

 

[Jufro]

Yeah, generating from the X_k 's is kind of nice but isn't really true.

But that is ok because if your generating sets are J_i then each J_i \subset \cup _{for some k's} X_k.

And then you could use that the fact that if A\subset 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.

 

[ArcanaNoir]

Holy crap I solved it.