A Property of set with finite measure

ntsivanidis
Messages
4
Reaction score
0

Homework Statement



If E has finite measure and \epsilon>0, then E is the disjoint union of a finite number of measurable sets, each of which has measure at most \epsilon.

Homework Equations



The Attempt at a Solution


I proceeded by showing that by definition of measure, there is a finite group of open sets O_i that contain E, whose union has the same measure (and contains E). By taking their closure, by compactness each has an open cover of \epsilon neighborhoods of a finite number of points. The union of these, within each O_i and then across all O_i, contains E.

My problem is i)to ensure the finite number of subsets are disjoint, and ii) to ensure that the union of these sets is equal to E.

Thanks!
 

Attachments

  • Screen shot 2010-11-14 at 3.15.25 PM.png
    Screen shot 2010-11-14 at 3.15.25 PM.png
    24.4 KB · Views: 533
Last edited:
Physics news on Phys.org
Then why don't you take the intersections of your sets with E and then differences and intersections of the sets themselves?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top