Measure Theory - The completion of R^2 under a point mass measure


by Sarcasticus
Tags: completion, mass, measure, point, theory
Sarcasticus
Sarcasticus is offline
#1
Oct14-09, 08:58 PM
P: 2
Hello;
1. The problem statement, all variables and given/known data
Let [tex]\mathcal{A}[/tex] be the [tex]\sigma[/tex]-algebra on [tex]\mathbb{R}^2[/tex] that consists of all unions of (possibly empty) collections of vertical lines. Find the completion of [tex]\mathcal{A}[/tex] under the point mass concentrated at (0,0).

2. Relevant equations

1st: Completion is defined as follows: Let [tex](X, \mathcal{A})[/tex] be a measurable space, and let [tex]\mu[/tex] be a measure on [tex]\mathcal{A}[/tex]. The completion of [tex]\mathcal{A}[/tex] under [tex]\mu[/tex] is the collection [tex]\mathcal{A}_{\mu}[/tex] of subsets A of X for which there are sets E and F in [tex]\mathcal{A}[/tex] such that
1) E [tex]\subset[/tex] A [tex]\subset[/tex] F, and
2) [tex]\mu[/tex](F-E) = 0.

2nd: A point mass measure concentrated at x is a measure [tex]\delta_x[/tex]defined on a sigma-algebra [tex]\mathcal{A}[/tex] such that, for any [tex]A \in \mathcal{A}[/tex], [tex]\delta_x(A) = 1[/tex] if [tex]x \in A[/tex] and [tex]\delta_x(A) = 0[/tex] otherwise.

3. The attempt at a solution

Here's my answer: Let [tex](\mathcal{A})_{\delta}[/tex] denote the completion of [tex]\mathcal{A}[/tex] under the pt. mass concentrated at (0,0) and let [tex]\delta[/tex] denote said measure. Then, for any set [tex]A \in \mathcal{A}[/tex], we have
[tex]A \subset A \subset A[/tex] and [tex]\delta(A-A)=0[/tex] always, so [tex]\mathcal{A} \in (\mathcal{A})_{\delta}.[/tex]
Consider any set [tex]A \in (\mathcal{A})_{\delta}[/tex]; then there exist sets E, F belonging to [tex]\mathcal{A}[/tex] such that [tex]E \subset A \subset F[/tex] and [tex]\delta(F - E) = 0[/tex]. Which means that either both E and F contain a line intersecting the origin, or neither does. This mean A will follow suit and, further, [tex]A \subset F[/tex] means that [tex]A \in \mathcal{A}[/tex] and hence [tex](\mathcal{A})_{\delta} \subset \mathcal{A}[/tex] and thus [tex]\mathcal{A} = (\mathcal{A})_{\delta}[/tex]

Except, this means the completion of any sigma algebra under a point mass measure will again be the sigma algebra. And, if this were the case, why wouldn't they just give us the general question in the first place, instead of a bunch of questions about it? (Only one displayed here.)
Hence, I think my answer mucks up somewhere.

Thanks in advance!
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Dick
Dick is offline
#2
Oct14-09, 09:40 PM
Sci Advisor
HW Helper
Thanks
P: 25,175
Think about a subset of R^2 that isn't a union of vertical lines but doesn't contain (0,0). Isn't that in the completion? I think the point is is to complete a sigma algebra of measureable sets by adding all sets of 'measure zero'.


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