Register to reply

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

by Sarcasticus
Tags: completion, mass, measure, point, theory
Share this thread:
Oct14-09, 08:58 PM
P: 2
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!
Phys.Org News Partner Science news on
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Oct14-09, 09:40 PM
Sci Advisor
HW Helper
P: 25,246
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'.

Register to reply

Related Discussions
Measure Theory General Math 17
So what if I was in motion and tried to measure my own mass? Special & General Relativity 9
Measure theory Calculus & Beyond Homework 2
Measure Theory Academic Guidance 5
Does time dilation occur even without a point of reference to measure motion? General Physics 9