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

1. Oct 14, 2009

### Sarcasticus

Hello;
1. The problem statement, all variables and given/known data
Let $$\mathcal{A}$$ be the $$\sigma$$-algebra on $$\mathbb{R}^2$$ that consists of all unions of (possibly empty) collections of vertical lines. Find the completion of $$\mathcal{A}$$ under the point mass concentrated at (0,0).

2. Relevant equations

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

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

3. The attempt at a solution

Here's my answer: Let $$(\mathcal{A})_{\delta}$$ denote the completion of $$\mathcal{A}$$ under the pt. mass concentrated at (0,0) and let $$\delta$$ denote said measure. Then, for any set $$A \in \mathcal{A}$$, we have
$$A \subset A \subset A$$ and $$\delta(A-A)=0$$ always, so $$\mathcal{A} \in (\mathcal{A})_{\delta}.$$
Consider any set $$A \in (\mathcal{A})_{\delta}$$; then there exist sets E, F belonging to $$\mathcal{A}$$ such that $$E \subset A \subset F$$ and $$\delta(F - E) = 0$$. 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, $$A \subset F$$ means that $$A \in \mathcal{A}$$ and hence $$(\mathcal{A})_{\delta} \subset \mathcal{A}$$ and thus $$\mathcal{A} = (\mathcal{A})_{\delta}$$

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.