The Unapologetic Mathematician

Mathematics for the interested outsider

The Measure Algebra of the Unit Interval

Let Y be the unit interval [0,1], let \mathcal{T} be the class of Borel sets on Y, and let \nu be Lebesgue measure. If \{\mathcal{Q}_n\} is a sequence of partitions of the maximal element Y of the measure algebra (\mathcal{T},\nu) into intervals, and if the limit of the sequence of norms \lvert\mathcal{Q}_n\rvert is zero, then \{\mathcal{Q}_n\} is dense.

If \epsilon is a positive number, then we can find some n so that \lvert\mathcal{Q}_n\rvert<\frac{\epsilon}{2}. If E is a subinterval of Y, then let E_1\in\mathcal{Q}_n be the unique interval containing the left endpoint of E. If E_1 also contains the right endpoint, then we can stop. Otherwise, let E_2 be the next interval of \mathcal{Q}_n to the right of E_1, and keep going (at most a finite number of steps) until we get to an interval E_k containing the right endpoint of E. The union U of all the \{E_i\} can overshoot E by at most \frac{\epsilon}{2} on the left and the same amount on the right, and so \rho(E,U)<\epsilon. Thus any interval can be approximated arbitrarily closely by some partition in the sequence. The general result follows, because we can always find a finite collection of intervals whose union is arbitrarily close to any given Borel set.

Now, say that every separable, non-atomic, normalized measure algebra (\mathcal{S},\mu) is isomorphic to the measure ring (\mathcal{T},\nu).

Since the metric space \mathfrak{S}(\mu) has a diameter of 1 — no two sets can differ by more than X, and \mu(X)=1 — we can find a dense sequence \{E_k\} in the space. For each n we can consider sets of the form

\displaystyle\bigcap\limits_{i=1}^n A_i

where either A_i=E_i or A_i=X\setminus E_i. The collection of all such sets for a given n is a partition \mathcal{P}_n. It should be clear that the sequence \{\mathcal{P}_n\} is decreasing, and the fact that \{E_n\} is dense implies that the sequence of partitions is also dense. We thus conclude that \lim\limits_{n\to\infty}\lvert\mathcal{P}_n\rvert=0.

The first partition \mathcal{P}_1 has two elements E_1 and X\setminus E_1. We can define t(E_1)=[0,\mu(E_1)] and t(X\setminus E_1)=(\mu(E_1),1], so that \nu(t(E_1))=\mu(E_1) and \nu(t(X\setminus E_1))=\mu(X\setminus E_1). This gives us a partition \mathcal{Q}_1 of Y that reflects the structure of \mathcal{P}_1. Similarly, we can carve up each of these intervals so that the resulting partition \mathcal{Q}_2 reflects the structure of \mathcal{P}_2. And we can continue, each time subdividing \mathcal{Q}_n into smaller intervals so that the next partition \mathcal{Q}_{n+1} reflects the structure of \mathcal{P}_{n+1}. Since this correspondence preserves measures, it follows that \lim\limits_{n\to\infty}\lvert\mathcal{Q}_n\rvert=0, and the above result then shows that the sequence \{\mathcal{Q}_n\} is dense.

Now, we can extend t from partition elements occurring in \{\mathcal{P}_n\} to finite unions of such elements by sending such a finite union to the finite union of corresponding elements of \{\mathcal{Q}_n\}. This gives an isometry from a dense subset of \mathfrak{S}(\mu) to a dense subset of \mathfrak{T}(\nu). Thus we can uniquely extend it to an isometry \bar{t}:\mathfrak{S}(\mu)\to\mathfrak{T}(\nu). Since t preserves unions and differences, and since these operations are uniformly continuous, it follows that \bar{t} gives us an isomorphism of measure algebras.

August 25, 2010 Posted by | Analysis, Measure Theory | Leave a comment

   

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