The Unapologetic Mathematician

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