## Lebesgue probability spaces, part II

This is a continuation of the previous post on the classification of complete probability spaces. Last time we set up the basic terminology and notation, and saw that for a complete probability space ${(X,\mathcal{A},\mu)}$, the ${\sigma}$-algebra ${\mathcal{A}}$ is countably generated mod 0 (which we denoted (CG0)) if and only if the pseudo-metric ${\rho(A,B) = \mu(A\Delta B)}$ makes ${\hat{\mathcal{A}} = \mathcal{A}/{\sim}}$ into a separable metric space. We also considered ${(\hat{\mathcal{A}},\mu)}$ as a measured abstract ${\sigma}$-algebra with non non-trivial null sets; this is the point of view we will continue with in this post.

Our goal is to present the result from [HvN] and [Ro] (see references from last time) that classifies such measured abstract ${\sigma}$-algebras up to ${\sigma}$-algebra isomorphism. To this end, let ${(\Sigma,\mu)}$ be a separable measured abstract ${\sigma}$-algebra with no non-trivial null sets; let ${X}$ be the maximal element of ${\Sigma}$, and suppose that ${\mu(X)=1}$. Note that ${X'}$ is the minimal element of ${\Sigma}$, which would correspond to ${\emptyset}$ if ${\Sigma}$ were a collection of subsets of some ambient space. In the abstract setting, we will write ${0=X'}$ for this minimal element.

An element ${E\in \Sigma}$ is an atom if it has no proper non-trivial subelement; that is, if ${F\in \Sigma, F\leq E}$ implies ${F=0}$ or ${F=E}$. By the assumption that ${\Sigma}$ has no non-trivial null sets, we have ${\mu(E)>0}$ for every atom. Note that for any two atoms ${E\neq F}$ we have ${E\wedge F = 0}$, and so ${\mu(E\vee F) = \mu(E) + \mu(F)}$.

Let ${\Sigma_a \subset \Sigma}$ be the set of atoms; then we have ${\sum_{E\in \Sigma_a} \mu(E) \leq 1}$, so ${\Sigma_a}$ is (at most) countable. Let

$\displaystyle \Sigma_{na} = \{ E\in \Sigma \mid E \wedge F = 0 \text{ for all } F\in \Sigma_a\},$

then ${\Sigma_{na}}$ is non-atomic; it contains no atoms. Thus ${\Sigma}$ can be decomposed as a non-atomic part together with a countable collection of atoms. In particular, to classify ${(\Sigma,\mu)}$ we may assume without loss of generality that ${\Sigma}$ is non-atomic.

Consider the unit interval ${[0,1]}$ with Lebesgue measure; let ${\mathop{\mathcal L}}$ denote the set of equivalence classes (mod 0) of measurable subsets of ${[0,1]}$, and ${m}$ denote Lebesgue measure, so ${(\mathop{\mathcal L},m)}$ is a measured abstract ${\sigma}$-algebra. Moreover, ${(\mathop{\mathcal L},m)}$ is separable, non-atomic, has no non-trivial null sets, and has total weight 1.

Theorem 1 Let ${(\Sigma,\mu)}$ be a separable non-atomic measured abstract ${\sigma}$-algebra with total weight 1 and no non-trivial null sets. Then ${(\Sigma,\mu)}$ is isomorphic to ${(\mathop{\mathcal L},m)}$.

The meaning of “isomorphism” here is that there is a bijection ${T\colon \Sigma \rightarrow \mathop{\mathcal L}}$ that preserves the Boolean algebra structure and carries ${m}$ to ${\mu}$. That is: ${A\leq B}$ iff ${T(A)\leq T(B)}$; ${T(A\vee B) = T(A) \vee T(B)}$; ${T(A\wedge B) = T(A)\wedge T(B)}$; ${T(A')=T(A)'}$; ${T(0)=0}$; and finally, ${m(T(A)) = \mu(A)}$. We are most used to obtaining morphisms between ${\sigma}$-algebras as a byproduct of having a measurable map between the ambient sets; that is, given measurable spaces ${(X,\mathcal{A})}$ and ${(Y,\mathcal{B})}$, a measurable map ${f\colon X\rightarrow Y}$ gives a morphism ${f^{-1} \colon \mathcal{B} \rightarrow \mathcal{A}}$. However, in the abstract setting we currently work in, there is no ambient space, so we cannot yet interpret ${T}$ this way. Eventually we will go from ${(\Sigma,\mu)}$ and (${\mathop{\mathcal L},m)}$ back to the measure spaces that induced them, and then we will give conditions for ${T}$ to be induced by a function between those spaces, but for now we stick with the abstract viewpoint.

We sketch a proof of Theorem 1 that roughly follows [HvN, Theorem 1]; see also [Ro, §1.3]. Given a measurable set ${A\subset [0,1]}$, we write ${[A]\in \mathop{\mathcal L}}$ for the equivalence class of ${A}$; in particular, we write ${[[a,b]]\in \mathop{\mathcal L}}$ for the equivalence class of the interval ${[a,b]}$.

Observe that if ${Q\subset [0,1]}$ is dense, then ${\Sigma}$ is generated by ${\{[0,q]] : q\in Q\}}$; thus we can describe ${T}$ by finding ${\{B_q \in \Sigma \mid q\in Q\}}$ such that

• ${\mu(B_q) = q = m[[0,q]]}$,
• ${B_q \leq B_{q'}}$ whenever ${q\leq q'}$,
• ${\{B_q\}_{q\in Q}}$ generates ${\Sigma}$,

and then putting ${T(B_q) = [[0,q]]}$. This is where separability comes in. Let ${A_1,A_2,\dots \in \Sigma}$ be a countable collection that generates ${\Sigma}$. Put ${T(A_1) = [[0,\mu(A_1)]]}$, so ${A_1 = B_q}$ for ${q=\mu(A_1)}$. Now what about ${A_2}$? We can use ${A_2}$ to define two more of the sets ${B_q}$ by noting that

$\displaystyle A_1 \wedge A_2 \leq A_1 \leq A_1 \vee A_2, \ \ \ \ \ (1)$

and so we may reasonably set ${T(E) = [[0,\mu(E)]]}$ for ${E\in \{A_1 \wedge A_2, A_1, A_2 \vee A_2\}}$.

To extend this further it is helpful to rephrase the last step. Consider

\displaystyle \begin{aligned} C_{00} &= A_1 \wedge A_2, \\ C_{01} &= A_1 \wedge A_2', \\ C_{10} &= A_1' \wedge A_2, \\ C_{11} &= A_1' \wedge A_2', \end{aligned}

and observe that (1) can be rewritten as

$\displaystyle C_{00} \leq (C_{00} \vee C_{01}) \leq (C_{00} \vee C_{01} \vee C_{10}).$

Writing ${\preceq}$ for the lexicographic order on ${\{0,1\}^2}$, we observe that each of these three is of the form ${D_x = \bigvee_{y\preceq x} C_y}$ for some ${x\in \{0,1\}^2}$.

Each ${C_x}$ above is determined as follows: ${x_1}$ tells us whether to use ${A_1}$ or ${A_1'}$, and ${x_2}$ tells us whether to use ${A_2}$ or ${A_2'}$. This generalises very naturally to ${n\geq 2}$: given ${x\in \{0,1\}^n}$, let

\displaystyle \begin{aligned} A_k^x &= \begin{cases} A_k & x_k = 0, \\ A_k' & x_k = 1. \end{cases}\\ C_x &= \bigwedge_{k=1}^n A_k^x. \end{aligned}

Now put ${D_x = \bigvee_{y\preceq x} C_y}$; this gives ${2^n}$ elements ${D_x}$ (the last one is ${X = \bigvee_{y\in \{0,1\}^n} C_y}$, and was omitted from our earlier bookkeeping). These have the property that ${D_x \leq D_w}$ whenever ${x\preceq w}$. Moreover, writing ${\{0,1\}^* = \bigcup_{n\in {\mathbb N}} \{0,1\}^n}$, the collection ${\{D_x \mid x\in \{0,1\}^*\}}$ generates ${(\Sigma,\mu)}$ because ${A_1,A_2,\dots}$ does. Let ${T(D_x) = [[0,\mu(D_x)]]}$; now we have all the pieces of the construction that we asked for earlier.

Putting it all together, let ${Q = \{ \mu(D_x) \mid x\in \{0,1\}^* \}}$. The following are now relatively straightforward exercises.

• ${Q}$ is dense in ${[0,1]}$ since ${(\Sigma,\mu)}$ is non-atomic.
• For each ${q\in Q}$ there is ${w\in \{0,1\}^*}$ with ${\mu(D_x) = q}$; if ${w,x\in \{0,1\}^*}$ such that this holds, then either ${w=x}$ or ${w = x11\cdots 1}$ (or vice versa). For this step we actually need to choose ${A_n}$ a little more carefully to guarantee that each ${C_x}$ is non-trivial.
• The previous step guarantees that ${T}$ is a bijection between ${\Sigma}$ and ${\mathop{\mathcal L}}$.
• Since ${T}$ preserves the order ${\leq}$ on the generating collections, it preserves the ${\sigma}$-algebra structure as well.
• Since ${T}$ carries ${\mu}$ to ${m}$ on the generating collections, it carries ${\mu}$ to ${m}$ on the whole ${\sigma}$-algebra.

Thus we have produced a ${\sigma}$-algebra isomorphism ${T}$, proving Theorem 1. Next time we will discuss conditions under which this can be extended to an isomorphism of the measure spaces themselves, and not just their abstract measured ${\sigma}$-algebras.
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