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.

About Vaughn Climenhaga

I'm an assistant professor of mathematics at the University of Houston. I'm interested in dynamical systems, ergodic theory, thermodynamic formalism, dimension theory, multifractal analysis, non-uniform hyperbolicity, and things along those lines.
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1 Response to Lebesgue probability spaces, part II

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