# Is every probability space a factor space of the Haar Measure on some group?

Let P be an arbitrary probability space.

I would like to find a compact topological group $G$ so that the Haar probability measure on $G$ admits a measurable map to the probability space $P$.

By a measurable map, I mean a function which lifts measurable sets to measurable sets of the same measure. That is, $f : Q \longrightarrow P$ is measurable when for all measurable $E \subseteq P,$ the set $f^{-1}(E)$ is measurable and satisfies

$$\mu_Q (f^{-1}(E)) = \mu_P (E).$$

I imagine that some enormous product space $(\mathbb{R} / \mathbb{Z})^{\kappa}$ will do. Anyone see a nice way to make this work?

Edit: My hunch that some power of the circle group will work is based on the game twenty questions. One player thinks of an object (a trampoline), and the other players ask a sequence of at most twenty yes/no questions (Can it fly? Is it legal to drive one on the highway? Would it hurt to swallow?). A strategy for twenty questions would consist of an enormous decision tree that says what question to ask next.

Here we're trying to guess a point of $P$ using at most $\kappa$ measurable yes/no questions. The answers come from the coordinates of $[0,1]^{\kappa}$ which may be interpreted according to the probability of a yes. The function $f$ could be built out of a suitable decision tree.

Perhaps a decision tree could be constructed out of a well-ordering of the sigma-algebra of measurable sets.

Update: Thanks for all the help!

Here's the story:

What I really want is to perform convolution in $L^2(P)$. I'll admit, $P$ is not a group, so I guess it's okay to switch to $L^2(G)$ provided that $G$ has $P$ as a factor space. But it turns out even this was too greedy.

Let $S \subseteq G$ be a dense subset with a measure structure inherited from $G$. Now we may perform convolution in $L^2(S)$, even though $S$ is not a group! The continuous functions are dense in $L^2(S)$, so it will suffice to convolve two of them. But any continuous function on $S$ extends uniquely to one on $G$. So we convolve in $L^2(G)$ and then restrict the result to $L^2(S)$.

Because of this, George Lowther's weaker result will suffice for my purposes. After all, a subset of full outer measure is certainly dense. I will accept his answer unless a full answer to the original question materializes.

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Why does any continuous function on $S$ extend (continuously) at all? Or does one know enough about $S$ to know that the space of continuously extendable functions is still dense in $L^2(S)$? –  L Spice Sep 11 '10 at 6:16
Good point. I bet a little point-set topology will fix this up, at least in the special case constructed by George Lowther below. –  John Wiltshire-Gordon Sep 11 '10 at 7:33
S has full outer measure, so the restriction to S gives an isomorphism $L^2(G)\cong L^2(S)$. In fact, if you are only interested in $L^n(P)$ then this is determined by the measure algebra, and Gerald Elgar's suggestion of using Maharam's structure theorem also works. –  George Lowther Sep 11 '10 at 11:37
For examples where it might not be possible to find a measure preserving map G->P you could consider the following nasty cases. (1) As suggested by Gerald Elgar, take P to be a highly nonmeasurable subset of [0,1] with outer measure 1 under the Lebesgue measure. Maybe a Vitali set or, supposing the CH fails, a set with cardinality strictly between $\aleph_0$ and $2^\aleph_0$. –  George Lowther Sep 11 '10 at 11:50
or, what I had in mind, (2) Take the space $2^{\mathbb{R}}$ of zero-one valued functions on the reals, with the 0-1 valued measure which simply assigns probability 1 to any set containing the function with constant value 1. Let P be the restriction of this to the (highly nonmeasurable but full outer measure) subset of $2^{\mathbb{R}}$ consisting of those functions which are zero everywhere outside of a countable set. This space consists of functions which are zero almost everywhere, but equal to 1 almost surely at each point. –  George Lowther Sep 11 '10 at 11:52

It is possible to find the following: A compact abelian group G with Haar measure $\mu_G$, a subset $S\subseteq G$ of full outer Haar measure and a measurable function $f\colon S\to X$ with $\mu_P(E)=\mu_S(f^{-1}(E))$ for measurable $E\subseteq P$. In fact, as you mention, G can be taken to be a large enough product of the circle group.

Here, I am implicitly referring to the sigma algebra $\mathcal{B}(S)\equiv\{S\cap E\colon E\in\mathcal{B}(G)\}$ and $\mu_S$ is the restriction of the Haar measure to $\mathcal{B}(S)$, $\mu_S(S\cap E)=\mu_G(E)$. Ideally, we would like to enlarge S so that it is actually of full measure, then it could be enlarged to all of G. I'm not sure if this is possible though (and suspect that it is not possible in ZFC). The problem is that if $f\colon P\to Q$ is a measure preserving map of probability spaces then $f(P)$ will be of full outer measure, but need not be measurable. If $f^{-1}\colon\mathcal{B}_Q\to\mathcal{B}_P$ is an onto map of their sigma algebras then P and Q are "almost isomorphic" probability spaces. If, however, $f(P)$ is not measurable then f does not have a right inverse (even up to zero probability sets), unless we restrict to the subset $f(P)\subseteq Q$.

In the following, I write 2={0,1}, so that, for a set I, 2^I is the set of {0,1}-valued functions from I. Letting $\pi_i\colon2^I\to\{0,1\}$ be the projection onto the i'th coordinate, the sets of the form $\pi_i^{-1}(S)$ generate a sigma algebra, which I will denote by $\mathcal{E}_I$.

First, we can reduce the problem to that of measures on $2^I$.

Step 1: Given a collection $\{A_i\}_{i\in I}$ of sets generating the sigma algebra on P, construct a probability measure $\mu_I$ on $2^I$ and a measure preserving map $f\colon P\to 2^I$ such that $f^{-1}\colon\mathcal{E}_I\to\mathcal{B}_P$ is onto.

Then, f will have a right inverse $g\colon f(P)\to P$ which is automatically measurable and measure preserving. To construct f, define $f(p)\in 2^I$ by $f(p)(i)=1$ if $p\in A_i$ and =0 otherwise. It can be checked that $\mu_I(S)=\mu(f^{-1}(S))$ for $S\in\mathcal{E}_I$ satisfies the required properties.

Now, I will use $G^I=(\mathbb{R/Z})^I$, which is a compact abelian group with Haar measure $\mu_{G^I}$, which is the product of the uniform measure on the circle $G=\mathbb{R/Z}$.

Step 2: Construct a measure preserving map $f\colon G^I\to 2^I$.

Once this map is constructed, putting it together with step 1 gives what I claimed.

For any $J\subseteq I$, use $\pi_J\colon 2^I\to2^J$ and $\rho\colon G^I\to G^J$ to denote restriction to J. Also use $\mu_J(S)=\mu_I(\pi_J^{-1}(S))$ for the induced measure on $2^J$. Zorn's lemma guarantees the existence of a subset $J\subset I$ and measure preserving map $f\colon G_J\to 2^J$ which is maximal in the following sense: for $J\subseteq K\subseteq I$ and measure preserving $g\colon G^K\to2^K$ with $\pi_J\circ g=f\circ\rho_J$ then K=J.

Then, J=I and we have constructed the required map. If not, choose any $k\in I\setminus J$, $K=J\cup\{k\}$ and define $h_k\colon G^K\to 2$ as follows (I let $A_k=\pi^{-1}_k(1)\subseteq2^I$).

$$h_k(x)=\begin{cases} 1,&\textrm{if }0\le x_k\le\mu_I$$A_k\mid\mathcal{E}_J$$\_{f(x)}\\\\ 0,&\textrm{otherwise}. \end{cases}$$

Defining $g\colon G^K\to2^K$ by $g(x)_i=f(x)_i$ for $i\in J$ and $g(x)_k=h_k(x)$ gives a measure preserving map extending f, and contradicting the maximality of J.

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It turns out that this answer is strong enough for my purposes (see the update above). I won't accept it right away, though, in case someone solves the actual problem. Thanks for your help! –  John Wiltshire-Gordon Sep 11 '10 at 5:54

I don't know about an arbitrary probability space, but if $S$ is a Borel space, and $\mu$ is a probability measure on $S$, then there exists a map on $f:((0,1],\lambda)\to S$ whose law is $\mu$. Here $(0,1]$ is the circle group and $\lambda$ the Lebesgue measure. This result follows, for instance, from the existence theorem (Theorem 3.19) in the 2nd edition of Kallenberg's Foundations of Modern Probability.

Most spaces of interest in probability are Borel, so maybe this partial result is still useful to you.

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I am aware of this result, and I may decide to use it. I'd like to avoid restricting myself to Borel spaces or standard probability spaces because 1) the theorems are (I believe) still true for arbitrary probability spaces, 2) mentioning standard or Borel probability spaces may make the results more opaque to my audience. My results are not about probability theory, and I'd rather avoid going into it. –  John Wiltshire-Gordon Sep 7 '10 at 20:11
OK, back to the drawing board for me! –  Byron Schmuland Sep 7 '10 at 20:17

I would consider Maharam's structure theorem...
But I guess that will get you only the measure algebra.

So, consider a set $P \subset [0,1]$ that is badly nonmeasurable... but outer measure $1$. Lebesgue measure induces a probability on its Borel sets, but is there a map $f$ as you require?

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Nice suggestion. I had to look up Maharam's structure theorem, but it seems to provide more or less what I was trying to construct more explicitly in my post. –  George Lowther Sep 11 '10 at 11:40

This is more of a guess: take a measurable cardinal $\kappa$ and look at the {0,1} probability measure defined on all of its subsets. Is it a factor of a Haar measure like you want?

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Is there a reason you think a measurable cardinal might be a counterexample? All quotients of a {0,1} measure remain {0,1} measures, but I doubt the converse holds. You make me wonder, though: is there a group with a {0,1} Haar measure? –  John Wiltshire-Gordon Sep 10 '10 at 23:06
Of course the converse does not hold - any 0-1 law shows that. But I am more interested in the fact that all of its subsets are measurable. If you look at, say, $[0,1]^\lambda$ for any infinite $\lambda$ you are bound to have some non(-borel)-measurable sets. Is there a nontrivial quotient space for which all subsets are measurable? –  Ori Gurel-Gurevich Sep 10 '10 at 23:32
I agree; the quotient map would have to be pretty remarkable to eliminate all those unmeasurable sets. –  John Wiltshire-Gordon Sep 11 '10 at 6:20
John, among Hausdorff groups only the trivial one supports such a measure. Indeed, suppose a Hausdorff group $G$ supports a $\{0, 1\}$ measure (which I guess means $\{0, 1\}$-valued?), and let $g$ be a non-identity element. Then there are disjoint neighbourhoods $U$ and $V$ of $1$ and $g$. We may, and do, replace $U$ by $U \cap g^{-1}U$, so that $g U \subseteq V$. Then $U$ has positive measure, necessarily $1$; so $U \sqcup g U$ has measure $2$. Do non-Hausdorff groups even necessarily support Haar measure? –  L Spice Sep 11 '10 at 6:22
(I meant $U \cap g^{-1}V$, not $U \cap g^{-1}U$.) –  L Spice Sep 11 '10 at 6:25