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.
Best Answer
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 $2^I$.
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 $G=\mathbb{R/Z}$.
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.