TITLE: Is every probability space a factor space of the Haar Measure on some group?
QUESTION [21 upvotes]: 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.

REPLY [3 votes]: 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?