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.

anycontinuous 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