# Can a probability measure with connected and compact support be realized as the pushforward of the uniform?

Suppose that $$\mu$$ is a Borel probability measure such that $$\text{supp}(\mu) \subseteq \mathbb{R}^n$$ is (locally) connected. Does there exist a continuous function $$f:[0, 1]^n \to \mathbb{R}^n$$ such that $$f_\#(\theta)=\mu$$? Here $$\theta$$ denotes the uniform distribution over $$[0, 1]^n$$ and $$f_\#(\theta)$$ is the pushforward measure of $$\theta$$ under $$f$$.

In the case where $$n=1$$ this is true by inverse transform sampling (we pick $$f$$ equal to the inverse cumulative distribution of $$\mu$$ which exists by connectedness of the support). In more dimensions this seems plausible but I haven't found a reference or been able to come up with a proof/counterexample.

Any hints or references are appreciated.

No, take the topologists sine-surve, that is, the set $$S=\{\langle x,\sin\frac1x\rangle:0. Now put a measure on this set by giving the vertical interval measure $$\frac12$$ (uniformly) and the graph of $$\sin\frac1x$$ too (uniform in $$x$$, say). A continuous function as desired would have to map the square onto $$S$$; but $$S$$ is connected but not locally connected, so not a continuous image of the unit square.
• I'm not sure; you can replace $[0,1]^n$ by the unit interval, which makes life a bit easier (or not).. None of the proofs of the Hahn-Mazurkiewicz theorem that I know take notion of measure into account. My guess is that the best you can get is a Borel isomorphism. – hartkp Sep 21 '18 at 17:55