# Must a measure on $2^{\mathbb{N}}$ be atomless to be a measure on $[0,1]$?

This question comes from section 4.4, page 17, of this paper.

Let $\mu$ be a Borel measure on Cantor space, $2^\mathbb{N}$. The authors say that

If the measure is atomless, via the binary expansion of reals we can view it also as a Borel measure on $[0,1]$.

Is it necessary that $\mu$ be atomless?

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I don't think so. What about a point-mass on $2^{\mathbb{N}}$ (leading to a point-mass on $\mathbb{R}$)? – Dirk Feb 25 '12 at 19:55

The existence of the measure on $[0,1]$ has nothing to do with atoms, per se.
Let $\varphi: 2^\mathbb{N}\to [0,1]$ be defined by $\varphi(x)=\sum_{n=0}^\infty {x(n)/2^n}$. This map is Borel measurable, and so for any Borel measure $\mu$ on $2^\mathbb{N}$, the image measure $\mu\circ\varphi^{-1}$ is a Borel measure on $[0,1]$.
The authors mention this condition, I think, so they can go back and forth between the two viewpoints. That is, for atomless measures the map $\mu\mapsto \mu\circ\varphi^{-1}$ is one-to-one.
The question is about the other direction. You write about transferring Borel measures from $[0,1]$ to $2^\mathbb N$, and the question asks whether or not the map takes a measure to measure, and why would this measure have to be atomless. – Asaf Karagila Feb 25 '12 at 20:15
@AsafKaragila $\mu\mapsto \mu\circ\varphi^{-1}$ sends measures on $2^\mathbb{N}$ to measures on $[0,1]$. This is the direction that the OP is interested in, I think. – Byron Schmuland Feb 25 '12 at 20:16
Is it true that the only time the map $\mu \mapsto \mu \circ \varphi^{-1}$ is not one-to-one is when $\mu$ has (the binary sequence corresponding to) a dyadic rational as an atom? – Quinn Culver Feb 25 '12 at 22:34
@Quinn Yes, the measure $\mu$ should not charge any sequence with only finitely many 0s or 1s. Under the map $\varphi$ these correspond to dyadic rationals. – Byron Schmuland Feb 25 '12 at 22:38