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?

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. 
