Consider the doubling map $g\colon [0,1]\to [0,1]$ given by $g(x)=2x \, {\rm mod}\; 1$. It is clearly discontinuous at 1/2. However, its counterpart $G$ on the circle $G(e^{2\pi i \theta}) = e^{4\pi i \theta}$ ($\theta\in [0,1)$) is continuous therefore $G$ has many invariant measures.
Very often it is claimed in ergodic theory books that one may remove discontinuity of $g$ by passing to the circle (by identifying the end-points of $[0,1]$). It is however not clear to me whether this operation affects the $g$-invariant measures or not.
Is there a one-to-one correspondence between $g$-invariant measures on $[0,1]$ and $G$-invariant ones on the circle?
It seems to me that this could be done if we took into account only those $g$-invariant measures whose probability distribution functions vanish at 0. Do I get this right?
Layman's explanation concerning passing from the interval to the circle in the case of the doubling map would be highly appreciated.
The difference is quite small and can at most affect measures that assign masses to $0$ and/or $1$.
If you define $g(x) = \lfloor 2x \rfloor$, then $g$ is a map from $[0,1]$ to $[0,1)$. Any invariant measure must assign zero measure to $\{1\}$ and the Borel-sigma algebra of $[0,1)$ is isomorphic (for any Borel measure) to $S^1={\Bbb R}/{\Bbb Z}$. So in this case you may uniquely identify measures in the two cases.
If, however, you define e.g. $g(x)=2x$ for $x\in [0,0.5]$ and $g(x)=2x-1$ for $x\in (0.5]$ then you do actually have two invariant distinct dirac masses at $0$ and $1$. You have a similar scenario for other variations of defining the map at the end-points.
But the difference will be minor. In particular, if you consider a measure that does not assign mass to 0 or 1 then from a measure point of view there is no difference between this system and the circle doubling map.