I have just begun studying $g$-measures, and thought it a good idea to study Keane's paper [1], where they were first introduced. However, I almost immediately stumbled upon a concept I am not familiar with, which is crucial to understanding the motivation for the rest of the paper.
The setting is as follows: Let $(X,T)$ be a dynamical system, where $X$ is a compact metric space $(X,d)$, and $T$ is a minimal covering transformation. By this we (read Keane) mean that:
- $T$ is everywhere $n$-to-1 ($n\geq 2$).
- $T$ is a local homeomorphism.
- There exists a $C>1$, such that for all $x\in X$ there exists a $\delta_x>0$, such that $d(x,y)<\delta_x\implies d(Tx,Ty)\geq Cd(x,y)$.
- For all $\varepsilon>0$ there exists a $N\in\mathbb N$, such that $T^{-N}(x)$ is $\varepsilon$ dense in $X$, for all $x$.
Now comes the concept I am not familiar with: Keane states that, for any measure $\mu$ on $X$ we can denote the measure on $X$ "obtained by lifting $\mu$ locally by $T^{-1}$ by $Q\mu$".
What exactly is this $Q\mu$?
My first thought was that $Q\mu$ was simply the pushforward $T_*\mu$ (defined by $T_\mu(A)=\mu(T^{-1}(A)))$, which agrees with his next line: "The total mass of $Q\mu$ is $n$ times that of $\mu$". However this is a global construction, and so does not lend itself to the use of the word "local".
My other thought is that $T$ being a minimal covering transformation allows us to construct a partition $X=\bigsqcup_{i=1}^nX_i$, where for each $i$, $T|_{X_i}$ is a homeomorphism. Let us denote each such restricted function by $T_i$. We can then define $Q\mu$ by $$Q\mu(A)=\sum_{i=1}^n\mu(T_i(A\cap X_i)).$$ This is most probably my best guess as to how $Q\mu$ is supposed to be defined, but I haven't proven rigorously yet that $T$ being a covering transformation allows such a decomposition (but it seems entirely natural that it should).
I will be very appreciative if someone can definitively tell me what the correct definition of $Q\mu$ should be.
[1] Keane, Michael, Strongly mixing g-measures, Invent. Math. 16, 309-324 (1972). ZBL0241.28014.
You have the right idea (probably the closest in spirit to what the author has in mind, although not the easiest to formalize - one need to construct the partition $(X_i)$, which is definitely possible). The problem is somewhat annoying, since measures are naturally pushed forward, and here we are trying to define some kind of pullback.
Here is another construction (which works well in this setting, but may fail if you only have measurable maps -- while the construction with partitions is more robust). By Riesz' theorem, there is a natural bijection between finite (signed) measures on $X$ and linear functionals on $\mathcal{C} (X, \mathbb{R})$, given by:
$$\mu (f) := \int_X f \ d \mu.$$
Let $S : X \to X$ be continuous. Another way of defining the pushforward of $\mu$ is by noticing that $L : f \mapsto f \circ S$ is a continuous operator from $\mathcal{C} (X, \mathbb{R})$ to itself. Then $\mu \circ L$ is a linear functional on $\mathcal{C} (X, \mathbb{R})$, so corresponds to a signed measure. This signed measure is $S_* \mu$. Indeed, for all continuous $f$,
$$\mu \circ L (f) = \int_X L(f) \ d \mu = \int_X f \circ S \ d \mu = (S_* \mu) (f).$$
This works for the composition operator $L$, but also for other operators. Let:
$$L(f) (x) := \sum_{y \in T^{-1} (x)} f(y).$$
Since $T$ is a covering map, we can write locally $Lf (x) = \sum_{i=1}^n f(T_i^{-1} (x))$, so $L(f)$ is also continuous. In addition, $L$ is obviously continuous as an operator from $\mathcal{C} (X, \mathbb{R})$ to itself (it has norm $n$). Hence, $\mu \circ L$ is a linear functional, which corresponds to a measure $Q \mu$ such that, for any continuous $f$,
$$Q \mu (f) = \int_X \left( \sum_{y \in T^{-1} (x)} f(y) \right) \ dx.$$
The measure $Q \mu$ is exactly what you want. Note that, since $f$ is $n$-to-$1$, we have $L (\mathbf{1}_X) = n \mathbf{1}_X$, so :
$$Q \mu (X) = Q \mu (\mathbf{1}_X) = \mu (L(\mathbf{1}_X)) = \mu (n \mathbf{1}_X) = n \mu (\mathbf{1}_X) = n \mu (X).$$