Prove or the disprove the existence of a limit of integrals Let $A \subset \mathbb{R}^2$ be the annulus defined, in polar coordinates, as
$$ A = \{ (r, \theta) \in [0, \infty) \times [0, 2\pi) : 2 \leq r \leq 3 \} \, .$$
Let $F: A \to A$, $F:(r, \theta) \mapsto (r, r \theta \text{ (mod } 2\pi))$.
Let $g : A \to \mathbb{R}$ be a continuous function. Discuss the existence of the limit
$$ \lim_{n \to \infty} \int_A g \circ F^{(n)}(x) \, \text{d}x \, ,$$
where $F^{(n)} = F \circ F \circ \dots \circ F$ denotes the composition of $F$ with itself $n$ times.
I cannot find a way to attack this problem. Here's what I tried: if $g \circ F^{(n)}$ converged, I think we could take the limit under the integral sign. So I tried to show the convergence of $F^{(n)}$. If it converged, then the limit function must send every point to a fixed point of $F$. So I determined the fixed points of $F$. But I don't think $F^{(n)}$ converges.
 A: This is probably overkill, as you are not asked to give the value of the limit.
But let $f_r(\theta)=r\theta \ (\mbox{mod } 2\pi )$ for $2\leq r \leq 3$. This is known in ergodic theory as the beta-transformation of the interval (here beta is $r$). It is known to be ergodic w.r.t. Lebesgue (for all $r>1$ in fact). Lebesgue is however not the invariant density. There is an $L^1$ function $h_r(\theta)$ so that for continuous $g$ as $n\rightarrow +\infty$
$$  \int_0^{2\pi} g(r,f_r^{(n)}(\theta)) d\theta \rightarrow \int_0^{2\pi} g(r,\theta) h_r(\theta) d\theta $$ for every $r$ (not just a.e.). The function $h_r(\theta)$ should be is measurable w.r.t. $(r,\theta)$ (but I have no reference for this). Modulo this, using Fubini and dominated convergence:
$$  \int_2^3\int_0^{2\pi} g(r,f_r^{(n)}(\theta)) r \;d\theta \;dr \rightarrow \int_2^3\int_0^{2\pi} g(r,\theta) h_r(\theta) r \; d\theta \; dr $$
But this is probably not the expected answer.
Note, however, that in general the limit is not $\int_A g \; dx$ which is not the correct answer.
A: If $f$ is continuous on the unit circle, then
$$\tag 1 \lim_{a\to \infty}\int_0^{2\pi} f(e^{iat})\,dt = \int_0^{2\pi} f(e^{it})\,dt.$$
As @Shalop said in the comments, that follows from periodicity. Now $F_n(re^{it}) = re^{ir^nt},$ so
$$\int_A g(F_n(x))\, dx = \int_2^3 \int_0^{2\pi}g(F_n(re^{it}))\,dt \,r\, dr=\int_2^3 \int_0^{2\pi}g(re^{ir^nt})\,dt \,r\, dr.$$
Call the inner integral $I_n(r).$ Because $r^n \to \infty$ for any $r\in [2,3],$ $(1)$ shows
$$\lim_{n\to \infty} I_n(r) = \int_0^{2\pi} g(re^{it})\,dt.$$
We have the easy estimate $|I_n(r)| \le 2\pi\sup_A|g|.$ Thus the domimated convergence theorem shows
$$\int_2^3I_n(r)\,r\, dr \to \int_2^3\int_0^{2\pi} g(re^{it})\,dt\,r\, dr = \int_A g(x)\, dx .$$
So $\int_A g(x)\, dx$ is the desired limit.
