$f$ preserves $\mu$ $\implies$ $f_A$ preserves $\mu_A$?

Question

I am struggling with seeing why this is true. Let $(X,\mathcal{B},f)$ be a dynamical system. Assume that $\mu$ preserves $f$ and that $\mu$ is a probability measure. Define $\mu_A=\frac{\mu|_A}{\mu(A)}$, $f_A=f^{n_A}$, where $n_A$ is the first return time to $A$. Here is how I started:

Let $B\subset A$. $$\mu_A(f_A^{-1}(B))=\frac{\mu|_A (f_A^{-1}(B))}{\mu(A)}=\frac{\mu (({f^{n_A}})^{-1}(B))}{\mu(A)}$$

Now the numerator of the last fraction should end up being $\mu(B)$, but I don't see why. I tried a few ways to look at it but none of them were fruitful.

Any ideas would be appreciated!

Answer 1

Small remark on what you've done so far, because I can't tell whether it's a point of confusion: Remember that the first return time depends on the point in question, so $n_A$ is really a function on $A$, and the definition of $f_A$ is written more precisely as $f_A(x) = f^{n_A(x)}(x)$. (This confusion seems to show up in the part of the proof you've started; $(f^{n_A})^{-1}(B)$ doesn't seem to make very much sense, apart from being a nonstandard way to write $(f_A)^{-1}(B)$.)

I'm going to be chatty in this answer instead of driving immediately to a proof because this problem is a good exercise.

In order to deal with this problem it's very natural to break $A$ up into disjoint subsets, depending on the first return time: $$ A_i = n_A^{-1}(i) = \lbrace x \in A : f^i(x) \in A, f^{j}(x) \notin A \text{ for } j < i\rbrace. $$ So you go and write $$ \mu_A(f_A^{-1}(B)) = \mu_A\Big( \bigcup_{i = 1}^{\infty} f_A^{-1}(B) \cap A_j \Big) = \sum \mu_A(f^{-i}(B) \cap A_i) = \mu(A) \sum \mu(f^{-i}(B) \cap A_i) $$ where the first equality follows from Poincaré recurrence, and the second is just applying countable additivity and the definition of $f_A^{-1}$.

My guess is this is where you got stuck (but this is a good reason it's helpful to include a bit more of what you've tried in posts). It's not immediately obvious what to do with this expression, since $f^{-i}(B) \cap A_i$ in general won't be the inverse image of anything.

In such cases it's important to be working with a concrete example, so that you can see better what's really going on. In this case (and many others in ergodic theory), the left shift on $\lbrace 0, 1 \rbrace^{\mathbb{N}}$ captures all the interesting behavior. Let $$ A = \lbrace (0, *, *, \cdots) \rbrace \qquad \text{and} \qquad B = \lbrace (0, 0, *, *, \cdots) \rbrace $$ where I'm writing $*$ to indicate elements of the sequence that can be either $0$ or $1$. Then the sets in our decomposition are \begin{align*} f^{-1}(B) \cap A_1 &= (0, 0, 0, *, *, \cdots) \\ f^{-2}(B) \cap A_2 &= (0, 1, 0, 0, *, *, \cdots) \\ f^{-3}(B) \cap A_3 &= (0, 1, 1, 0, 0, *, *, \cdots) \\ f^{-4}(B) \cap A_3 &= (0, 1, 1, 1, 0, 0, *, *, \cdots) \\ \end{align*} and so on. It certainly looks like these will add to $\mu(B)$ (which is what we want to show), and maybe you can begin to see a way of writing these that will make things tractable.

Does this give any thoughts? I'll supply a few progressive hints for if you continue to be stuck:

Note that you can write \begin{align*} A_i = A \cap f^{-1}(A^c) \cap \cdots \cap f^{-(i-1)}(A^c) \cap f^{-i}(A) = A \cap \bigcap_{k=1}^{i-1} f^{-k}(A^c) \cap f^{-i}(A) \end{align*}

It follows from this that:

\begin{align*} f^{-i}(B) \cap A_i = A \cap \bigcap_{k=1}^{i-1} f^{-k}(A^c) \cap f^{-i}(B) \end{align*}

The result now follows from a lemma, which looks like a mess but should be relatively clear from the above discussion:

For all measurable sets $B, A$, and all $n \geq 1$, \begin{align*} \mu(B) = \sum_{i = 1}^n \mu \Big( A \cap \bigcap_{k=1}^{i-1} f^{-k}(A^c) \cap f^{-i}(B) \Big) + \mu \Big( \bigcap_{i=0}^{n-1} f^{-i}(A^c) \cap f^{-n}(B) \Big) \end{align*} The proof is by induction; see for instance Lemma 6.15 in this book for an explicit writeup.