Measure preserving transformation and ergodic implies unbounded return time for some measurable set Suppose that $(X,\mu)$ is a non-atomic borel probability space and that $T:X \to X$ is measure preserving, ergodic and invertible.  Show that there is a measurable set $A$ with positive measure and unbounded return time.
So far I showed that for each $n$ and $\epsilon>0$ there is a set $B$ such that $B$, $T^{-1}(B)$, ..., $T^{-(n-1)}(B)$ are disjoint and cover $X$ up to a set of measure less that $\epsilon$ by using the return times.  This would imply that there are sets with arbitrarily large return time, but I'm not sure how to use this to show that there is a set with unbounded return time.  Any help would be appreciated.
 A: First we'll show that there are non periodic points, otherwise we could decompose $X = \underset{n \geq 1}{\bigcup}P_n$ into it's periodic points with same period, where $P_n$ is the set of points which are $n$-periodic and not $j$-periodic for any $j < n$. By ergodicity, one of those sets has full measure, say $P_n$. Since $X$ has no atoms, we may find a subset $A_n \subset P_n$ such that $0 < \mu(A_n) < \frac{1}{n}$, but then
$$ \underset{0\leq i \leq n-1}{\bigcup}T^i(A_n) $$
is $T$-invariant with measure strictly between $0$ and $1$ which is impossible by ergodicity.
Now take any non periodic point $p \in X$, $O(p)$ its orbit that has zero measure, consider $A = O(p)^c \cup \{p\}$ which has total measure. Notice that $p$ never returns to $A$, therefore its first return time is equal to $\infty$, in particular it's unbounded.
A: In case someone is interested, I wanted to prove a stronger version of this exercise where $T$ need not be invertible (see exercise 9 of page 56 in Petersen's Ergodic Theory), and interpreting "$A$ has unbounded return time" to mean that for all $n$, $A\setminus(T^{-1}A\cup\dots\cup T^{-n}A)$ has positive measure. My proof is a bit complicated so it would be great if there was a simpler one.
Exercise. Prove that for any ergodic measure preserving transformation $T$ on a nonatomic space $(X,\mu)$ there is a set $A$ of positive measure for which the return time is unbounded.
Proof.
We want to find a set $A$ with unbounded return time, in the sense that $\forall n$ we have
$$\mu(A\setminus(T^{-1}A\cup\dots\cup T^{-n}A))>0.$$
Lemma 1.
Any subset $C\subseteq X$ with $0<\mu(C)<\frac{1}{n}$ satisfies that $T^{-n}(C)\setminus(C\cup T^{-1}C\cup\dots\cup T^{-n+1}C)$ has positive measure.
Proof: If we had $T^{-n}C\subseteq \bigcup_{i=0}^{n-1}T^{-i}C$ (up to measure $0$), then the set $\bigcup_{i=0}^{n-1}T^{-i}C$ would be $T$-invariant, which is not possible because $0<\mu(\bigcup_{i=0}^{n-1}T^{-i}C)<1$. $\square$
Lemma 2.
Suppose a subset $Y\subseteq X$ has return time $>n$, that is, $\mu(Z)>0$ with $Z:=Y\setminus(T^{-1}Y\cup\dots\cup T^{-n}Y)$. Let $Y'$ be such that $\mu(Y\Delta Y')<\frac{\mu(Z)}{n+1}$. Then $Y'$ also has return time $>n$.
Proof: Let $Z'=Y'\setminus\bigcup_{i=1}^nT^{-i}Y'$. We want to prove $\mu(Z')>0$. One way to do this is proving that $\mu(Z\setminus Z')<\mu(Z)$. But for any $x\in Z\setminus Z'=(Y\setminus\bigcup_{i=1}^nT^{-i}Y)\setminus(Y'\setminus\bigcup_{i=1}^nT^{-i}Y')$, then either $z\in Y\setminus Y'$ or $z\in Y\cap Y'$, in which case $z\in T^{-i}Y'$ for some $i$, so as $z\not\in T^{-i}Y$, we have $z\in T^{-i}(Y'\setminus Y)$. So $Z\setminus Z'\subseteq(Y\setminus Y')\cup\bigcup_{i=1}^nT^{-i}(Y'\setminus Y)$, and thus as we wanted:
$$\mu(Z\setminus Z')\leq\mu(Y\setminus Y')+\sum_{i=1}^n\mu(T^{-i}(Y'\setminus Y))<\mu(Z).\square $$
Now that we are done with the lemmas, let's define the set $A$. It will be the limit of a sequence $A_n$ of sets with return time $>n$. Remember that the limit of a sequence $A_n$ is a measurable set well defined up to measure $0$ whenever $\sum_n\mu(A_n\Delta A_{n+1})<\infty$. In fact, if we let $B_n=A_n\setminus(T^{-1}A_n\cup\dots\cup T^{-n}A_n)$, then we will ensure that $\frac{\mu(B_n)}{n+1}>\sum_{i=n}^\infty\mu(A_i\Delta A_{i+1})\geq\mu(A_n\Delta A)$. So by lemma 2 with $Y=A_n$, $Z=B_n$ and $Y'=A$, $A$ will also have return time $>n$, and as this will be satisfied $\forall n$, we will conclude that $A$ has unbounded return time.
Let's construct the sequence $A_n$. Let $A_1$ be some subset of $X$ with $0<\mu(A_1)<1$, and let $B_1=A_1\setminus T^{-1}A_1$ be as above. As $T$ is ergodic, $\mu(B_1)>0$. Now suppose by induction hypothesis that we have a finite sequence $A_1,\dots,A_n$ such that, defining $B_n$ as above, we have for $k=1,\dots,n$ that $\frac{\mu(B_k)}{2k}>\sum_{i=k}^{n-1}\mu(A_i\Delta A_{i+1})$.
By lemma $1$ (and using that $(X,\mu)$ is non atomic) we can find a set $C_n$ of measure $\varepsilon$, for some small $\varepsilon$ to be determined, such that $D_n:=T^{-n-1}C_n\setminus (C_n\cup T^{-1}C_n\cup \dots\cup T^{-n}C_n)$ has positive measure. Now we define $A_{n+1}=(A_n\cup T^{-n-1}C_n)\setminus (C_n\cup T^{-1}C_n\cup \dots\cup T^{-n}C_n)$. Then $D_n\subseteq A_{n+1}$, and the points of $D_n$ don't return to $A_{n+1}$ until at least $n+2$ steps, so $D_n\subseteq B_{n+1}$, so $\mu(B_{n+1})>0$. Moreover, $\mu(A_n\Delta A_{n+1})\leq (n+2)\mu(C_n)$, so if we make $\mu(C_n)$ small enough we can ensure that for $k=1,\dots,n+1$ we have $\frac{\mu(B_k)}{2k}>\sum_{i=k}^{n}\mu(A_i\Delta A_{i+1})$.
Our sequence $A_n$ satisfies $\frac{\mu(B_n)}{n+1}>\frac{\mu(B_n)}{2n}\geq\sum_{i=n}^\infty\mu(A_i\Delta A_{i+1})$, as we wanted.
