Return to Answer

The statement is false in general, I added a counterexample at the end of my answer to show that some separability condition like countable separability (or the stronger condition of being Lebesgue from KhashF's answer) is necessary. The counterexample is just a glorified identity map.

Here is a proof of the statement in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.

Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.

So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.

Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.

Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, a contradiction.$\square$

In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$, see ($*$) below for more details), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.

But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.

($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$, that implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.

Counterexample: Consider the space $X:=\mathbb{Z}\times\mathbb{S}^1$, with $\sigma$-algebra $B:=\{\mathbb{Z}\times A;A\text{ Lebesgue measurable in }\mathbb{S}^1\}$, and consider the map $f:X\to X;f((n,p))=(n+1,p)$.

The statement is false in general, I added a counterexample at the end of my answer to show that some separability condition like countable separability (or the stronger condition of being Lebesgue from KhashF's answer) is necessary. The counterexample is just a glorified identity map.

Here is a proof of the statement in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.

Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.

So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.

Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.

Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, a contradiction.$\square$

In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$, see ($*$) below for more details), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.

But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.

($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$, that implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.

Counterexample: Consider the space $X:=\mathbb{Z}\times\mathbb{S}^1$, with $\sigma$-algebra $\mathcal{B}:=\{\mathbb{Z}\times A;A\text{ Lebesgue measurable in }\mathbb{S}^1\}$, with probability measure $\mu(\mathbb{Z}\times A)=m(A)$ ($m$ being Lebesgue measure) and consider the map $f:X\to X;f((n,p))=(n+1,p)$.

As requested in the comments, here is a proof in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.

Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.

So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.

Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.

Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, a contradiction.$\square$

In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$, see ($*$) below for more details), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.

But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.

($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$, that implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.

The statement is false in general, I added a counterexample at the end of my answer to show that some separability condition like countable separability (or the stronger condition of being Lebesgue from KhashF's answer) is necessary. The counterexample is just a glorified identity map.

Here is a proof of the statement in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.

Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.

So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.

Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.

Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, a contradiction.$\square$

In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$, see ($*$) below for more details), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.

But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.

($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$, that implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.

Counterexample: Consider the space $X:=\mathbb{Z}\times\mathbb{S}^1$, with $\sigma$-algebra $B:=\{\mathbb{Z}\times A;A\text{ Lebesgue measurable in }\mathbb{S}^1\}$, and consider the map $f:X\to X;f((n,p))=(n+1,p)$.

As requested in the comments, here is a proof in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.

Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.

So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.

Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.

Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, a contradiction.$\square$

In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$, see ($*$) below for more details), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.

But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.

($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$. That implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.

As requested in the comments, here is a proof in the case that $(X,\mathcal{B})$ is countably separated, i.e, there exists a sequence $(A_n)_n$ of measurable sets such that if $x,y\in X$ are distinct then there is some $n\in\mathbb{N}$ such that $x\in A_n$ but $y\not\in A_n$ or viceversa.

Now as in the question, let $A\in\mathcal{B}$ satisfy $\mu(A)>0$, we want to find $E\subseteq A$ such that $\mu(E)>0$ and $T^{-k}E\cap E=\varnothing$. It will be enough to find $E$ such that $\mu(E)>0$ and $\mu(T^{-k}E\cap E)=0$, and then we can remove a null set from $E$.

So suppose for the sake of contradiction that for all $E\in\mathcal{B}$ with $E\subseteq A$ and $\mu(E)>0$, we have $\mu(E\cap T^{-k}E)>0$.

Claim: Any measurable subset $E\subseteq A$ must satisfy $\mu(E\Delta T^{-k}E)=0$.

Proof: If $\mu(E\Delta T^{-k}E)>0$, then as $T$ is measure preserving we have $\mu(E\setminus T^{-k}E)>0$, and we also have $\mu((E\setminus T^{-k}E)\cap T^{-k}(E\setminus T^{-k}E))=\varnothing$, a contradiction.$\square$

In particular, $\mu(A\cap T^{-k}A)=0$ and letting $B_n:=A_n\cap A$, we have $\mu(B_n\Delta T^{-k}B_n)=0$. After removing a zero measure set from $X$ (remove $\bigcup_{i,j=0,1,\dots}T^{-ki}A\Delta T^{-kj}A$ from $X$ and similarly with each of the $B_n$, see ($*$) below for more details), we may assume that $T^{-k}A=A$ and $T^{-k}B_n=B_n$ for all $n$.

But then for any point $x\in A$, $T^kx\in A_n$ iff $x\in A_n$ for all $n$. Thus, $T^kx=x$. That is, the entire set $A$ is fixed pointwise by $T^k$, a contradiction.

($*$) Note that if $T:X\to X$ is a measure preserving transformation, $A$ is measurable and $\mu(A\Delta T^{-1}A)=0$, that implies that $\mu(T^{-n}(A\Delta T^{-1}A))=0$ for all $n$, that is, $\mu(T^{-n}A\Delta T^{n-1}A)=0$. Thus, all the sets $A,T^{-1}A,T^{-2}A,\dots$ have pairwise measure $0$ differences, that is, $\mu(T^{-n}A\cap T^{-m}A)=0$ for all $n,m\in\mathbb{N}$. In fact, consider the set $Y=X\setminus\bigcup_{n,m=0,1,\dots}T^{-n}A\Delta T^{-m}A$. Then $T$ maps $Y$ to $Y$: indeed, if $Tx\not\in Y$ for some $x\in X$, then $Tx\in T^{-n}A\Delta T^{-m}A$ for some $m,n$. Thus, $x\in T^{-n-1}A\Delta T^{-m-1}A$, so $x$ is not in $Y$ either. So we may define $S:Y\to Y$ as the restriction of $T$, and then $S$ is measure preserving. Moreover, $S^{-1}(Y\cap A)$ is exactly $Y\cap A$: indeed, as $Y$ contains no points of $A\Delta T^{-1}A$, any $x\in Y$ satisfies $x\in A$ iff $Tx\in A$ iff $Sx\in A$.