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Week 8 Worksheet Solutions
-
- Since $T$ is a bijection the sets $E$ and $T^{-1}(E)$ always have the same cardinality and therefore the same measure.
- Fix a set $E \subset \Z/p\Z$ with positive measure. This means there is a point $e \in E$. Since $E$ is invariant it contains the orbit of $e$. But the orbit of $e$ is all of $\Z/p\Z$ so $\mu(E) = 1$.
- As $X$ is finite we will have ergodicity if and only if all orbits are equal to all of $X$. This happens when and only when $r$ and $s$ are coprime.
- The collection
\[
\{ [a,b) : 0 \le a \l b \le 1 \} \cup \{ \emptyset \}
\]
is a π-system so it suffices to check
\[
\lambda([a,b)) = \lambda(T^{-1}([a,b)))
\]
for all $0 \le a \l b \le 1$. But $\lambda([a,b)) = b-a$ and
\[
T^{-1}([a,b)) = [\tfrac{a}{2},\tfrac{b}{2}) \cup [\tfrac{a+1}{2},\tfrac{b+1}{2})
\]
is a union of disjoint intervals whose lengths add to $b-a$.
- This is covered in the solutions to the Week 8 Tutorial.
- Fix $k \in \N$. For each $N \in \Z$ define
\[
I_k(N) = \left\{ x \in \R : \dfrac{N}{2^k} \le f(x) \l \dfrac{N+1}{2^k} \right\}
\]
and note that $T^{-1}(I_k(N)) = I_k(N)$ for all $N \in \Z$. By ergodicity each one has either measure 1 or measure 0. As their union is all of $X$ there must be exactly one $N(k) \in \Z$ for which $\mu(I_k(N(k))) = 1$. The sets in the decreasing sequence
\[
k \mapsto I_k(N(k))
\]
all have full measure so their intersection - which consists of those $x \in X$ at which $f$ assumes a specific value - has measure one as well. Therefore $f$ is equal on a set of full measure to a constant function.