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Let $(X,\mathcal{F},\mu)$ be a prob. space and $T:X \rightarrow X$ be measure preserving, and $f \in \mathcal{L}^1(X,\mathcal{F},\mu)$. Assume that $f \geq 0$ and suppose the set $A = \{f > 0\}$ has positive measure. Show that for $\mu$ almost all $x \in A$ we have that $\sum_{n=1}^\infty f(T^nx) = \infty$.

My current proof is this ugly beast:

Define $A_q = \{x \in X: f(x) > q\}$ and note that $A = \bigcup_{q \in \mathbb{Q}_{>0}} A_q$ and that if $q_1 > q_2$ then we have that $A_{q_2} \subset A_{q_1}$. Due to the fact that $\mu(A) = \sum_{q \in \mathbb{Q}_{>0}}\mu(A_q)$, there exists a number of sets $A_q$ of non-zero measure. Now consider the set $U_x = \bigcap_{x \in A_q}A_q$ for $x \in A$. Since for all $A_q,A_p$ we have either $A_p \subset A_q$ or $A_q \subset A_p$, this implies there must be a $q' \in \mathbb{Q}_{>0}$ such that $A_{q'} = U_x$.

Now let us assume that such an $U_x$ has non-zero measure. This means that for almost all $x \in U_x$ we have that there exists a $k \in \mathbb{N}_{>0}$ such that $f(T^kx) \in U_x$. And since $U_x = A_{q'}$ we have that $f(T^{nk}x) > q'$ for all $n$. Therefore: $$ \sum_{n=1}^\infty f(T^nx) \geq \sum_{n=1}^\infty f(T^{nk}x) > \sum_{n=1}^\infty nq' = \infty.$$

Now if $U_x$ has measure zero, then we have $\mu\{x \in A: \mu(U_x) = 0\} \leq \sum_{\mu(U_x) = 0}\mu(U_x) = 0$. This of course means that $\mu\{x \in A: \mu(U_x) >0\} = \mu(A)$.

Any ideas on a better approach to this? I'm not even sure if this way actual shows what I want to.

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Hm ... first you haven't $\mu(A) = \sum_q \mu(A_q)$, but $\mu(A) = \sup_q \mu(A_q)$. And $U_x = A_{q'}$ only if $f(x)$ is rational, with $q' = f(x)$ then, otherwise $f(x) = \{y \in X \mid f(y) \ge f(x)\}$. –  martini Nov 13 '12 at 19:06
    
Oh dear, I had the feeling a lot of the argumentation wasn't quite kosher. I'm having a lot of difficulty constructing a set $\{f>c\}$, for some $c \in \mathbb{R}$, of positive measure. My feeling is this would be the way to go, then use Poincare's recurrence theorem to show the sum is infinite. –  BallzofFury Nov 13 '12 at 19:17
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Since $A = \bigcup_{q > 0} A_q$, $\mu(A_q) > 0$ for some $q$. Now use Poincare recurrence on $A_q$. –  Robert Israel Nov 13 '12 at 19:41
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Thats exactly the problem I had in my reasoning, many thanks Robert. New and hopefully improved:

Define $A_q:= \{f > q\}$ and note that $A = \bigcup_{q \in \mathbb{Q}_{>0}}A_q$, since $f\geq0$. If for some $q \in \mathbb{Q}_{>0}$ we have that $\mu(A_q) > 0$, then by Poincare's recurrence theorem we have that there exists a $k \in \mathbb{N}$ such that $f(T^nkx) > q$ for all $n$. Therefore:

$$ \sum_{n=1}^\infty f(T^nx) \geq \sum_{n=1}^\infty f(T^{nk}x) > \sum_{n=1}^\infty q = \infty.$$

Consider the collections of sets $A_q$ for which $\mu(A_q) = 0$. The union of all these sets also has measure zero, since it is the union of countable null sets. This means that $\mu\left(\bigcup_{\mu(A_q)>0}A_q \right) = \mu(A)$ holds and thus $\sum_{n=1}^\infty f(T^nx) = \infty$ for almost all $x \in A$.

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