$\newcommand{\N}{\mathbb{N}}$ $\newcommand{\Z}{\mathbb{Z}}$

Let $E = \{0, 1\}^\Z$, with the usual product topology. Let $A$ be a meagre set of $E$, and let $\mu$ be a measure on $E$ with the following properties:

  1. $\mu$ has full support.
  2. $\mu$ is shift-invariant, i.e. if we let $\sigma$ be the shift on $E$, $\sigma((x_i)_{i \in \N}) = (x_{i+1})_{i \in \N}$, then for any $F \subset E, \;\mu(F) = \mu(\sigma(F))$.

Do we have $\mu(A) = 0$ ?

I know that meagre sets need not be of measure $0$ in general, but the usual counterexamples are given for the Lebesgue measure on $\mathbb R$. Is it still true for full-support measures on a Cantor Space ?

If this still does not hold in this setting, are there any "easy" properties that, if satisfied by either $A$ or $\mu$, would ensure that $\mu(A) = 0$ ?


1 Answer 1


Let $D$ be a countable dense subset of the separable complete metric space $E$, and let $U(d,k,n)$ $(k \ge 1, k,n \in \Bbb N$) be an open neighbourhood of $d \in D$ such that $\mu(U(d,k,n)) \le \frac{1}{2^{k+n}}$, for a Borel probability measure $\mu$ that is $0$ on all singletons (then $\mu$ is regular, and such open sets exist). Then $U_n = \bigcup_{k,d} U(d,k,n)$ is open and dense (as it contains $D$) and has measure $\le \frac{1}{2^n}$, so $N:=\bigcap_n U_n$ has measure $0$ and is co-meagre (as $E$ is complete), so $E\setminus D$ is meagre of measure $1$ (or measure $\mu(E)$ in case of a finite measure, more generally).

So as the Cantor set is compact metric, it's certainly complete and separable, and so for finite Borel measures without atoms, we always have meagre subsets that have full measure, using the above ideas.

But of course there are plenty of meagre sets of measure $0$ too (all countable subsets, e.g.) and it's a set by set thing which case we have.


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