# Measure of meagre/nowhere dense sets in the Cantor Space

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

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 $$\mu(A) = 0$$ ?

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).
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.