# A set of positive-measure not being a countable union of cylinder sets and zero-measure sets?

Let $(A^\mathbb{N}, \mathcal{B}(A^\mathbb{N}), \mu)$ be a measure space, where $A^\mathbb{N}$ is a set of one-sided sequences over a finite alphabet $A \subset \mathbb{N}$, $\mathcal{B}(A^\mathbb{N})$ is the Borel sigma-algebra generated by cylinder sets and $\mu$ is a measure with full support (say, Bernoulli or Markov).

Is it possible in this case to construct an analogue of the fat Cantor set, namely a set of positive measure that cannot be represented as a union of an open set and a set of measure zero?

It seems to me that the answer is "no", but I could neither prove it myself nor find any reference.

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Well by binary expansion, one can get a dynamical system which is isomorphic to the one-torus with say the "times 2" action, hence one can push-forward the Lebesgue measure to this system and hence get a "fat cantor set" in this system also. I think you should at-least to limit the measures $\mu$ in the question not to be a.c. wrt to Lebesgue. –  Asaf Aug 26 '13 at 8:36
Thank you, I've corrected the question a bit. –  Anton Aug 26 '13 at 8:44
If the support of $\mu$ has empty interior then it is such a set. Do you want $\mu$ to be fully supported? –  Ian Morris Aug 26 '13 at 9:14
Yes, exactly, The word "fully supported" seems to be the most simple description of what I want to have, I add it to the question –  Anton Aug 26 '13 at 9:20

Let $\mu$ be a non-atomic Borel probability measure on a Polish space $X$, and let $\delta>0$. Then there exists a closed set $K \subset X$ such that $\mu(K)>1-\delta$ and $K$ has empty interior.
To see this we argue as follows. Let $(x_n)_{n=1}^\infty$ be a sequence of distinct points which is dense in $X$. For each $n \geq 1$ let $U_n$ be an open ball around $x_n$ which is so small that $\mu(U_n)<\delta/2^n$, which can be found since $\mu$ is non-atomic and since every Borel probability measure on a Polish space is outer regular. Let $K:=X \setminus \bigcup_{n=1}^\infty U_n$, then $K$ is closed since it is the complement of an open set, and satisfies $\mu(K)\geq 1-\sum_{n=1}^\infty \mu(U_n)>1-\delta$. Since $K$ does not contain any element of the dense sequence $(x_n)$ it cannot contain an open set.