# Does ultrafilter have measure one?

Define a new product measure on cantor space as follows:u({0})=a,u({1})=1-a,where a$\in$(0,1/2].

Does any ultrafiter U hasn't measure one?

When a=1/2,I know U hasn't measue one.I guess neither when a$\in$(0,1/2), but I don't know how to prove.

-
Cantor space: $\{0,1\}^I$ with $I$ countably infinite. Product measure with each factor as shown. Sometimes known as a "biased coin" measure. But I don't know what characteristics this ultrafilter is supposed to have. – Gerald Edgar Apr 21 '12 at 12:39
OK, I now guess an ultrafilter on $I$ defines a subset of $X=\{0,1\}^I$ and the question is whether such an ultrafilter can have measure one. Is this what you want to ask? – Gerald Edgar Apr 21 '12 at 12:43
Complementation in $I$ induces a measure-class preserving automorphism of $2^I$. So any subset of $2^I$ with full measure contains two elements of the form $A, I \setminus A$ and in particular isn't an ultrafilter. – Clinton Conley Apr 21 '12 at 13:23
@Benjamin: A subset of $I$ amounts to a single point in $\{0,1\}^I$, but an ultrafilter, being a family of subsets, amounts to a subset of $\{0,1\}^I$. – Andreas Blass Apr 21 '12 at 13:54
Ramiro, I unfortunately can't, since the claim I made is false. The set of strings with density $1-a$ will have measure $1$, while its image under complementation (the strings with density $a$) will have measure $0$. Sorry about that! – Clinton Conley Apr 21 '12 at 20:47

This question is a bit more subtle than I had originally thought (in the comments), but anyway here's an argument that seems to work. I will assume for notational convenience that the ultrafilter is on the first infinite ordinal $\omega$.

Fix $k \in \omega$ with $1/k < a$. The main claim is that any conull subset of $2^\omega$ with respect to the $(a, 1-a)$-product measure $\mu$ contains elements $x_0, \ldots, x_{k-1}$ such that $\bigcap_{i < k} x_i = \emptyset$ (that is, for each $n \in \omega$ there is $i < k$ with $x_i(n) = 0$). The trick is to make the situation "continuous" by introducing the function $f: [0,1)^\omega \to 2^\omega$ given by $f(y)(n) = 0$ if $y(n) < a$ and $f(y)(n) = 1$ if $y(n) \geq a$. It should be straightforward to check that if $\nu$ is the usual Lebesgue product measure on $[0,1)^\omega$, then $\nu(f^{-1}(A)) = \mu(A)$ for basic open (and thus all measurable) sets $A \subseteq 2^\omega$.

So suppose $A \subseteq 2^\omega$ is $\mu$-conull, thus $B = f^{-1}(A)$ is $\nu$-conull. Consider the $\nu$-preserving automorphism of simultaneous rotation by $1/k$, i.e., $g_k: [0,1)^\omega \to [0,1)^\omega$ given by $g_k(y)(n) = y(n) + 1/k$ (mod $1$). Since $B$ is $\nu$-conull, there is some point $y \in \bigcap_{i < k} g_k^{-i}(B)$. Then the points $x_i = f(g_k^i(y))$ are as desired, since for any $y \in [0,1)^\omega$ and any $n \in \omega$, at least one of $g_k^i(y)(n) = y(n) + i/k$ (mod $1$) is less than $a$.

In particular, any $\mu$-conull set closed under finite intersection contains the empty set, so it can't be an ultrafilter.

-
thanks for your answer,its a nice proof. – Jialiang He Apr 25 '12 at 14:32