Define a new product measure on cantor space as follows:u({0})=a,u({1})=1a,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.
Define a new product measure on cantor space as follows:u({0})=a,u({1})=1a,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. 

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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 $(a, 1a)$product measure $\mu$ contains elements $x_0, \ldots, x_{k1}$ 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. 


$\{0,1\}^I$
, but an ultrafilter, being a family of subsets, amounts to a subset of$\{0,1\}^I$
. – Andreas Blass Apr 21 '12 at 13:54