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. 



Does ultrafilter have measure one?
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.


