Consider $2^\omega$ with the usual measure, consider a countable family of strings of the same lenght $(s_i)_{i<\omega}$ such that the sets of coordinates they specify is pairwise disjoint, for example each $s_i$ specifies coordinates from $in+1$ to $(i+1)n$.
I am trying to prove that the set of points $x$ of $2^\omega$ such that $x\in[s_i]$ for some $i$ has measure $1$. When the strings are of length $1$ then it is clear because looking at the complementary is obvious that it has measure $0$, and taking the union of such kind of conditions we have a family of events which is strictly increasing in probaility, but I can't show that it goes to $1$. Is it even true?
I think in the easier case where each $s_i$ specifies coordinates from $in+1$ to $(i+1)n$ there is an easy proof. This is because the measure of the set $ \{x : x\notin [s_i] \text{ for } i=1,\dots,m\}$ is $(\frac{2^{n-1}-1}{2^{n-1}})^{m-1}$ and this converges to $0$ as $m$ goes to $\infty.$ But what if the $s_i$ do not cover all the possible coordinates?
Let me try to generalize the previous argument by adding some 'padding', divide the space in such a way that $s_i$ says something about $n$ coordinates of a set of $n+m_i$ coordinates still disjoint pairwise ( here $m_i$ is the 'padding' ), with this trick we can now assume that we cover every single coordinate. Now the measure of $x\notin [s_i]$ is $\frac{(2^{n}-1)2^{m_i}}{2^{n+m_i}}=\frac{2^{n}-1}{2^{n}}$ and the measure of $ \{x : x\notin [s_i] \text{ for } i=1,\dots,m\}$ is $(\frac{2^{n}-1}{2^{n}})^{m-1}$.