I want to show, that a filter $\mathcal{F}$ on $\omega$ (considered as a subset of $2^\omega$), which is small, is measurable.
I found a lemma (without proof), that every small set is null. So, if $\mathcal{F}$ is small, then $\mu(\mathcal{F})=0$. To show, that $\mathcal{F}$ is measurable, one must show that $\mu(\mathcal{F}\triangle B)=0$ for a Borel set B. Now, since the empty set is a Borel set, it follows $\mu(\mathcal{F}\triangle \emptyset)=\mu(\mathcal{F})=0$.
Is this correct? I also want to know why every small set is null?
A set $H\subseteq 2^\omega$ is small, if there exists a partition $\langle I_n:n\in\omega\rangle$ of $\omega$ into disjoint finite intervals and a sequence $\langle J_n:n\in\omega\rangle$ such that
$J_n\subseteq 2^{I_n}$
$\sum_{n\in\omega}|J_n|*2^{-|I_n|}<\infty$
$H\subseteq \{x\in 2^\omega:\exists^{\infty}n\ x\restriction I_n \in J_n\}$
A measure on a Polish $X$ space will be refered as a function $\mu:BOREL(X)\rightarrow [0,1]$ s.t.
$\mu(\emptyset)=0,\mu(X)=1$
If $\{A_n:n\in\omega\}\subseteq BOREL(X)$ is a sequence of pairwise disjoint sets, then $\mu(\cup_{n\in\omega}A_n)=\sum_{n\in\omega}A_n$
$\mu$ is nonatomic
$\mu$ is translation invariant
for every $A\in BOREL(X)$ and $\epsilon>0$, there exists a compact set and an open set $U$ such that $K\subseteq A \subseteq U$ and $\mu(U\cap K^{c})<\epsilon$
Measurable sets are defined as $MEASURABLE(X)=\{A:\exists B\in BOREL(X) \mu(A\triangle B)=0\}$
For a measure to be translation invariant, you should have an underlying group structure, not just a Polish space. In the case of $2^\omega$, this is the countable power of $\mathbb{Z}_2$, so indeed you have one. And there is only one Borel probability measure on $2^\omega$ that is translation invariant. This Haar measure $\mu$ is characterized by $$ \mu(\{x\in 2^\omega : x \restriction I = f \})=2^{-|I|},\qquad (*) $$ for every finite $I\subset \omega$ and $f\in 2^I$.
Now it is enough to see that $$ \{x\in 2^\omega:\exists^{\infty}n\ x\restriction I_n \in J_n\}=\bigcap_m\bigcup_{n>m}\{x\in 2^\omega:x\restriction I_n \in J_n\} $$ has arbitrarily small measure. For this, first fix $m$ and calculate: $$ \mu\left(\bigcup_{n>m}\{x\in 2^\omega:x\restriction I_n \in J_n\}\right)\leq \sum_{n>m}\mu(\{x\in 2^\omega: x\restriction I_n \in J_n\}) $$ and this last term equals $ \sum_{n>m}|J_n|\cdot 2^{-|I_n|} $ by $(*)$. Since the full sum converges, you only have to choose $m$ big enough.