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

  1. $J_n\subseteq 2^{I_n}$

  2. $\sum_{n\in\omega}|J_n|*2^{-|I_n|}<\infty$

  3. $H\subseteq \{x\in 2^\omega:\exists^{\infty}n\ x\restriction I_n \in J_n\}$

A measure on a Polish $X$ space will be refered as a function $\mu:BOREL(X)\rightarrow [0,1]$ s.t.

  1. $\mu(\emptyset)=0,\mu(X)=1$

  2. If $\{A_n:n\in\omega\}\subseteq BOREL(X)$ is a sequence of pairwise disjoint sets, then $\mu(\cup_{n\in\omega}A_n)=\sum_{n\in\omega}A_n$

  3. $\mu$ is nonatomic

  4. $\mu$ is translation invariant

  5. 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 $MEASURABLE(X)=\{A:\exists B\in BOREL(X) \mu(A\triangle B)=0\}$

  • $\begingroup$ What is a "small" filter? In your desired conclusion, does "measurable" mean that the corresponding set of reals is Lebesgue measurable? If you already know that $\mu(\mathcal F)=0$ doesn't that immediately imply that $\mathcal F$ is measurable (since "measurable" ought to mean simply that $\mu(\mathcal F)$ is defined at all)? $\endgroup$ – Henning Makholm May 25 '16 at 11:00
  • $\begingroup$ The filter on $\omega$ will be identified as a subset of $\omega$ via characteristic functions. The definition of a small set can be found in "Bartoszynski, Tomek & Judah, Haim Set theory: On the structure of the real line" in chapter 4. I will add it here later. $\endgroup$ – peer May 25 '16 at 14:15
  • $\begingroup$ So I take it you only want answers from people who have that book? $\endgroup$ – Henning Makholm May 25 '16 at 15:48
  • $\begingroup$ Of course not. I added the definitions. $\endgroup$ – peer May 26 '16 at 2:01
  • $\begingroup$ It seems that the two $|A|$ notations mean something different (for subsets of $\omega$ and of functions, respectively). Could you elucidate that? $\endgroup$ – Pedro Sánchez Terraf May 26 '16 at 13:34

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.

  • $\begingroup$ How did you derive equation $(*)$? When introducing the measure $\mu$ as the product measure on $2^\omega$ of the measures $\mu_i$ on $i\in\omega$, $(*)$ becomes clear to me. Finally you apply the continuity proberty $\lim\mu(A_n)=\mu(A)$ for $A_n\uparrow A$ using $\mu\left(\bigcup_{n>1}\{x\in 2^\omega:x\restriction I_n \in J_n\}\right)<\infty$ $\endgroup$ – peer Jun 4 '16 at 22:31
  • $\begingroup$ And the $\mu_i$ are defined as $\mu_{s(i)}(\{i\})=\mu_{s(i)}(\{\{i\}\})=\frac{1}{2}$ in my comment above. $\endgroup$ – peer Jun 5 '16 at 7:50
  • $\begingroup$ @peer I'm not sure what the subscript $s(i)$ means there. Nevertheless, you can find the whole treatment of such measures in Kechris' Classical Descriptive Set Theory, chap.17. There, property (*) is stated for $I$ an initial segment of $\omega$, but it should be easy to derive this form. For the last displayed inequality I'm just using subadditivity. $\endgroup$ – Pedro Sánchez Terraf Jun 5 '16 at 21:06
  • $\begingroup$ $s(i)$ just means the successor of $i$ defined as $s(i)=\{i,\{i\}\}$, since we want to define a measure for each element of $\omega$ and $\omega$ is defined as the smallest set containing the empty set and with $i$ also $s(i)$. What i wrote might not be correct. I meant $\mu_i(\{0\})=\mu_i(\{1\})=\frac{1}{2}$ for $i\in\omega$. $\endgroup$ – peer Jun 6 '16 at 17:08
  • $\begingroup$ @peer Your last statement is sensible. Btw, the successor of $i$ is $i\cup\{i\}$, which is usually different from you wrote $\endgroup$ – Pedro Sánchez Terraf Jun 6 '16 at 22:36

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