Amoeba forcing adds a null set covering all old null sets

Question

I am currently reading a paper by Goldstern, Kellner and Shelah, in which they, pretty nonchalantly, state "Amoeba forcing will add a null set covering all old null sets", without proving this fact or giving a reference. The only thing I could find that would prove this statement was in the Bartoszynski book "Set Theory: On the structure of the Real Line", where he states that any two Amoeba forcings are equivalent, which would mean that, even when forcing with just one Amoeba forcing, for any $n\in\omega$, there would be a set of measure $<\frac{1}{n}$ (by a standard density argument), covering all old null sets, which would imply the statement. But I cannot seem to understand his proof, therefore I am wondering if perhaps there is an easier way, i.e. simply constructing an $\mathbb{A}_{1/2^{n+1}}$-generic Filter from a $\mathbb{A}_{1/2^n}$-generic filter.

Answer 1

The conditions in Amoeba forcing are open sets of measure less than $1/2$. (Say, in $2^\omega$.)

A condition $q$ is stronger than $p$ iff $q \supseteq p$. (Alternatively, use closed sets of measure greater than $1/2$. Then stronger conditions will be smaller.)

For a generic filter $G$ let $U_G$ be the union of all sets in $G$. An easy density argument shows that $U$ is an open set of measure $1/2$. (I can elaborate if necessary)

Another easy density argument shows that every ground model null set is contained in $U_G$. More precisely, whenever $c$ is a (natural) code for a $G_\delta$ null set, then the $G_\delta$ null set $B_c$ described by $c$ (in $V[G]$) is a subset of $U_G$.

Every "rational" infinite $01$-sequence (i.e., with only finitely many $1$'s) defines a measure-preserving translation map $x \mapsto x+t$ on $2^\omega$. Every rational translate of $U_G+t $ has the same property as $U_G$: measure $1/2$, and it covers every ground model $G_\delta$ null set.

(Proof: given such a null set $N$, also the set $N-t$ is null, hence covered by $U_G$, so $N$ is covered by $U_G+t$.)

Now we get to the main point: How to get from $1/2$ down to $0$?

Let $U'_G:= \bigcap_t (U_G + t)$, where $t$ ranges over all rational sequences. Then $U'_G$ is a $G_\delta$ set, its measure is at most $1/2$, but by Kolmogorov's 0-1-law, its measure must be $0$. But $U'_G$ still covers all ground model null sets.

Remark: If you do an iteration of Amoeba forcing (of limit length) rather than a single forcing, you could (but why would you?) replace the above argument by the following: the first Amoeba forcing gives you an open set of measure 1/2, the second a new open set of measure 1/3, etc. Now take the intersection of all these sets.

Answer 2

I believe the following works:

Let $C$ be Cantor space and for $\sigma\in 2^{<\omega}$ let $C_\sigma=\{f\in C: \sigma\prec f\}$. There is a canonical bijection $i_\sigma:C_\sigma\rightarrow C$ given by cutting off the initial $\sigma$.

Suppose $G$ is amoeba-generic over $V$. For $\sigma\in 2^{<\omega}$ let $G_\sigma=i_\sigma[G\cap C_\sigma]$. The point is:

• By "spending measure elsewhere," for each $\epsilon>0$ there will be some $\sigma$ with $m(G_\sigma)<\epsilon$.

• But by the usual "engulfing" argument, we'll have $N\subseteq G_\sigma$ whenever $N$ is null in the ground model.

So the $\Pi^0_2$ set $\bigcap_{\sigma\in 2^{<\omega}}G_\sigma$ is null and covers all ground model null sets.

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EDIT: Of course the first bulletpoint above is the heart of the argument, so let me explain why it's true.

First, note that by genericity it's enough to prove the following:

$(*)$ Suppose $A\subseteq C$ is open with $m(A)<{1\over 2}$. Then for all $\delta>0$ there is some $\sigma\in 2^{<\omega}$ such that ${m(A\cap C_\sigma)\over m(C_\sigma)}<\delta$.

This implies the first bulletpoint, and is what "spending measure elsewhere" refers to: supposing we have a condition $A$ and an $\epsilon>0$, let $\sigma$ be the string gotten by applying $(*)$ with $\delta={\epsilon\over 2}$. Then we consider some larger open $A'$ with $A'\cap C_\sigma=A\cap C_\sigma$ and ${1\over 2}-m(A')<{m(C_\sigma)\epsilon\over 2}$. We'll have that if $G$ extends $A'$ then $m(i_\sigma[G_\sigma])<\epsilon$ as desired.

So it just remains to prove $(*)$. For this, look at the complement $A^c$ of our set and note that it is a non-null set and hence on some interval has relative measure arbitrarily close to $1$.

Note that we really do need to think about intervals specifically - or at least some canonical countable collection of opens - since at the end we need a countable intersection to get the desired result. The variation of $(*)$ gotten by shifting from $C_\sigma$s to arbitrary open sets is trivial but unhelpful.