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Consider two random reals $x, y$ over a transitive model $V$ of ZFC. More specifically, if $\mathcal C^V={}^\omega2$ is the Cantor space, composing the canonical homeomorphism with the projections $\mathcal C^V\stackrel\cong\longrightarrow \mathcal C^V\times \mathcal C^V\longrightarrow \mathcal C^V$ we obtain two continuous maps inducing complete embeddings $h_i:\mathcal B^V/\mathcal N\longrightarrow \mathcal B^V/\mathcal N$, where $\mathcal B^V$ is the Borel $\sigma$-algebra of $\mathcal C$ and $\mathbb N$ is the ideal of null sets with regard to the usual product measure ($i=1,2$). If $G$ is a generic ultrafilter, then $h_i^{-1}[G]$ are also generic ultrafilters determining two random reals $x,y$ (over $V$ in principle).

I would like to know a proof that $y$ is in fact random over $V[x]$. All the texts I have seen pass over this fact very quickly.

Let me indicate what I have tried to do:

I tried to show that $y$ does not belong to any null Borel subset $B$ of $\mathcal C^{V[x,y]}$ with code in $V[x]$. On the contrary, if $y\in B$, I use that $B$ can be represented as $B=C^{V[x]}_x$, where $C$ is a Borel subset of $\mathcal C^V\times \mathcal C^V$ with code in $V$ (Lemma 3.1.6 in Bartoszynsky and Judah Set Theory. On the Structure of the Real Line). Here $C^{V[x]}$ represents the Borel set with the same code as $C$ and $C^{V[x]}_x=\{y\in\mathcal C^{V[x]}:(x,y)\in C^{V[x]}\}$.

If $C$ is null, then $(x,y)\in C^{V[x,y]}$ and a contradiction follows, but, unfortunately, the fact that $C_x^{V[x]}$ is null does not imply that $C$ is necessarily null (see here). I do not know if the Borel set $C$ can be taken null anyway.


I edit the question after the Andreas Blass' answer:

I thought I have completed the proof, but I have realized that my reasoning was circular. Namely:

According to Blass' indication, we must prove that $(P'\times {}^\omega2)\cap C$ has measure zero. On the contrary, there exists a Borel set $E\subset P'$ of positive measure in $V[x]$ such that $m(((P'\times {}^\omega2)\cap C)_u)>0$ for each $u\in E$.

We choose a random real $x'\in E$ and then we obtain a contradiction, since $C_{x'}$ should be null and non-null.

The problem is why there exists a random real in $E$. I had used that the set $R(V)$ of random reals in $V[x]$ over $V$ has outer measure $1$. But now I have realized that the proof I knew of this fact uses the fact we are trying to prove, namely, that null Borel sets of $V[x]$ can be represented as sections of null Borel sets of $V$.

The proof I know is that appearing in Bartoszynsky and Judah's book (Lemma 3.2.16). The argument is as follows: $R(V)$ is a tail set, so its outer measure must be $0$ or $1$. Hence, it suffices to show that it is not $0$. Since $x+\mathcal C^V\subset R(V)$, it is enough to show that $\mathcal C^V$ is not null (in $\mathcal C^{V[x]}$). This is shown in Lemma 3.2.39, which starts by representing a null set of the random extension as a section of a null set of the base model $V$.

This is stated without any comment, so I wonder if it is obvious for some reason.


I have found a different argument for proving that $\mathcal C^B$ is not null avoiding circularity. Mi doubt is definitely solved. Thank you again.

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Since $x$ is random over $V$, the fact that $C_x^{V[x]}$ is null is (like any fact about $x$ in $V[x]$) forced by some condition in $V$. This condition is the equivalence class, modulo the null ideal, of some positive-measure Borel set $P$ in $V$. Let $P'$ be the Borel set in $V[x]$ with the same code. Then $x\in P'$, so $(x,y)\in (P'\times{}^\omega2)\cap C$, and this has measure zero because $P$ forces $C_x$ to have measure zero.

In other words, although $C$ need not have measure zero, intersecting it with a suitable cylinder containing $(x,y)$ (namely the cylinder over $P'$) does, and that's all you need.

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  • $\begingroup$ Sorry. I thought it was all clear, but I have an (I hope) last doubt. Since it is a bit long I have edited the question instead of using a comment. $\endgroup$ – Carlos Apr 5 '16 at 12:04

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