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Suppose $M$ is a model of ZFC, and $\mu^M$ is the Lebesgue measure on $\mathbb R^M$ such that $\mu^M(\mathbb R^M)=1$. It is known that if $r$ is a Cohen real over $M$ and $N=M[r]$ then $\mu^N(\mathbb R^M)=0$.

This is a very strong transition from having sets with a full measure being annihilated into nullity. Is it possible that for some[every?] $x\in(0,1)$ there exists $N=M[G]$ a generic extension of $M$ such that $\mu^N(\mathbb R^M)=x$?

If the answer is negative in its full generality ($M$ is just any model of ZFC) can we add some assumptions for a positive answer? (e.g. $M\models CH$)

This is really just idle curiosity which could not be satisfied via Google, references to a possible answer would be just as welcomed as a complete answer

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The answer is no. If $\mathbb{R}^V$ is measurable in a forcing extension $V[G]$ having new reals, then the measure must be $0$. The point is that every new real $x$ in $V[G]$ but not in $V$ is transcendental over $\mathbb{R}^V$, since one cannot add algebraic numbers by forcing. It follows that the translates of $\mathbb{R}^V$ by the powers of $x$ are disjoint. Thus, a Vitali-style argument with wrapped translations of the unit interval of $V$ shows that if $\mathbb{R}^V$ is measurable, it must have measure zero.

Gunter Fuchs and I had made this observation in connection with Sebastian's question Probabilities independent of ZFC?, which is very much related to your question.

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  • $\begingroup$ Ah, so simple! Seeing how this is a Vitali style argument, my next question would have to be obvious: what happens when we remove the axiom of choice so the Vitali argument fails? $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 16, 2012 at 1:27
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    $\begingroup$ The argument doesn't seem to use AC, since I'm not picking representatives here, but using only the transcendence of the new reals over the ground model field. So perhaps it works fine as-is? $\endgroup$
    – Joel David Hamkins
    Commented Feb 16, 2012 at 1:30
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    $\begingroup$ Instead of powers, you could also use multiples of the new real. Or, you can appeal to the Hewitt-Savage 01-1-law, which says that any measurable tail set must have measure 1 or measure 0. (A tail set in $X \subseteq 2^\omega$ is one that is closed under rational translations, i.e., $X+s=X$ for all finite sequences $s$.) The same argument shows that any filter extending the Frechet filter must have measure 0, or be non-measurable. $\endgroup$
    – Goldstern
    Commented Feb 16, 2012 at 7:54

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