# Can we change the Lebesgue measure by forcing?

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.
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. –  Goldstern Feb 16 '12 at 7:54