It is well-known that if $V[G]$ is a generic extension by adding a Cohen real, then the set $\{r \in V[G]: r$ is Cohen generic over $V\}$ has measure zero.

On the other hand, if $V[G]$ is a generic extension by adding a random real, then the set $\{r \in V[G]: r$ is random over $V\}$ has outer measure one.

Question. I am wondering what is know about other generic reals? Are there any cases where the answer for them is not known?

Please provide references or proofs for each case.

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    $\begingroup$ Since a Mathias real (viewed as a subset of $\omega$) has asymptotic density $0$, the set of Mathias reals in any forcing extension will have measure $0$. The same trivial observation will handle some other cases of your question, but I'll be interested to see nontrivial information. $\endgroup$ – Andreas Blass Dec 13 '18 at 15:43
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    $\begingroup$ Another observation: If $g$ is a Sacks real over $V$, then the set of Sacks reals in $V[g]$ cannot have measure $0$. This is because every real in $S = \mathbb R^{V[g]} \setminus \mathbb R^V$ is a Sacks real, so if $S$ were null then $S \cup (S+g)$ would be too; however, $S \cup (S+g) = \mathbb R^{V[g]}$. Because $S$ is closed under rational translations, its being non-null implies it has full outer measure. My guess would be that $\mathbb R^V$ is null in $V[g]$, but I'm not sure right now. $\endgroup$ – Will Brian Dec 13 '18 at 21:44
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    $\begingroup$ Actually, now that I think about it, I think that both $S$ and its complement should have full outer measure. The reason is that if $\mathbb R^V$ were null in $V[g]$, then iterating the Sacks forcing (in the usual way -- length $\omega_2$, countable supports) would produce a model where, e.g., every non-measurable set has size $\aleph_2$. This doesn't happen, so adding a Sacks real can't make the ground model reals null. So I guess this shows that after adding a Sacks real, the set of Sacks reals is non-measurable with full outer measure. $\endgroup$ – Will Brian Dec 13 '18 at 21:50
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    $\begingroup$ Let $I$ be a $\sigma$-ideal on the reals and suppose that there is $X\in I$ and a Lebesgue null set $Y$ such that $X\cup Y$ is a partition of the reals (such as the case when $I$ is the meager ideal), then every $\mathbb{P}_I$ generic real must be in $Y$, hence the set of $\mathbb{P}_I$-generic reals has measure zero. My guess is that the above condition should be equivalent to "the set of $\mathbb{P}_I$-generics is null", but I haven't thought about this carefully. $\endgroup$ – Haim Dec 13 '18 at 22:30
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    $\begingroup$ If $r$ is random, then in $L[r]$, the set of random reals is not measureable $\endgroup$ – 喻 良 Dec 13 '18 at 23:22

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