I hope I can answer this question in a reasonable way. The natural way to define the iteration $\mathbb P*\dot{\mathbb Q}$ is to consider all pairs $(p,\dot q)$ with
$p\in\mathbb P$ and $p\Vdash\dot q\in\dot{\mathbb Q}$.
Unfortunately this is a proper class. (The definition in Jech's book also gives a proper class.)
The problem in Kunen's definition of the iteration $\mathbb P*\dot{\mathbb Q}$ is not so much that he defines
$(p,\dot q)\in\mathbb P*\dot{\mathbb Q}$ if $p$ forces $\dot q$ to be in $\dot{\mathbb Q}$,
but that he only considers names $\dot q$ in the domain of $\dot{\mathbb Q}$.
He does this to make sure that the iteration $\mathbb P*\dot{\mathbb Q}$ is a set rather than a proper class.
But there are other ways to make sure that the iteration is a set, for example by allowing
arbitrary names for $\dot q$ that are in some sufficiently large $H_\chi$ (sets with transitive closure of size $<\chi$).
"Sufficiently large" depends on $\mathbb P$ and $\dot{\mathbb Q}$.
Now Jech only considers pairs $(p,\dot q)$ such that $\dot q$ is forced by $1_{\mathbb P}$ to be in $\dot{\mathbb Q}$. But this doesn't really make a difference since by the existential completeness lemma, if $p\Vdash\dot q\in\dot{\mathbb Q}$, then there is a name $\dot r$ which is forced by $1_{\mathbb P}$ to be a condition in $\dot{\mathbb Q}$ and such that $p\Vdash\dot q=\dot r$.
Now in the case of a two step iteration, the definitions by Jech and Kunen are equivalent,
and both are equivalent to the one that I suggested (arbitrary names in the second coordinate, but cutting off somewhere).
The right way to see this is to first show that if you cut off Jech's iteration at a sufficiently large $H_\chi$, then you get a dense subset of Jech's iteration.
This cut off version of Jech's iteration is dense in my version (with the same sufficiently large $H_\chi$).
Now Kunen's iteration is equivalent to mine since given $(p,\dot q)$ in my iteration,
there are $s\leq p$ and $\dot t$ in the domain of $\dot Q$ such that
$s\Vdash\dot q=\dot t$. We have $(s,\dot t)\leq(p,\dot q)$, showing that Kunen's iteration is dense in mine.
Things start to fall apart if we go to longer iterations, and iterations with infinite support. My argument above still shows that Kunen's iteration of arbitrary length, with finite supports, is equivalent to my version for long iterations.
This is because given a finitely supported condition in the long iteration in my sense,
say with support of length $n$, you can decrease the first $n-1$ coordinates to force the last coordinate to be something given by a name in the domain of the respective name of a partial order. Then you decrease the first $n-2$ coordinates to take care of the $n-1$-th coordinate and so on.
This argument clearly does not go through with countable supports. Hence the countable support iteration in Kunen's sense is not necessarily equivalent to my version or Jech's version. It is equivalent, however, if all the iterands are given by full names.
Now, what is the problem in terms of properness? The proof of properness of countable support iterations of proper forcing (and I am referring to the proof in Goldstern's
"Tools for your forcing construction") makes frequent use of the existential completeness lemma to cook up names for conditions. Now in the Kunen version of the iteration
it may happen that these names cannot be used since they may not be in the domain of the respective name for a partial order.
And in fact, as the exercise shows, even the countable support iteration of Cohen forcing
of length $\omega$ in Kunen's sense depends on which names you choose for the iterands:
Full names yield a proper forcing (which would be equivalent to the iteration in my sense),
but taking canonical names yields a forcing notion that collapses $\omega_1$.
I hope this helps.
(An attempt at an answer, and also my first posting here. Thanks to Andres Caicedo for the reformatting.)
I claim that a single Cohen real makes the set of old reals strong measure zero.
Reals are functions from $\omega$ to 2.
Let ${\mathbb C}$ be Cohen forcing, and let $c$ be the name of the generic real.
Let $(n_k)$ be a sequence of ${\mathbb C}$-names for natural numbers.
I will find a sequence $(s_k)$ of names for finite $01$-sequences
($s_k$ of length $n_k$)
such that ${\mathbb C}$ forces: every old real is in some $[s_k]$.
Let $D_k$ be a dense open set deciding the value of $n_k$ and
containing only conditions of length at least $k$.
Say, each $q$ in $D_k$ decides that the value of $n_k$ is $f_k(q)$, where
$f_k$ is a function in the ground model defined on $D_k$.
Each $f_k$, and also the sequence $(f_k)$, is in $V$.
Now we work in the extension.
(The point is that even though we now know the actual values of
$n_k$, we will play stupid and use the names only, plus the minimal
amount of information that we need from the generic real.
This lets us gauge exactly how much information from the
generic we need.)
In the extension I will define a sequence $(i_k)$ of natural numbers.
Let $i_k$ be the minimal $i$ such that $c \mathord\upharpoonright i$ is in $D_k$,
where $c\mathord\upharpoonright i = c$ restricted to $i$.
(So $i_k$ is at least $k$.)
For each $k$ we now define a $01$-sequence $s_k$ of length $n_k$ as follows: Take $n_k$ successive bits from the Cohen real $c$, starting at position $i_k$. (Formally: $s_k(j) = c(i_k+j)$ for all $j\lt n_k$.)
I claim that "every old real is in some $[s_k]$" is forced.
Assume not, so let $p$ force that $x$ is not covered.
Let $k$ be larger than the length of $p$. So $p$ not in $D_k$.
Extend $p$ to $q$ so that $q$ is in $D_k$, $q$ minimal.
Let $l$ be the length of $q$.
So $q$ forces that $i_k$ is exactly $l$. Also $q$ forces that $n_k = f_k(q)$.
Now extend $q$ to $q'$, using the first $f_k(q)$ bits of $x$.
So $q'$ is stronger than $q$, and $q'$ forces that $s_k$ is an initial
segment of $x$.
mg*
Best Answer
In your extension $V_1$, there are actually continuum many, that is $\aleph_{\omega+1}$ many $V$-generic Cohen reals. In general, whenever you add even a single Cohen real, then the extension wills have continuum many $V$-generic Cohen reals, because if $c$ is a $V$-generic Cohen real and $x$ is any real in the ground model, then the bit-wise sum $c\oplus x$ will be a $V$-generic Cohen real, since this induces an automorphism of the forcing in the ground model, and these are all different. (In your case, as Goldstern mentions in the comments, we may view the forcing as adding all but one of the Cohen reals, and then a final Cohen real, to achieve $\aleph_{\omega+1}$ many $V$-generic Cohen reals by this reasoning.)
But it is true that not every new real of $V[G]$ is a $V$-generic Cohen real. For example, one may easily construct reals that obey some regular pattern, repeating their digits in pairs, for example, which prevent them from being Cohen reals, even if they are not in the ground model.
Nevertheless, if $V[G]$ is a forcing extension obtained by adding any number of Cohen reals and $z$ is any real in $V[G]$, then I claim that $z\in V[c]$ for some $V$-generic Cohen real in $V[G]$. This is because the countable chain condition of the forcing means that one needs only countably much information from $G$ in order to construct $z$, and restricting the generic sequence of Cohen reals to any countable domain is isomorphic again to adding a single Cohen real.
In general, if $V\subset V[G]$ is any forcing extension and $W$ is a model in between $V\subset W\subset V[G]$, such as $W=V[z]$ for some real $z$, then $W$ is also a forcing extension of $V$ by a complete subalgebra of the Boolean algebra giving rise to $G$.
In the case of adding a Cohen real, forcing which has a countable dense set, every subalgebra of its Boolean algebra also has a countable dense set, and all nontrivial forcing notions with a countable dense set are isomorphic to adding a Cohen real. In this sense, every real added in a Cohen real forcing extension (and this includes every real in your model by my observations above) is generic over $V$ for forcing that is isomorphic to the forcing to add a Cohen real.
So the answer to your final question is that yes, these extra reals are $V$-generic for some forcing notion, and that forcing notion is isomorphic to the forcing to add a single Cohen real!
Let me conclude with an interesting tidbit:
Theorem. In the forcing extension $V[c]$ obtained by adding a single $V$-generic Cohen real, there is a family of continuum many pairwise mutually generic Cohen reals. Indeed, there is a perfect set $P$ in $V[c]$, all of whose finite subsets are mutually $V$-generic Cohen reals.
Proof. Consider the forcing to add such a perfect set $P$. We want to force to create a tree, all of whose branches are $V$-generic Cohen reals, and such that any finitely many branches are mutually generic Cohen reals. Let conditions be finite binary trees, ordered by end-extension. It is dense for the leaves to be extended into any given dense subset of Cohen forcing. And for finite products of Cohen forcing with itself, it is dense to extend the tree so that all pairs (or triples etc.) of branches are inside any given dense set in the product forcing. Thus, this forcing will create such a tree and hence such a perfect set.
Finally, observe that our tree forcing has only countably many conditions, and thus it is isomorphic to adding a single Cohen real. So the forcing extension $V[c]$ already has such a pefect set. QED
Of course, the branches through the perfect set will not be $V$-generic for the forcing to add continuum many Cohen reals, since that forcing is not a subalgebra of the forcing to add only one. The reals in the perfect set are only mutually generic when taken finitely many at a time, but not fully mutually generic for infinite collections. For example, the perfect set contains reals that are the limits of other of its elements, and this violates mutual genericity for those infinite families.