Let $cov(null)$ be the minimum cardinality of a set of nullsets of $\mathbb{R}$ so that their union equals $\mathbb{R}$. In a model $M[G]$ obtained from a ctm $M$ with $M\models CH$ by adding $\aleph_2$ many Cohen reals with forcing, we have $M[G]\models 2^{\omega}=\aleph_2$. Is $cov(null)^{M[G]}$ equal to $\aleph_1$ or to $\aleph_2$ ?
edit: Sorry, I just saw this is an exercise in Kunen's Set Theory (with a hint). So case closed, I suppose.
On
So as I am going to do this exercise, I can as well write it all down here.
Kunen gives this hint: He defines nullsets $W_f=\bigcap_n\bigcup_{m>n}(f(m)-2^{-m},f(m)+2^{-m})\in M[G]$ for $f:\omega\to\mathbb{Q}$ and asks to show a property (+) that every real in $M[G]$ is an element of a $W_f$ for some $f\in M$. This shows that $cov(null)^{M[G]}=\aleph_1$, since $M\models CH$ and the forcing is ccc.
Every real in $M[G]$ is already in an extension by $<\aleph_2$ many of the Cohen reals, because the forcing is ccc and every real in $M[G]$ thus has a countable nice name, which can be interpreted in a partial of the forcing. Similarly, for an extension by $\aleph_1$ many Cohen reals, every real in the extension is in an extension by countably many Cohen reals. So it is enough to show (+) for an extension by countably many of the $\aleph_2$ Cohen reals. Let $P$ denote such a forcing.
Let $G$ be $P$-generic, and assume $\dot x\in M$, $p\in G$ with $p\Vdash\dot x: \omega\to 2\land \forall f\in (\mathbb{Q}^{\omega})^M \dot x \not \in W_{f}$. Choose for each $f\in ({\mathbb{Q}}^{\omega})^M$ some $p_f\leq p$ and $n_f<\omega$ with $p_f\Vdash \dot x\not \in \bigcup_{m>n_f}(\check f(m)-2^{-m},\check f(m)+2^{-m})$. Since there are only countably many such pairs, by Baire's theorem there is a nonmeager set $A\subseteq({\mathbb{Q}}^{\omega})^M$ (with ${\mathbb{Q}}^{\omega}$ topologized simply like $\omega^{\omega}$ by a bijection of $\mathbb{Q}$ and $\omega$), $A\in M$, so that for all $f,g\in A$, $p':=p_f=p_g$ and $n':=n_f=n_g$.
Now for each $k<\omega$, let $D_k=\{ q\leq p'\, |\, \exists a\in \{0,1\} q\Vdash \dot x (k)=a\}$. It is dense below $p'$. There is a filter $F\subseteq P$, $F\in M$ intersecting all the $D_k$ (because these are only countably many dense sets). $F$ defines a $y:\omega\to 2$, $y\in M$, by setting $y(k)=a$ iff there is $q\in F$ with $q\Vdash \dot x(k)=a$.
Now in $M$, $y\not \in \bigcup_{m>n'}(f(m)-2^{-m},f(m)+2^{-m})$ for all $f\in A$, because if $y$ was in any of these intervals, we could obtain a condition in $F$ which would force $\dot x$ to be in that interval too. Also, the set of $f\in ( \mathbb{Q}^{\omega})^M$ with $y\in \bigcup_{m>n'}(f(m)-2^{-m},f(m)+2^{-m})$ is open (changing $f$ starting from a large enough $m$ will preserve $y$ being in any particular of these intervals). Its complement in $(\mathbb{Q}^{\omega})^M$ is thus closed and contains $A$, so being nonmeager closed it has nonempty interior. But this means that there is some $f$ with $y\not \in \bigcup_{m>n'}(f(m)-2^{-m},f(m)+2^{-m})$ so that changing the values of $f$ arbitrarily at places larger than some fixed $m$ preserves $y$ not being in that union, which is impossible, because $\mathbb{Q}$ is dense in $\mathbb{R}$.
Here's a better fact: Adding one Cohen real makes covering of null ideal $\omega_1$.