Please fix this incorrect proof from Kunen To entertain myself, I've been (very, very slowly) working my way through Kunen's Set Theory:  An Introduction to Independence Proofs (1980 edition).  I've convinced myself that one of his proofs in the Martin's Axiom section of Chapter 2 is incorrect in a couple of places.  I think I've figured how how to patch one of the mistakes, but not the other.
First, the definition relevant to the mistakes:


Definition 2.4:  Let $\langle \Bbb P, \leq \rangle$ be a partial order.  $D \subset \Bbb P$ is dense in $\Bbb P$ iff $\forall p \in \Bbb P \exists q \leq p(q \in D)$.  $G \subset \Bbb P$ is a filter in $\Bbb P$ iff




(a)  $\forall p, q \in G \exists r \in G(r \leq p \land r \leq q)$, and




(b)  $\forall p \in G \forall q \in \Bbb P (q \leq p \implies q \in G)$.


Now the mistakes:


Theorem 2.21:  Assume MA($\kappa$).  Let $M_\alpha$, for $\alpha \lt \kappa$, be subsets of $\Bbb R$, each of Lebesgue measure $0$.  Then $\bigcup \{M_\alpha~\mid ~ \alpha \lt \kappa \}$ has Lebesgue measure $0$.




Proof:  Let $\mu$ be Lebesgue measure.  Recall that a subset $M \subseteq \Bbb R$ has measure $0$ iff $\forall \varepsilon \gt 0$ there is an open $U \subseteq \Bbb R$ such that $M \subseteq U$ and $\mu(U) \leq \varepsilon.$  Fix $\varepsilon \gt 0$.  We shall find an open $U$ with $\mu(U) \leq \varepsilon$ and $\bigcup_{\alpha \lt \kappa} M_\alpha \subseteq U$.  Define




$$\Bbb P = \{ p \subset \Bbb R~\mid ~ p \text{ is open } \land \mu(p) \lt \varepsilon \}.$$




Define $p \leq q$ iff $q \subset p$; then $p$ and $q$ are compatible iff $\mu(p \cup q) \lt \varepsilon$ in which case $p \cup q$ is a common extension of $p$ and $q$.




Intuitively, "$p$ forces $p \subset U$."  Formally, if $G$ is a filter in $\Bbb P$, let $U_G=\bigcup G$.  $U_G$ is clearly open.  We must also check that $\mu(U_G) \leq \varepsilon$.  First, note that if $p, q \in G$, they have a common extension $r \in G$ and since $r \leq p \cup q,$ $\color{red} {\text{ we have }p \cup q \in G.}$  Next, by induction on $n$, if $p_1, \ldots , p_n \in G$, then $p_1 \cup \ldots \cup p_n \in G$ and hence in $\Bbb P$, so $\mu(p_1 \cup \ldots \cup p_n) \lt \varepsilon$.  Thus, by the countable additivity of $\mu$, whenver $A$ is a countable subset of $G, \mu(\bigcup A) \leq \varepsilon.$




$G$ is uncountable, but we now show that $\bigcup A = \bigcup G$ for some countable $A \subset G$; this will follow from the fact that the topology of $\Bbb R$ has a countable base.  Let $\mathscr B$ be the countable set of open intervals with rational endpoints.  If $x \in p$ for some $p \in G$, then there is a $q \in \mathscr B$ with $x \in q \subset p$.  Then $q \geq p$, $\color{red}{\text{ so } q \in G \text{ since } G \text{ is a filter.}}$  Thus, if $A = G \cap \mathscr B$, then $\bigcup A = \bigcup G = U_G$, so $\mu(U_G) \leq \varepsilon$.


The proof proceeds from there.  I can patch the first mistake.  Since the common extension $r \leq p \cup q$, it follows that $p \cup q \subseteq r$ so $\mu(p \cup q) \leq \mu(r) \lt \varepsilon,$ which is the result we need.  But I can't figure out how to patch the second mistake.
I'd appreciate any help.  Thanks.
(Incidentally, this seems sufficiently clear, and the book was in such widespread use, that I can't believe I'm the first person to find this, but I haven't been able to find any on-line discussion of these errors.)
 A: There is a mistake in the definition of filter. Condition (b) has to read:

$\forall p\in G\forall q\in\Bbb P(\color{orange}{p\leq q} \implies q\in G)$

That is, a filter is closed upwards. I would personally add a condition (c) that says that a filter is nonempty, but this is not really important.
This should clarify the problems with the proof, but in case it doesn't here is some more:

In the proof, $\Bbb P$ is defined to be the set of opens in $\Bbb R$ with measure $<\epsilon$ ordered reversely by inclusion. A filter $F$ of $\Bbb P$ is a set of opens such that that any finite union $\bigcup p_i$ of opens $p_i\in F$ is also in the filter (this is what condition (a) gives us), and such that for any open subset $p\subset q$ (with $p\in \Bbb P$) of some open $q\in F$ we also have $p\in F$ (this is what (b) gives us).
So a stronger condition $q\leq p$ gives us a larger open set than $p$. This is what we want, since we want a large open set (so that it covers the union of all $M_\alpha$) that still has measure $\leq\epsilon$.

So the first mistake in the proof is not a mistake:
If $p,q\in G$, then they have a common $r\leq p$ and $r\leq q$. This is the same as saying $r\supset p$ and $r\supset q$. Clearly then $r\supset p\cup q$ as well, which means $r\leq p\cup q$. By point (b) from the definition of filter, then $p\cup q\in G$. 
The second mistake follows directly from the corrected condition (b). Note that the set $\mathscr{B}$ of open rational intervals should be the set of open rational intervals with measure $<\epsilon$, since otherwise we cannot guarantee that $q\in \Bbb P$.
