[proofplan]
We first replace each null set by a Borel null superset, so it is enough to cover the Borel null sets by open sets of arbitrarily small [Lebesgue measure](/page/Lebesgue%20Measure). For a fixed $\varepsilon > 0$, we define a forcing poset whose conditions are finite pieces of the index set together with one [open set](/page/Open%20Set) of measure below $\varepsilon$ covering the corresponding null sets. The dense sets force every index to appear, and the ccc property follows by thinning an uncountable family using the $\Delta$-system lemma and finite rational-interval approximations of the open sets. Martin's axiom then gives a filter whose directed union is an open set covering all the null sets, and directedness gives the global measure bound.
[/proofplan]
[step:Replace the given null sets by Borel null supersets]
Let $\mathcal{L}^1$ denote one-dimensional Lebesgue outer measure on all subsets of $\mathbb{R}$, whose restriction to Lebesgue measurable subsets is Lebesgue measure. For each $i \in I$, since $N_i \subseteq \mathbb{R}$ is Lebesgue null, for every $m \in \mathbb{N}$ there exists an open set $O_{i,m} \subseteq \mathbb{R}$ such that $N_i \subseteq O_{i,m}$ and
\begin{align*}
\mathcal{L}^1(O_{i,m}) < \frac{1}{m}.
\end{align*}
Define
\begin{align*}
B_i := \bigcap_{m=1}^{\infty} O_{i,m}.
\end{align*}
Then $B_i$ is a Borel subset of $\mathbb{R}$, $N_i \subseteq B_i$, and monotonicity of Lebesgue measure gives
\begin{align*}
\mathcal{L}^1(B_i) \leq \mathcal{L}^1(O_{i,m}) < \frac{1}{m}
\end{align*}
for every $m \in \mathbb{N}$. Hence $\mathcal{L}^1(B_i)=0$. It is therefore enough to prove that $\bigcup_{i \in I} B_i$ is Lebesgue null.
[/step]
[step:Define the forcing poset that records finite covered subfamilies]
Fix $\varepsilon > 0$. Define $P_\varepsilon$ to be the set of all pairs $(F,U)$ such that $F \subseteq I$ is finite, $U \subseteq \mathbb{R}$ is open,
\begin{align*}
\mathcal{L}^1(U) < \varepsilon,
\end{align*}
and
\begin{align*}
\bigcup_{i \in F} B_i \subseteq U.
\end{align*}
For $(F,U),(F',U') \in P_\varepsilon$, define
\begin{align*}
(F',U') \leq (F,U)
\end{align*}
if and only if $F \subseteq F'$ and $U \subseteq U'$. This is a partial order: reflexivity and transitivity follow from inclusion of finite sets and inclusion of open subsets of $\mathbb{R}$, and antisymmetry follows because mutual inclusion forces equality of both coordinates.
For each $i \in I$, define
\begin{align*}
D_i := \{(F,U) \in P_\varepsilon : i \in F\}.
\end{align*}
We claim that each $D_i$ is dense in $P_\varepsilon$. Let $(F,U) \in P_\varepsilon$ and fix $i \in I$. Since $\mathcal{L}^1(U)<\varepsilon$, define
\begin{align*}
\eta := \varepsilon - \mathcal{L}^1(U) > 0.
\end{align*}
Because $B_i$ is Lebesgue null, there exists an open set $V \subseteq \mathbb{R}$ such that $B_i \subseteq V$ and
\begin{align*}
\mathcal{L}^1(V) < \eta.
\end{align*}
Then $F \cup \{i\}$ is finite, $U \cup V$ is open, and
\begin{align*}
\mathcal{L}^1(U \cup V) \leq \mathcal{L}^1(U) + \mathcal{L}^1(V) < \mathcal{L}^1(U) + \eta = \varepsilon.
\end{align*}
Also
\begin{align*}
\bigcup_{j \in F \cup \{i\}} B_j \subseteq U \cup V.
\end{align*}
Thus $(F \cup \{i\},U \cup V) \in P_\varepsilon$, it belongs to $D_i$, and it extends $(F,U)$. Hence $D_i$ is dense.
[/step]
[step:Prove that the forcing poset has the countable chain condition]
We prove that $P_\varepsilon$ has no uncountable antichain. Let
\begin{align*}
\mathcal{A} := \{(F_\alpha,U_\alpha) : \alpha \in A\}
\end{align*}
be an uncountable family of conditions in $P_\varepsilon$, where $A$ is an uncountable index set and the displayed map $\alpha \mapsto (F_\alpha,U_\alpha)$ is an enumeration of the family.
By the $\Delta$-system lemma applied to the uncountable family of finite sets $(F_\alpha)_{\alpha \in A}$, there exist an uncountable set $A_0 \subseteq A$ and a finite set $R \subseteq I$ such that
\begin{align*}
F_\alpha \cap F_\beta = R
\end{align*}
whenever $\alpha,\beta \in A_0$ are distinct.
Since each condition has $\mathcal{L}^1(U_\alpha)<\varepsilon$, the sets
\begin{align*}
A_q := \{\alpha \in A_0 : \mathcal{L}^1(U_\alpha)<q\},
\end{align*}
where $q \in \mathbb{Q}$ and $0<q<\varepsilon$, cover $A_0$. Since there are only countably many such rational numbers, choose $r \in \mathbb{Q}$ with $0<r<\varepsilon$ such that
\begin{align*}
A_1 := \{\alpha \in A_0 : \mathcal{L}^1(U_\alpha)<r\}
\end{align*}
is uncountable.
Choose $\delta \in \mathbb{Q}$ with $\delta>0$ and
\begin{align*}
r + 3\delta < \varepsilon.
\end{align*}
For each $\alpha \in A_1$, since $U_\alpha$ is an open subset of $\mathbb{R}$ of finite Lebesgue measure, choose a finite union $W_\alpha$ of open intervals with rational endpoints such that
\begin{align*}
\mathcal{L}^1(U_\alpha \triangle W_\alpha) < \delta
\end{align*}
and
\begin{align*}
\mathcal{L}^1(W_\alpha) < r+\delta.
\end{align*}
There are only countably many finite unions of open intervals with rational endpoints, so there exist an uncountable set $A_2 \subseteq A_1$ and one such finite union $W \subseteq \mathbb{R}$ such that
\begin{align*}
W_\alpha = W
\end{align*}
for every $\alpha \in A_2$.
Choose distinct $\alpha,\beta \in A_2$. We estimate the measure of $U_\alpha \cup U_\beta$. Since
\begin{align*}
U_\alpha \cup U_\beta \subseteq W \cup (U_\alpha \setminus W) \cup (U_\beta \setminus W),
\end{align*}
subadditivity gives
\begin{align*}
\mathcal{L}^1(U_\alpha \cup U_\beta)
\leq \mathcal{L}^1(W) + \mathcal{L}^1(U_\alpha \setminus W) + \mathcal{L}^1(U_\beta \setminus W).
\end{align*}
Because $U_\alpha \setminus W \subseteq U_\alpha \triangle W$ and $U_\beta \setminus W \subseteq U_\beta \triangle W$, we get
\begin{align*}
\mathcal{L}^1(U_\alpha \cup U_\beta) < (r+\delta)+\delta+\delta = r+3\delta < \varepsilon.
\end{align*}
Therefore
\begin{align*}
(F_\alpha \cup F_\beta, U_\alpha \cup U_\beta) \in P_\varepsilon.
\end{align*}
Indeed, $F_\alpha \cup F_\beta$ is finite, $U_\alpha \cup U_\beta$ is open, its measure is less than $\varepsilon$, and it contains every $B_i$ with $i \in F_\alpha \cup F_\beta$. This condition extends both $(F_\alpha,U_\alpha)$ and $(F_\beta,U_\beta)$, so those two conditions are compatible. Hence every uncountable family in $P_\varepsilon$ contains two compatible conditions, and $P_\varepsilon$ is ccc.
[guided]
The only delicate point in the proof is the ccc verification. We must show that an uncountable collection of conditions cannot be pairwise incompatible. A condition has two parts: a finite set $F_\alpha$ of indices and an open set $U_\alpha$ of measure less than $\varepsilon$. Two conditions will be compatible if the union of their finite index sets and the union of their open sets still form a valid condition. The finite index part causes no measure problem; the real issue is proving
\begin{align*}
\mathcal{L}^1(U_\alpha \cup U_\beta) < \varepsilon
\end{align*}
for some two distinct conditions in the uncountable family.
Let
\begin{align*}
\mathcal{A} := \{(F_\alpha,U_\alpha) : \alpha \in A\}
\end{align*}
be an uncountable family of conditions. We first apply the $\Delta$-system lemma to the finite sets $(F_\alpha)_{\alpha \in A}$. This produces an uncountable subset $A_0 \subseteq A$ and a finite root $R \subseteq I$ such that
\begin{align*}
F_\alpha \cap F_\beta = R
\end{align*}
whenever $\alpha,\beta \in A_0$ are distinct. The purpose of this step is to make the finite combinatorial part uniform; after this thinning, taking unions of two finite sets creates no hidden conflict.
Next we make the measure sizes uniformly bounded away from $\varepsilon$. For each rational number $q$ with $0<q<\varepsilon$, define
\begin{align*}
A_q := \{\alpha \in A_0 : \mathcal{L}^1(U_\alpha)<q\}.
\end{align*}
Since every $\mathcal{L}^1(U_\alpha)$ is strictly less than $\varepsilon$, the countable family $(A_q)_{q \in \mathbb{Q},\,0<q<\varepsilon}$ covers the uncountable set $A_0$. Therefore at least one of these sets is uncountable. Choose $r \in \mathbb{Q}$ with $0<r<\varepsilon$ such that
\begin{align*}
A_1 := \{\alpha \in A_0 : \mathcal{L}^1(U_\alpha)<r\}
\end{align*}
is uncountable.
Choose a rational number $\delta>0$ satisfying
\begin{align*}
r+3\delta<\varepsilon.
\end{align*}
For every $\alpha \in A_1$, approximate $U_\alpha$ in measure by a finite union $W_\alpha$ of open intervals with rational endpoints. More precisely, choose $W_\alpha$ so that
\begin{align*}
\mathcal{L}^1(U_\alpha \triangle W_\alpha)<\delta
\end{align*}
and
\begin{align*}
\mathcal{L}^1(W_\alpha)<r+\delta.
\end{align*}
This approximation is available because an open subset of $\mathbb{R}$ with finite Lebesgue measure is a countable disjoint union of open intervals, and finitely many bounded intervals with rational endpoints can approximate that union in measure as closely as desired.
There are only countably many finite unions of open intervals with rational endpoints. Since $A_1$ is uncountable, one approximation must occur uncountably often. Thus there are an uncountable set $A_2 \subseteq A_1$ and one finite rational interval union $W$ such that
\begin{align*}
W_\alpha = W
\end{align*}
for all $\alpha \in A_2$.
Now take distinct $\alpha,\beta \in A_2$. The common approximation $W$ controls both $U_\alpha$ and $U_\beta$. Since every point of $U_\alpha \cup U_\beta$ either lies in $W$, or lies in the part of $U_\alpha$ outside $W$, or lies in the part of $U_\beta$ outside $W$, we have
\begin{align*}
U_\alpha \cup U_\beta \subseteq W \cup (U_\alpha \setminus W) \cup (U_\beta \setminus W).
\end{align*}
By subadditivity of Lebesgue measure,
\begin{align*}
\mathcal{L}^1(U_\alpha \cup U_\beta)
\leq \mathcal{L}^1(W) + \mathcal{L}^1(U_\alpha \setminus W) + \mathcal{L}^1(U_\beta \setminus W).
\end{align*}
The inclusions $U_\alpha \setminus W \subseteq U_\alpha \triangle W$ and $U_\beta \setminus W \subseteq U_\beta \triangle W$ give
\begin{align*}
\mathcal{L}^1(U_\alpha \cup U_\beta) < (r+\delta)+\delta+\delta = r+3\delta < \varepsilon.
\end{align*}
Therefore the pair
\begin{align*}
(F_\alpha \cup F_\beta, U_\alpha \cup U_\beta)
\end{align*}
is a condition in $P_\varepsilon$. It extends both original conditions because $F_\alpha \subseteq F_\alpha \cup F_\beta$, $F_\beta \subseteq F_\alpha \cup F_\beta$, $U_\alpha \subseteq U_\alpha \cup U_\beta$, and $U_\beta \subseteq U_\alpha \cup U_\beta$. Thus $(F_\alpha,U_\alpha)$ and $(F_\beta,U_\beta)$ are compatible. Since every uncountable family contains such a compatible pair, $P_\varepsilon$ satisfies the countable chain condition.
[/guided]
[/step]
[step:Apply Martin's axiom to obtain one directed open cover]
Since $P_\varepsilon$ is ccc and the family of dense subsets
\begin{align*}
\mathcal{D} := \{D_i : i \in I\}
\end{align*}
has cardinality at most $\kappa$, $MA_\kappa$ yields a filter $G \subseteq P_\varepsilon$ meeting every $D_i$.
Define
\begin{align*}
O_\varepsilon := \bigcup\{U \subseteq \mathbb{R} : (F,U) \in G\}.
\end{align*}
Then $O_\varepsilon$ is open, being a union of open subsets of $\mathbb{R}$. For each $i \in I$, choose a condition $(F_i,U_i) \in G \cap D_i$. Since $i \in F_i$, the definition of $P_\varepsilon$ gives $B_i \subseteq U_i$, and therefore
\begin{align*}
B_i \subseteq O_\varepsilon.
\end{align*}
Hence
\begin{align*}
\bigcup_{i \in I} B_i \subseteq O_\varepsilon.
\end{align*}
[/step]
[step:Use directedness of the filter to prove the measure bound]
We prove that $\mathcal{L}^1(O_\varepsilon) \leq \varepsilon$. Let $K \subseteq O_\varepsilon$ be compact. Since $O_\varepsilon$ is the union of the open sets appearing in conditions of $G$, compactness gives finitely many conditions
\begin{align*}
(F_1,U_1),\dots,(F_n,U_n) \in G
\end{align*}
such that
\begin{align*}
K \subseteq \bigcup_{m=1}^{n} U_m.
\end{align*}
Because $G$ is a filter, it is directed downward with respect to the ordering $\leq$. Thus there exists $(F^*,U^*) \in G$ such that
\begin{align*}
(F^*,U^*) \leq (F_m,U_m)
\end{align*}
for every $m \in \{1,\dots,n\}$. By definition of the order, $U_m \subseteq U^*$ for every $m$, so
\begin{align*}
K \subseteq U^*.
\end{align*}
Since $(F^*,U^*) \in P_\varepsilon$, we have $\mathcal{L}^1(U^*)<\varepsilon$. Monotonicity gives
\begin{align*}
\mathcal{L}^1(K) \leq \mathcal{L}^1(U^*) < \varepsilon.
\end{align*}
Now $O_\varepsilon$ is open in $\mathbb{R}$. Writing $O_\varepsilon$ as a countable disjoint union of open intervals, the Lebesgue measure of $O_\varepsilon$ is the supremum of the measures of compact subsets obtained by taking finitely many closed subintervals inside finitely many of those components. Since every such compact subset has measure at most $\varepsilon$, it follows that
\begin{align*}
\mathcal{L}^1(O_\varepsilon) \leq \varepsilon.
\end{align*}
[/step]
[step:Let the measure tolerance tend to zero]
For every $\varepsilon>0$, we have constructed an open set $O_\varepsilon \subseteq \mathbb{R}$ such that
\begin{align*}
\bigcup_{i \in I} B_i \subseteq O_\varepsilon
\end{align*}
and
\begin{align*}
\mathcal{L}^1(O_\varepsilon) \leq \varepsilon.
\end{align*}
Therefore, by the definition of Lebesgue outer measure,
\begin{align*}
\mathcal{L}^1\left(\bigcup_{i \in I} B_i\right)=0.
\end{align*}
Thus $\bigcup_{i \in I} B_i$ is Lebesgue null. Since $N_i \subseteq B_i$ for every $i \in I$, monotonicity of Lebesgue outer measure gives
\begin{align*}
\mathcal{L}^1\left(\bigcup_{i \in I} N_i\right)
\leq
\mathcal{L}^1\left(\bigcup_{i \in I} B_i\right)
=0.
\end{align*}
Thus $\bigcup_{i \in I} N_i$ is Lebesgue null.
[/step]
