$A \subset \mathbb R^n$ is open, and $B \subset A$ is a null set, meaning:
$\forall \epsilon > 0$ there are open boxes $Q_1, Q_2, \dots \subset \mathbb R^n$ such that $$B \subset \bigcup_{j=1}^{\infty}Q_j$$
and $$\sum_{j=1}^{\infty}v(Q_j) < \epsilon$$
I want to show that not only do such boxes exist, but we can even find such boxes that also have the property that $Q_j \subset A$ for all $j$, and it's quite difficult for me.
Since $A$ is open, it can be represented as a countable union of closed boxes $A = \bigcup_{j=1}^{\infty}\overline{Q_j}$
So $ B \subset \bigcup_{j=1}^{\infty}\overline{Q_j}$, but we don't know that the sum of the volumes of these boxes is less than epsilon, and these boxes are not open.
My intuition tells me that there is a very very small $\epsilon$ such that if the volumes of the covers of $B$ is less than $\epsilon$, then they are all in $A$, but I can't prove it.
Suppose $A$ is not open. Would that make a difference?
In $\mathbb{R}^n$ every open set can be expressed as a countable union of open boxes. You can get the required sequence of boxes the folowing way:
1) Take $Q_j$ such that $B\subset\bigcup\limits_j Q_j$ and $\sum\limits_j \nu(Q_j)<\varepsilon$.
2) Express $Q_j \bigcap A$ as a countable union $\bigcup\limits_i \tilde{Q}_j^i$.
3) Since a countable union of countable sets is countable you can reindex all $\tilde{Q}_j^i$ as $\tilde{Q}_k$.