I was taught that a set $S$ is said to be null set if $\forall \epsilon > 0$ there are open boxes $Q_1, Q_2, \dots$ such that $S \subset \bigcup_{i=1}^{\infty} Q_i$ and $\sum_{i=1}^{\infty}v(Q) < \epsilon$
The emphasis was on open boxes, the professor stressed it many times.
This seems counter-intuitive to me, in my opinion it does not matter whether the boxes are open or closed, as one follows from the other.
Proof:
If the boxes are open, then $Q_i \subset \overline{Q_i}$ and of course the volumes are equal, so $S \subset \bigcup_{i=1}^{\infty}Q_i \subset\bigcup_{i=1}^{\infty}\overline{Q_i}$, so we found a cover using closed boxes.
If the boxes are closed: Consider the closed box $Q = [a_1,b_1] \times[a_2,b_2]\times \dots\times[a_n,b_n]$.
We can inflate it to create a new open box, like so $Q^* = (2a_1-b_1,2b_1-a_1)\times(2a_2-b_2,2b_2-a_2)\times\dots\times(2a_n-b_n,2b_n-a_n)$
Obviously now $Q \subset Q^*$.
And furthermore, $v(Q^*) = 3(b_1-a_1)\cdot3(b_2-a_2)\dots 3(b_n-a_n) = 3^nv(Q)$
So if $Q_1,Q_2, \dots $ are closed boxes that cover $S$ with sum of volumes less than epsilon, then $Q_1^*,Q_2^*,\dots$ are open boxes that cover $S$, with sum of volumes less than $3^n \epsilon$, which is arbitrarily small.
Is this correct? I know it seems trivial but the professor kind of messed with my head.
Edit:
If this is true and this proof is correct, then it also doesn't matter if the boxes are neither closed nor open. Like $(0,1] \times [0,1)$ or $(0,1) \times [0,1]$. As we can inflate them like I did to create strictly open boxes, and from there it's a solved case. If I am right.
I assume you're working in $\mathbb{R}^n$.
A null set is one of volume 0. Since volume is first defined on open boxes, we extend it to any set $S \subset \mathbb{R}^n$ by defining $$v(S) := \inf_{\cup_{n \geq 1} Q_n \supset S} \sum_{n \geq 1} v(Q_n),$$ where the $Q_n$ are open boxes. When $S$ is a null set, i.e. $v(S)=0,$ this means exactly that for all $\epsilon > 0$ there exist open $\cup_{n \geq 1} Q_n \supset S$ with $$\sum_{n \geq 1} v(Q_n) < \epsilon.$$ In the case that $S$ is null, you are right that we have $$\inf_{\cup_{n \geq 1} Q_n \supset S} \sum_{n \geq 1} v(Q_n)=0$$ for all closed $Q_n$. We could define volume with closed boxes, but it is more natural to define volume using open boxes mostly because the of "volume of a union/sum of the volumes" relationship.