# Simple proof check - in null sets, doesn't matter if cover is closed

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.
for example if each box has only one same point $P$ , then the total content is zero which is obviously less than $\epsilon$ but the S={P} is not the null set.
Note that the box $[a,a]$ is not empty but its length is $0$
• But if $S$ is countable, isn't it a null set because we can enclose each point $s_k$ with an open box of volume $\epsilon/2^k$? Commented Jun 9, 2018 at 22:27
• If the boxes are open and the inequality holds for every \epsilon $then it is the null set. Commented Jun 9, 2018 at 23:02 • @MohammadRiazi-Kermani But$S=\{P\}$is a null set. You can cover it with$(P-\frac{\epsilon}{3}, P + \frac{\epsilon}{3})$, which is a "box" with "volume"$\frac{2\epsilon}{3} < \epsilon$. Furthermore, if each box is a single point, and we have countably many of them that cover$S$, then$S\$ is a null set (since countable union of null sets is null) Commented Jun 10, 2018 at 6:50