Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero?

Question

There is an elementary theory of subsets of $\Bbb{R}^n$ of measure zero, namely one defines the volume of a cube in the obvious way and one says that a subset $A$ has measure zero if given any $\epsilon>0$ there exists a countable number of cubes that cover $A$ and such that the sum of the volumes of the cubes is $\leq \epsilon$. One can show, with modest effort, that this notion is invariant under diffeomorphisms and thus leads to the notion of subsets of measure zero on a smooth manifold. This notion shows up in Sard's Theorem which says that the set of critical values has measure zero.

Is there an elementary argument why non-empty open subsets do not have measure zero? Evidently it follows from standard measure theory, but for my topology course I would appreciate it if there was an elementary argument, but I can't think of one and I can't find one.

This is stated as an exercise in Lee's book on smooth manifolds, but it's not obvious to me. Note that even $n=1$ seems tricky.

Answer 1

The first part of my argument is borrowed from Iosif Pinelis, and the second part is different.

Every open set contains a closed cube, so it suffices to show the closed cube does not have measure $0$. In fact, we will show that for a cube of side length $r$, the total volume of open cubes covering it is at least $r^n$.

To do this, assume that it is covered by open cubes of total volume $<r^n$, and note that we can assume there are finitely many cubes in the covering by compactness. Let $\ell_1,\dots, \ell_m$ be the side lengths of the cubes.

Say our cube is $[0,r]^n$. Consider only the points of the form $(x_1,\dots, x_n)$ where $x_i \in \{0 , r/N, 2r/N, 3r/N, \dots, r\}$. There are $(N+1)^n$ such points. The maximum number of points of the form $\{0 , r/N, 2r/N, 3r/N, \dots, r\}$ contained in an interval of length $\ell$ is $\leq N \ell+1$, so the number of such points contained in a cube of side length $\ell$ is $\leq (N\ell/r+1)^n$.

Since every point is in one cube, we obtain $$ \sum_{i=1}^m (N \ell_i/r+1)^n \geq (N+ 1)^n$$

Divide both sides by $N^n$ and take the limit as $N$ goes to $\infty$ to obtain

$$ \sum_{i=1}^m \ell_i^n / r^n \geq 1$$ or $$ \sum_{i=1}^m \ell^n \geq r^n,$$ as desired.

Answer 2

Any nonempty open set in $\mathbb R^n$ contains a compact cube $C$ of volume $v:=|C|>0$. (All our cubes will be assumed to have edges parallel to the coordinate axes.)

So, it is enough to show that $C$ cannot be covered by a set $S$ of (say) open cubes with total volume $<v$. By compactness, without loss of generality the cardinality (say $k$) of the set $S$ is finite.

Replace now the compact cube $C$ by the corresponding left-open cube contained in $C$ of the same volume $v$, and still denote this left-open cube by $C$. A left-open cube here means the Cartesian product of left-open intervals. Also, replace each open cube $c\in S$ by the corresponding left-open cube containing $c$ of the same volume $|c|$, and still denote the resulting (covering $C$) finite set of left-open cubes by $S$. Moreover, by a semiring argument, without loss of generality the left-open cubes $c\in S$ are nonempty and pairwise disjoint, and their union is $C$.

Take any (left-open) cube $c\in S$. If all the hyperplanes through all the faces of $c$ do not intersect the interior of $C$, then $c$ contains $C$, and we are done.

Otherwise, there is some $c_*\in S$ such that the hyperplane $H$ through some one of the faces of $c_*$ intersects the interior of $C$. This hyperplane $H$ cuts $C$ into two disjoint left-open cubes, say $C_1$ and $C_2$. Also, $H$ cuts each $c\in S$ into two disjoint left-open cubes. Moreover, for each $j\in\{1,2\}$, the number of nonempty left-open cubes of the form $C_j\cap c$ for $c\in S$ is $<|S|$. Therefore and because $|C|=|C_1|+|C_2|$, the desired result follows by induction on $k=|S|$.

Answer 3

Let me mention another proof of the $\sigma$-additivity of the lenght of intervals, that uses transfinite induction instead of HB compactness. Say $n=1$, and we wish to prove that the set-function $$\lambda:[a,b)\mapsto b-a$$ is $\sigma$-additive on the family $\mathcal J$ of all left-closed, right-open intervals of $\mathbb{R}$ The key observation is that if $J:=[c,d)\in\mathcal J$ is partitioned into a countable family $\mathcal{A}:=\{[a,a')\}_{a\in A}$, the set $A\subset J$ of their left extrema is well-ordered by the order induced by $\mathbb R$ (Indeed, if $\emptyset\neq B\subset A$ one has $c\le\beta:=\inf B<d$ therefore $\beta$ is in some of the $[a,a')\in\mathcal{A}$, that is $a\le \beta<a'$. Since $\beta$ is the infimum of $B$, there are elements of $B$ in $[\beta,a')$, whence $\beta=a\in B$ because the only element of $A$ in $[a,a')\supset(\beta,a')$ is $a$). This allow to prove the thesis $$\lambda([c,d))= \sum_{x\in A} \lambda([x,x'[)$$ by (countable) transfinite induction on the well-ordered set $A':=A\cup\{d\}$: suppose the equality $$\lambda([c,a))= \sum_{x\in A\atop x<a} \lambda([x,x'[)$$
holds true for all $a\in A$, $a<b\in A'$: it then easily follows (try it!) that it is also true for $b$.

The same argument also work to show the $\sigma$-additivity on $$\mu :\mathcal{J}\ni[a,b)\mapsto f(b)-f(a),$$ with an increasing, left continuous function $f$. (and then the Carathéodory extension theorem extends it to the whole Borel $\sigma$-algebra).

Answer 4

This is a bit of a long comment and perhaps a useful complement to the other answers.

Having quickly read the question, I almost proceeded to downvote. But on second thought, I think it is not as trivial as it seems and, in fact, lies at the very heart of the standard construction of Lebesgue measure even for $n=1$ using Carathéodory's Theorem. After some easy simplifications it boils down to showing: if the finite semiopen interval $(a,b]$ is the disjoint union of $(a_i,b_i]$, $i\in\mathbb{N}$, then $$ b-a\le\sum_{i=1}^{\infty}(b_i-a_i)\ . $$ Now this is close to what the OP asks. Suppose the set contains a nonflattened cube, i.e., $b-a>0$. Show that the sum on the right or outer measure of the set is $>0$, e.g., $\ge b-a$.

After one more round of simplification (give yourself an $\varepsilon$ of room by shortening $(a,b]$ into a compact interval $[a,b]$, enlarging the $(a_i,b_i]$ into open intervals $(a_i,b_i)$ and finally extracting a finite cover), we arrive at the following purely combinatorial lemma.

Combinatorics Lemma: Suppose $[a,b]\subset\cup_{i=1}^{p}(a_i,b_i)$ then $$ b-a\le\sum_{i=1}^{p}(b_i-a_i)\ . $$

The lemma is not really difficult, but it is not trivial. I think the best way to see that is to imagine a game of Tetris where the covering intervals fall from the sky in the order in which they are numbered from $1$ to $p$ and they pile up (a bit also like in Xavier Viennot's theory of heaps of pieces), possibly in very complicated structures. The lemma can be proved by an easy induction on $p$: look at the interval $(a_i,b_i)$ which covers the endpoint $b$ and argue according to the relative position of $a_i$ and $a$. If $a_i<a$ that interval alone does all the work. If $a_i\ge a$, then the remaining $p-1$ intervals cover $[a,a_i]$, etc. etc.

Now the Lemma has an obvious $n$-dimensional generalization with finitely many open boxes covering a closed box. I suppose one could prove it by double induction on $n$ and $p$. Trying to go around that is perhaps why we typically use some Fubini integration argument instead.

Answer 5

It should also be mentioned that $[0,1]$ being not a null Lebesgue set can also be derived by the Lebesgue-Vitali characterization theorem for Riemann integrable function. If $[0,1]$ were negligible, any bounded function on it would be Riemann integrable, but we know that for instance the Dirichlet function $\chi_{[0,1]\cap\mathbb{Q}}$ is not.

Answer 6

One can show that a closed cube is not even negligible using (hyper)rectangles in your definition of "negligible".

Indeed, given a cover with rectangles (w.l.o.g. initially open up to enlarging them a bit, w.l.o.g. finite by compactness of the cube, and w.l.o.g. contained in the cube), up to subdividing them you can assume that whenever a side of a rectangle is $[a,b]$ (or $(a,b)$, $[a,b)$, $(a,b]$) there are no rectangles whose side (along the same direction) has an endpoint strictly between $a$ and $b$.

Now replace the rectangles with their closure. Now two rectangles are either disjoint or identical (up to borders) and, discarding duplicate rectangles, the sum of the volumes is the volume of the cube.