# Measure of Compact Set with Empty Interior

For my Integration course I've been proposed the following problem with which I have been struggling:

Prove that there exists a compact set $$K \subset \mathbb{R}^n$$ with empty interior and measure $$0 \leq \alpha < +\infty$$.

My approach:

My first idea was to use an appropiate real continuous non-negative function such that $$f(x_0)=0, \lim_{x \to \infty} f(x) = +\infty$$ and use the intermediate value theorem. Following this reasoning, let $$A \subset \mathbb{R}^n$$ Lebesgue measurable which will be explicitly constructed later. Define, $$f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}, f(x) = m(A \cap [-\underline{x},\underline{x}])$$ where $$\underline{x} = (x,\ldots,x)^t$$ is a $$\mathbb{R}^n$$ vector. I claim that $$f$$ a continuous function such that $$f(0)=0, \lim_{x \to \infty} f(x) = m(A)$$.

Proof:

Given $$x_0 \in \mathbb{R}, \varepsilon > 0$$. Choose $$\delta = (\frac{\varepsilon}{2})^{1/n}$$, if $$x \in \mathbb{R}_{\geq 0} : |x-x_0| < \delta$$, we may suppose that $$x < x_0$$ (the other case is completely analogous), then $$|f(x)-f(x_0)| = |m(A \cap [-\underline{x},\underline{x}])-m(A \cap [-\underline{x_0},\underline{x_0}])| =$$ $$= m(A \cap ([-\underline{x_0},-\underline{x}) \cup (\underline{x},\underline{x_0}])) = 2 m(A \cap (\underline{x},\underline{x_0}]) = 2 |x-x_0|^n < 2 \delta^n < \varepsilon$$

$$f(0) = 0$$ is trivial. Finally, in order to compute $$\lim_{x \to \infty}$$, note that the function is monotonous increasing hence the limit exists (either finite or infinite limit). Let $$x_n$$ be a sequence such that $$x_n \to +\infty$$. Therefore, there exists a monotonous increasing subsequence $$\{x_{n_k}\}$$. Therefore, taking the limit to the sequence of $$m(A \cap [-\underline{x_{n_k}},\underline{x_{n_k}}])$$ is the measure of increasing sets, hence is the measure of the union of the sets, that is, $$\lim_{x \to \infty} f(x) = m(\bigcup_{k \in \mathbb{N}} A \cap [-\underline{x_{n_k}},\underline{x_{n_k}}]) = m(A)$$

Note that, in particular, if $$A$$ is a closed set with empty interior, then the set $$A \cap [-x,x]$$ would be compact and with empty interior. Therefore, it will be enough to find a convenient set $$A$$ such that (i) is closed, (ii) has empty interior (iii) has infinite measure in order to complete the exercise.

In doing so, let $$C$$ be the fat Cantor Set of measure $$\frac{1}{2}$$ in $$\mathbb{R}$$, which is a closed set with empty interior. Now, let $$C' = \bigcup_{n=0}^{\infty} (C+n)$$ hence $$m(C')=+\infty$$. Finally, define $$A = \mathbb{R} \times \cdots \times \mathbb{R} \times C'$$. Then, $$A$$ is closed as it is product of closed sets, has empty interior and has infinite measure. Therefore, such a set has been found verifying (i),(ii),(iii).

Proof of C' being closed:

Let $$\{x_n\} \subset C'$$ be an arbitrary sequence that converges, $$x_n \to x$$. Then, $$x_n$$ must be bounded, that is, there exist a finite number $$\{n_1,\ldots,n_k\}$$ such that $$\{x_n\} \subset \bigcup_{i=1}^k (C+n_i)$$ which is a closed set, hence the limit $$x \in \bigcup_{i=1}^k (C+n_i) \subset C'$$, that is, $$C'$$ is closed as desired.

Another idea for constructing A:

Let $$\{x_n\}$$ be an enumeration of $$\mathbb{Q}^n$$, let $$A' = \bigcup_{n \in \mathbb{N}} B(x_n:1/2^n)$$. Then, $$A'$$ is open and has finite measure. Therefore, let $$A=(A')^c$$. Then $$A$$ is closed, has infinite measure and empty interior (as it contains no rational points).

I would like to know whether if my reasoning is correct and if there are any other alternative (simpler) ways to handle the problem. Another idea I also had was to construct a set as the union of several fat Cantor Sets which add up to the given $$\alpha$$, which will be called $$A'$$. Then, define $$K=A' \times [0,1] \times \cdots \times [0,1]$$ and I would like to use that $$m(K)=m(A')$$ though I'm not sure about this fact (we haven't studied how measures behave under cartesian product).

• Explain "measure $0 \leq \alpha < +\infty$. " Mar 18, 2022 at 17:49
• What is all this stuff about the continuous function for? Mar 18, 2022 at 17:52
• Which set is your $K$? Mar 18, 2022 at 17:56
• (i) When I say measure $0 \leq \alpha < + \infty$, I refer to a set such that $m(K) = \alpha$. (ii) The continuity reasoning is in order to use the Intermediate Value Theorem conveniently as $f(0)=0, \lim_{x \to \infty} f(x) = m(A)$ hence given an adecuate set $A$ you can generate the rest of the sets from this function. Mar 18, 2022 at 18:46
• I'm going to explain how: Let $\alpha \in (0,\infty)$ be given, $A$ a set that has empty interior, is closed and infinite measure (which has been shown it exists) . Then, by the intermediate value theorem there exists $x_0$ such that $f(x_0) = \alpha$. Let $K=A \cup [-\underline{x_0},\underline{x_0}]$, then it clearly follows that $K$ is closed, bounded and has empty interior. By construction, $m(K) = f(x_0) = \alpha$ hence such a set has been found verifying the conditions of the exercise. Mar 18, 2022 at 18:48

Rescale The Set

An easier approach is to find a compact set $$A$$ with empty interior and positive finite measure, and rescale it to have measure $$\alpha$$.

For example the product $$A = C^n$$ of fat Cantor sets has measure $$(1/2)^n$$.

In general the power of a set $$B^n$$ has measure $$\mu(B)^n$$ if using any measure on $$\mathbb R$$ and the corresponding product measure on $$\mathbb R^n$$.

Another option is to construct a Menger Sponge by removing hypercubes from $$R^n$$ and keep track of how the volume remains positive.

Once you have $$A$$ just define $$k =\frac{\alpha}{ \mu(A)}$$ and $$K = kA = \{k a: a \in A\}$$.

It is straightforward to show compactness and empty-interior-ness transfers to $$K$$ and also that $$\mu(K) = k \mu(A) = \alpha$$.

• As I have already said, I haven't proved yet that $\mu(B^n) = \mu(B)^n$ (which obviously would make the statement much simpler). We haven't defined the Lebesgue measure on $\mathbb{R}^n$ as a product measure but extending the measure defined on the open sets of $\mathbb{R}^n$. Mar 18, 2022 at 18:57
• Hmm. . . Then it is probably easiest to construct a Menger Sponge as $A$. This is actually simpler if you do it in the abstract. Just let $U_1,U_2,\ldots$ be a base for the open sets on the cube, and iteratively remove an open piece of each $U_n$ of size at most $1/2^{n+1}$. That leaves a Sponge with volume $1/2$ or more. Mar 18, 2022 at 19:13