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 $f(0)=0, \lim_{x \to \infty} f(x) = m(A)$.


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).

  • 1
    $\begingroup$ Explain "measure $0 \leq \alpha < +\infty$. " $\endgroup$
    – Daron
    Mar 18, 2022 at 17:49
  • $\begingroup$ What is all this stuff about the continuous function for? $\endgroup$
    – Daron
    Mar 18, 2022 at 17:52
  • $\begingroup$ Which set is your $K$? $\endgroup$
    – Daron
    Mar 18, 2022 at 17:56
  • $\begingroup$ (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. $\endgroup$
    – gal127
    Mar 18, 2022 at 18:46
  • $\begingroup$ 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. $\endgroup$
    – gal127
    Mar 18, 2022 at 18:48

1 Answer 1


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.

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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 $\mu(K) = k \mu(A) = \alpha$.

  • $\begingroup$ 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 $\mathbb{R}^n$. $\endgroup$
    – gal127
    Mar 18, 2022 at 18:57
  • 1
    $\begingroup$ 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. $\endgroup$
    – Daron
    Mar 18, 2022 at 19:13

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