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).
