How to construct an unbounded set $A$ with outer measure $m^*(A)=3 ?$

Question

I'm trying to construct a set $A\subseteq\mathbb R$ with the following property:

$A$ is unbounded with outer measure $m^*(A)=3$.

The outer measure is defined as follows:

Let $A\subseteq\mathbb R$ and define the outer measure of $A$ by $$ m^*(A) := \inf\left\{ \sum_{n=1}^\infty \ell(I_n)\,\bigg|\,I_n \text{ is an open interval with } A\subseteq\bigcup^\infty_{n=1} I_n \right\}, $$ where $\ell(I_n)$ denotes the length of the interval $I_n$.

I am aware that unbounded open sets like e.g., $$ A_1 = \bigcup^\infty_{n=1}\left(n,n+\frac{1}{n^2}\right)$$ and $$ A_2 = \bigcup^\infty_{n=1}\left(2^n,2^n+2^{-n}\right)$$ have finite outer measure. In fact, I know that $m^*(A_2)=1$. I have also noticed that there are bounded open sets satisfying this property. (e.g., $m^*((0,3))=3$).

But how to construct an unbounded open set satisfying a specific value (e.g., $3$) for outer measure? TIA.

Answer 1

Ok, well you've mentioned: $$\bigcup_{n\ge1}(2^n,2^n+2^{-n})$$

Just take: $$\bigcup_{n\ge1}(2^n,2^n+3\cdot 2^{-n})$$Because its outer measure equals its measure equals (countable additivity, easy to check disjointness of the intervals): $$\sum_{n\ge1}3\cdot2^{-n}=3$$

Answer 2

You can just scale up or down any given set. In other words, for $k>0$, we can define $kA=\{x|\frac{x}{k}\in A\}$. Then, $kA$ is unbounded iff $A$ is and $m(kA)=k \cdot m(A)$, so you can turn your favorite unbounded set of nonzero finite measure into an unbounded set of any arbitrary nonzero finite measure.