A measure $\mu$ on $(\mathbb{Z^+}, 2^{\mathbb{Z^+}})$ such that $\{\mu(E):E\subset\mathbb{Z^+}\}=[0,1]$

Question

I have been trying to come up with a measure $\mu$ on $(\mathbb{Z^+}, 2^{\mathbb{Z+}})$ such that $\{\mu(E):E\subset\mathbb{Z^+}\}=[0,1]$ and I have thought about the following:

(EDIT: my proof is wrong since $\mu$, as defined, is not a measure because it doesn't satisfy the criterion that $\mu(\bigcup_{k=1}^{\infty}E_k)=\sum_{k=1}^{\infty} E_k$ if the $E_k$s are disjoint: I suppose it could be somehow patched by redefining $$\mu(E):=\begin{cases}
0 & \text{ if } E=\emptyset\\
\sup\{x\in [0,1]:x\geq q_k, k\in E\}-\inf\{x\in [0,1]:x\leq q_k, k\in E\} & \text{ if } E\neq\emptyset
\end{cases}$$ but I fear that there could be some Vitali set-like counterexamples showing that this too is not a measure so given how complicated this proof looks I think it is just better to use Kavi Rama Murthy's approach.)

Let $\{q_n\}_{n\in\mathbb{Z^+}}$ be an enumeration of the rational numbers in $[0,1]$:
then if we take $E\subset\mathbb{Z^+}$ by such enumeration this set will correspond to a subset $\{q_k:k\in E\}\subset\mathbb{Q}\cap[0,1]$ so if we set $$\mu(E):=\begin{cases}
0 & \text{ if } E=\emptyset\\
\sup\{x\in [0,1]:x\geq q_k, k\in E\} & \text{ if } E\neq\emptyset
\end{cases}$$ we have a measure on $(\mathbb{Z^+}, 2^{\mathbb{Z+}})$ .

It is clear from the definition of $\mu$ that $\{\mu(E):E\subset\mathbb{Z^+}\}\subset [0,1]$.
Now, if $x\in [0,1]$ and if $x$ is rational then there is $k\in\mathbb{Z^+}$ such that $q_k=x$ and being $\{k\}\in 2^\mathbb{Z^+}$, $\mu(\{k\})=q_k=x\in[0,1]$; if $x$ is irrational then, by Dedekind cut construction it can be seen as the supremum of a set of rational numbers in $[0,1]$ which will correspond via the enumeration above to some subset $E\in 2^{\mathbb{Z^+}}$ such that $\mu (E)=x$ and this concludes the proof that $\{\mu(E):E\subset\mathbb{Z^+}\}=[0,1]$.

Is this proof correct (and is there a way to improve it)? Are there other (possibly simpler) measures that one can define on $(\mathbb{Z^+},2^{\mathbb{Z^+}})$?

Thanks.

Answer 1

Answer for the last part: Define $\mu (A)=\sum_{n \in A} \frac 1 {2^{n}}$. Any point $t$ between $0$ and $1$ has an expansion $t=\sum \frac {t_n} {2^{n}}$ where $t_n=0$ or $1$ for each $n$. If you take $A=\{n: t_n=1\}$ then $\mu(A)=t$.