# Measure space $(X,\mathcal{S},\mu)$ such that $\{\mu(E):E\in\mathcal{S}\}=\{\infty\}\cup\bigcup_{k=1}^{\infty}[3k,3k+1]$

I am trying to find a measure space $$(X,\mathcal{S},\mu)$$ such that $$\{\mu(E):E\in\mathcal{S}\}=\{\infty\}\cup\bigcup_{k=0}^{\infty}[3k,3k+1]$$ and I have come up with the following:

$$(X,\mathcal{S},\mu)=(\mathbb{R},\mathcal{B},\mu)$$, where $$\mathcal{B}$$ is the set of all Borel subsets of $$\mathbb{R}$$ and $$\mu:\mathcal{B}\to [0,\infty], \mu(E):=|E\cap [0,1]|+\sum_{j=1}^{\infty}3j\cdot\delta_{3j}(E)$$, where $$|\cdot|$$ is the outer measure and $$\delta_{3j}$$ is the Dirac measure.

Now, $$\mu(\emptyset)=0$$ and if $$E_1,E_2,\dots$$ are disjoint sets in $$\mathcal{B}(\mathbb{R})$$ then $$\mu(\bigcup_{k=1}^{\infty}E_k)=|\bigcup_{k=1}^{\infty}E_k\cap [0,1]|+\sum_{j=1}^{\infty}3j\cdot \delta_{3j}(\bigcup_{k=1}^{\infty}E_k)=\sum_{k=1}^{\infty}|E_k\cap [0,1]|+\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}3j\cdot \delta_{3j}(E_k)=\sum_{k=1}^{\infty}[|E_k\cap [0,1]|+\sum_{j=1}^{\infty}3j\cdot\delta_{3j}(E_k)]=\sum_{k=1}^{\infty}\mu(E_k)$$ so $$\mu$$ is a measure and by the way it is built it gives us $$[0,1]$$ translated by $$3k$$ for $$k\geq 1$$, exactly as we wanted.

Is this correct? Is there another (perhaps simpler) measure space that has the same property? Thanks.

We know how to obtain a measure in $$(\mathbb{Z^+}, 2^{\mathbb{Z+}})$$ such that $$\{\mu(E):E\subset\mathbb{Z^+}\}=[0,1]$$ as shown in this answer.
The idea is of that answer is that we can asign each positive integer the following measure $$\mu (\{i\}) = \frac{1}{2^i}$$, such that $$\mu (A)=\sum_{n \in A} \frac 1 {2^{n}}$$. This covers all the values from $$[0,1]$$. We only need to add $$[3, 4]$$, $$[6, 7]$$..., and we would only need to have values in our measure such as $$3, 6, 9, ...$$. So a measure in $$(\mathbb{Z^+}, 2^{\mathbb{Z+}})$$ which bijects with $$\{\frac{1}{2^n} \} \cup \{3n\}$$ will work.
Your first term is Lebesgue measure on $$[0,1]$$. Instead of your complicated second term, I would just use $$3$$ times counting measure on any countably infinite subset of $$\mathbb R$$.