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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.

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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.

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  • $\begingroup$ Can you please elaborate your answer a bit more please, @Jordi ? $\endgroup$ Sep 10, 2023 at 12:13
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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 $\mathbb R$.

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  • $\begingroup$ Can you please elaborate your answer a bit more please, @GEdgar ? $\endgroup$ Sep 10, 2023 at 12:15

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