Compact submeasure without Darboux property
Posted: Tue Nov 06, 2018 4:08 pm
On one of the seminars a question whether every compact submeasure has Darboux property was posed.
One of the examples from the paper Paolo Leonetti, Salvatore Tringali: Upper and lower densities have the strong Darboux property, http://arxiv.org/abs/1510.07473 https://doi.org/10.1016/j.jnt.2016.11.005 can be used to get a counterexample.
It is Example 2. I will briefly summarize it here.
We start with any monotone and subadditive function $f\colon \mathcal P(\mathbb N)\to [0,1]$.
Then we define
$$\mu^*(X)=
\begin{cases}
1 & \text{if }f(x)>0, \\
0 & \text{otherwise}.
\end{cases}
$$
It is quite easy to check that $\mu^*$ defined in this way is monotone and subadditive.
If we take for $f$ the upper asymptotic density, then $\mu^*(2\mathbb N)=\mu^*(2\mathbb N+1)$ and for every $\varepsilon>0$ we have a decomposition of $\mathbb N$
$$\mathbb N = (2\mathbb N) \cup (2\mathbb N+1)$$
into finitely many sets such that each of them has measure smaller than $\varepsilon$.
One of the examples from the paper Paolo Leonetti, Salvatore Tringali: Upper and lower densities have the strong Darboux property, http://arxiv.org/abs/1510.07473 https://doi.org/10.1016/j.jnt.2016.11.005 can be used to get a counterexample.
It is Example 2. I will briefly summarize it here.
We start with any monotone and subadditive function $f\colon \mathcal P(\mathbb N)\to [0,1]$.
Then we define
$$\mu^*(X)=
\begin{cases}
1 & \text{if }f(x)>0, \\
0 & \text{otherwise}.
\end{cases}
$$
It is quite easy to check that $\mu^*$ defined in this way is monotone and subadditive.
If we take for $f$ the upper asymptotic density, then $\mu^*(2\mathbb N)=\mu^*(2\mathbb N+1)$ and for every $\varepsilon>0$ we have a decomposition of $\mathbb N$
$$\mathbb N = (2\mathbb N) \cup (2\mathbb N+1)$$
into finitely many sets such that each of them has measure smaller than $\varepsilon$.