Proof. It is clear that (G1) $\implies $ (G2) $\implies $ (G3) $\implies $ (G4).

(G4) $\implies $ (G3) See [Reference Borodulin-Nadzieja, Farkas and Plebanek5, Proposition 5.4].

(G3) $\implies $ (G1) Suppose that $\mathcal {I}=\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ for some sequence $\boldsymbol {\mu }=(\mu _n)$ of lscsms such that $(S_n) \in \mathcal {F}_{\mathrm {disj}}$ , where $S_n:=\mathrm {supp}(\mu _n)$ for each n. Note that we can assume without loss of generality that $S:=\bigcup _n S_n=\omega $ . Indeed, in the opposite, if $S^c$ is finite then it is sufficient to replace $\mu _0(A)$ with $\mu _0(A)+|A\cap S^c|$ for all $A\subseteq \omega $ . Otherwise, let $(x_n)$ be the infinite increasing enumeration of $S^c$ and replace every $\mu _n(A)$ with $\mu _n(A)+\frac {1}{n}|A\cap \{x_n\}|$ . This is possible, considering that

$$ \begin{align*} \textstyle \mathcal{I}=\mathrm{Exh}(\varphi_{\boldsymbol{\mu}})=\mathrm{Exh}(\limsup_{n}\mu_n). \end{align*} $$

At this point, let $(T_n)\in \mathrm {Fin}^\omega $ be the sequence defined recursively as it follows: set $T_0:=[0,\max S_0]$ and, for each $n\in \omega $ , set

$$ \begin{align*} \textstyle T_{n+1}:=\left(\max \bigcup_{i\le n}T_i,\, \max\left(S_{n+1}\cup \bigcup \{S_k: \min S_k \le \max \bigcup_{i\le n}T_i\}\right)\right]. \end{align*} $$

Observe that $(T_n)$ is a sequence of (possibly empty) pairwise disjoint finite intervals such that $\bigcup _{n}T_n=\omega $ . Moreover, for each $n \in \omega $ there exists $j=j(n)\in \omega $ with $S_n\subseteq T_{j(n)}\cup T_{j(n)+1}$ : indeed, if $j(n)$ is the minimal integer such that $S_n\cap T_{j(n)}\neq \emptyset $ (so that $T_{j(n)}\neq \emptyset $ and $\min S_n \le \max T_{j(n)}$ ), then

$$ \begin{align*}\textstyle \max(T_{j(n)+1})\geq\max\left(\bigcup\{S_k: \min S_k \le \max \bigcup_{i\le j(n)}T_i\}\right)\geq\max(S_n). \end{align*} $$

Let $(V_n)$ be the biggest subsequence of $(T_n)$ with nonempty elements, so that $(V_n) \in \mathcal {F}_{\mathrm {int}}$ , and define the sequence $\boldsymbol {\nu }=(\nu _n)$ of lscsms by

$$ \begin{align*} \textstyle \forall n \in \omega, \forall A\subseteq \omega, \quad \nu_n(A):=\sup_k \mu_k(A \cap V_n). \end{align*} $$

Note that $\nu _n(A)=\sup _{k\geq n}\mu _k(A\cap V_n)$ , since $S_k\cap V_n=\emptyset $ whenever $k<n$ (indeed $S_k\subseteq \bigcup _{n\le k} T_n\subseteq \bigcup _{n\le k}V_n$ for all $k \in \omega $ ). Moreover, it follows by construction that $\text {supp}(\nu _n)=V_n$ for each $n \in \omega $ , hence it is sufficient to show that $\mathcal {I}=\mathrm {Exh}(\varphi _{\boldsymbol {\nu }})$ .

On the one hand, it is clear that if $\mu _n(A)\to 0$ then

$$ \begin{align*} \textstyle \nu_n(A)=\sup_{k\ge n}\mu_k(A\cap V_n) \le \sup_{k\ge n}\mu_k(A) \to 0, \end{align*} $$

hence $\mathcal {I}\subseteq \mathrm {Exh}(\varphi _{\boldsymbol {\nu }})$ .

On the other hand, suppose that $\nu _n(A)\to 0$ and fix $\varepsilon>0$ . Then there exists $n_0\in \omega $ such that for all $n> n_0$ . Let $k_0$ be the minimal integer such that $S_k\cap \bigcup _{n\le n_0}V_n=\emptyset $ for all $k\ge k_0$ . Fix $k\ge k_0$ and $n\in \omega $ such that $S_k\subseteq V_n\cup V_{n+1}$ (hence, in particular, $n>n_0$ ). We conclude that

$$ \begin{align*} \begin{split} \mu_k(A)&=\mu_k(A\cap S_k)\leq\mu_k(A\cap (V_n\cup V_{n+1}))\\ &\textstyle \leq\mu_k(A\cap V_n)+\mu_k(A\cap V_{n+1}) \leq\nu_n(A)+\nu_{n+1}(A)\le \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon, \end{split} \end{align*} $$

which shows that $\mu _k(A)\to 0$ , therefore $\mathrm {Exh}(\varphi _{\boldsymbol {\nu }}) \subseteq \mathcal {I}$ .⊣

Proof. It is clear that (D1) $\implies $ (D2) $\implies $ (D3) $\implies $ (D4).

(D3) $\Longleftrightarrow $ (D5) See [Reference Solecki and Todorcevic24, Lemma 6.7].

(D4) $\implies $ (D1) See [Reference Solecki and Todorcevic25, Lemma 3.1].⊣

Proof. (F1) $\implies $ (F2) This is obvious.

(F2) $\implies $ (F3) See [Reference Solecki and Todorcevic24, Lemma 6.7].

(F3) $\implies $ (F4) See [Reference Solecki and Todorcevic24, Proposition 6.8(b)].

(F4) $\implies $ (F1) If $\mathcal {I}=\mathrm {Fin}$ then $\mathcal {I}=\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ , where $\boldsymbol {\mu }=(\mu _n)$ and each $\mu _n$ is the Dirac measure on $n \in \omega $ . If $\mathcal {I}=\mathrm {Fin}\oplus \mathcal {P}(\omega )$ is represented on $\omega $ as $\{A\subseteq \omega : A \cap 2\omega \in \mathrm {Fin}\}$ , then $\mathcal {I}=\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ , where $\mu _n$ is the Dirac measure on $2n$ , for each $n \in \omega $ .⊣

Proof. It is sufficient to see that $h[\{0\}\times \omega ]$ is an infinite set which does not contain any infinite subset in $\mathcal {I}$ .⊣

Proof. Let $\varphi $ be a lscsm such that $\mathcal {I}=\mathrm {Exh}(\varphi )$ . We may assume that $\mathrm {supp}(\varphi )=\omega $ (indeed, it is easy check that $\mathcal {I}=\mathrm {Exh}(\tilde {\varphi })$ , where $\tilde {\varphi }$ is the lscsm defined by $\tilde {\varphi }(A):=\varphi (A)+\sum _{a \in A\setminus \mathrm {supp}(\varphi )}1/(a+1)^{2}$ for all $A\subseteq \omega $ ).

Let $\nu $ be the submeasure defined by

(3) $$ \begin{align} \forall B\subseteq \omega^2,\quad \nu(B):=\varphi\left(\left\{m \in \omega: B_{(m)} \neq \emptyset\right\}\right). \end{align} $$

To conclude the proof, we claim that $\widehat {\mathcal {I}}=\mathrm {Exh}(\lambda )$ , where $\lambda $ is the submeasure defined by

(4) $$ \begin{align} \forall A\subseteq \omega,\quad \lambda(A):=\nu(h^{-1}[A]). \end{align} $$

(Note that $\lambda $ is a lscsm and that $\mathrm {supp}(\lambda )=\omega $ .)

$\mathrm {Exh}(\lambda )\subseteq \widehat {\mathcal {I}}$ : Fix $A \in \mathrm {Exh}(\lambda )$ and set $B:=h^{-1}[A]$ . Then

(5) $$ \begin{align} \textstyle 0=\|A\|_\lambda=\inf_{F \in \mathrm{Fin}}\lambda(A\setminus F)=\inf_{G \in [\omega^2]^{<\omega}}\nu(B\setminus G). \end{align} $$

First, we want to prove that $B \in \emptyset \times \mathrm {Fin}$ . Indeed, in the opposite, there would exist $m \in \omega $ such that $B_{(m)}\notin \mathrm {Fin}$ . However, we would obtain

$$ \begin{align*} \textstyle \nu(B\setminus G) \ge \nu((\{m\}\times B_{(m)})\setminus G)=\varphi(\{m\})>0, \end{align*} $$

for every finite set $G\subseteq \omega ^2$ , which contradicts (5). Secondly, we show that $B \in \mathcal {I}\times \emptyset $ . Thanks to (5), for every $\varepsilon>0$ , there exists a finite set $G\subseteq \omega ^2$ such that $\nu (B\setminus G)<\varepsilon $ . Let $F=\{m\in \omega :G_{(m)} \neq \emptyset \}\in \mathrm {Fin}$ . Then

$$ \begin{align*} \varphi\left(\left\{m\in\omega:B_{(m)} \neq \emptyset\right\}\setminus F\right)\leq \varphi\left(\left\{m\in\omega:(B\setminus G)_{(m)} \neq \emptyset\right\}\right)=\nu(B\setminus G)<\varepsilon. \end{align*} $$

By the arbitrariness of $\varepsilon $ , we have $B \in \mathcal {I}\times \emptyset $ . To sum up, we have $B \in (\emptyset \times \mathrm {Fin}) \cap (\mathcal {I}\times \emptyset )$ , so that $h^{-1}[\mathrm {Exh}(\lambda )] \subseteq (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}\times \emptyset )$ .

$\widehat {\mathcal {I}}\subseteq \mathrm {Exh}(\lambda )$ : Suppose now that $B \in (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}\times \emptyset )$ and fix $\varepsilon>0$ . Since $B \in \mathcal {I}\times \emptyset $ , there exists $F\in \mathrm {Fin}$ such that $\varphi \left (\left \{m\in \omega : B_{(m)} \neq \emptyset \right \}\setminus F\right )<\varepsilon $ . However, since $B \in \emptyset \times \mathrm {Fin}$ , the set $G:=B \cap (F\times \omega )$ is finite. Hence

$$ \begin{align*} \nu(B\setminus G)=\varphi\left(\left\{m\in\omega:(B\setminus G)_{(m)} \neq \emptyset\right\}\right)=\varphi\left(\left\{m\in\omega:B_{(m)} \neq \emptyset\right\}\setminus F\right)<\varepsilon. \end{align*} $$

Therefore $\widehat {\mathcal {I}}=\mathrm {Exh}(\lambda )$ , which concludes the proof.⊣

Proof. Let $\varphi $ be a density-like lscsm such that $\mathcal {I}=\mathrm {Exh}(\varphi )$ and consider the lscsm $\lambda $ defined as in the proof of Theorem 3.2. Fix $\varepsilon>0$ . By Proposition 2.2, there is $\delta =\delta (\varepsilon )>0$ such that for all $(E_n) \in \mathrm {Fin}^\omega \cap \mathcal {G}_{\varphi ,\delta }$ there exists $I \in [\omega ]^\omega $ with $\varphi \left (\bigcup _{i \in I}E_i\right )<\varepsilon $ . We claim that the same $\delta $ witnesses the fact that $\lambda $ is density-like.

Fix $(F_n) \in \mathcal {F}_{\mathrm {disj}} \cap \mathcal {G}_{\lambda ,\delta }$ . Define $E_n:=\left \{m \in \omega : h^{-1}[F_n]_{(m)}\neq \emptyset \right \} \in \mathrm {Fin}$ for each $n \in \omega $ and note that

$$ \begin{align*}\varphi(E_n)=\varphi\left(\left\{m \in \omega: h^{-1}[F_n]_{(m)}\neq \emptyset\right\}\right)=\lambda(F_n)<\delta. \end{align*} $$

Thus, $(E_n) \in \mathrm {Fin}^\omega \cap \mathcal {G}_{\varphi ,\delta }$ and there exists $I \in [\omega ]^\omega $ with $\varphi \left (\bigcup _{i \in I}E_i\right )<\varepsilon $ . Then

$$ \begin{align*}\textstyle \lambda(F)=\varphi(\{m \in \omega: h^{-1}[F]_{(m)} \neq \emptyset\})=\varphi(\bigcup_{i \in I}E_i)<\varepsilon, \end{align*} $$

where $F:=\bigcup _{i \in I}F_i$ . This concludes the proof.⊣

Proof. Let us suppose that $\widehat {\mathcal {I}}=\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ , where $\boldsymbol {\mu }=(\mu _n)$ is a sequence of lscsms such that $(G_n) \in \mathcal {F}_{\mathrm {disj}}$ , where $G_n:=\mathrm {supp}(\mu _n)$ for each $n \in \omega $ .

Fix a strictly increasing sequence $(x_n) \in \omega ^\omega $ such that

$$ \begin{align*}\forall n \in \omega, \quad x_n \in h[\{n\} \times \omega] \,\,\,\text{ and }\,\,\, |G_n \cap X| \le 1, \end{align*} $$

where $X:=\{x_k: k \in \omega \}$ (it is easy to see that such sequence exists). It follows that $X\notin h[\mathcal {I}\times \emptyset ]$ , hence $X\notin \widehat {\mathcal {I}}=\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ . This implies that there exists $\varepsilon>0$ and a strictly increasing sequence $(m_t)\in \omega ^\omega $ such that $\mu _{m_t}(X) \ge \varepsilon $ for all $t \in \omega $ . However, by construction, each $G_{m_t}$ contains at most one element from X; hence, exactly one since $\mu _{m_t}(X)\neq 0$ , let us say $\{y_t\}:=G_{m_t} \cap X$ for all $t \in \omega $ . It follows that $\mu _{m_t}(Z) \not \to 0$ , where Z stands for any infinite subset of $Y:=\{y_t: t \in \omega \}$ , therefore $\mathcal {P}(Y) \cap [\omega ]^\omega \cap \mathrm {Exh}(\varphi _{\boldsymbol {\mu }})=\emptyset $ . This implies that every infinite subset of Y does not belong to $\widehat {\mathcal {I}}$ . Considering that $Y \cap h[\{n\}\times \omega ]$ is finite for all $n \in \omega $ , this contradicts the hypothesis that $\mathcal {I}$ is tall.⊣

Proof of Theorem 1.3. Let $\mathcal {I}_d$ be the ideal of density zero sets, which is a tall generalized density ideal. By Lemma 3.1 and Theorems 3.4 and 3.5, we get that $\widehat {\mathcal {I}}_d$ is a (nontall) density-like ideal which is not a generalized density ideal.⊣

Question 2. How many pairwise nonisomorphic ideals $\widehat {\mathcal {I}}$ are there, with $\mathcal {I}$ tall density-like ideal?

Proof. Thanks to [Reference Kwela, Popławski, Swaczyna and Tryba15, Theorem 3], there exists a family of simple density ideals $\{\mathcal {I}_\alpha : \alpha <2^\omega \}$ such that $\mathcal {I}_\alpha \not \leq _{\mathrm {K}}\mathcal {I}_\beta $ for all distinct $\alpha ,\beta <2^\omega $ .

Hence, given distinct $\alpha ,\beta <2^{\omega }$ , we claim that $\widehat {\mathcal {I}_\alpha }$ is not isomorphic to $\widehat {\mathcal {I}_\beta }$ , i.e., there is no bijection $f: \omega ^2\to \omega ^2$ such that $f[A] \in (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}_\alpha \times \emptyset )$ if and only if $A \in (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}_\beta \times \emptyset )$ for all $A\subseteq \omega ^2$ .

Suppose that $f: \omega ^2 \to \omega ^2$ is a bijection and suppose that there exist an infinite set $A\subseteq \omega $ and $k\in \omega $ such that $f[A\times \{0\}]\subseteq \{k\}\times \omega $ . Since $\mathcal {I}_\beta $ is tall, there is an infinite $B\subseteq A$ such that $B\in \mathcal {I}_\beta $ . Thus, $B\times \{0\}\in (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}_\beta \times \emptyset )$ , but $f[B\times \{0\}]\notin (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}_\alpha \times \emptyset )$ as $f[B\times \{0\}]\cap (\{k\}\times \omega )$ is infinite. This implies that the function $g:\omega \to \omega $ defined by $f(n,0) \in \{g(n)\} \times \omega $ for all $n \in \omega $ is finite-to-one.

Since $\mathcal {I}_\alpha \not \leq _{\mathrm {K}}\mathcal {I}_\beta $ , there exists a (necessarily infinite) set $X\in \mathcal {I}_\alpha $ such that $g^{-1}[X]\notin \mathcal {I}_\beta $ . Define $Y:=f[\omega \times \{0\}] \cap (X\times \omega )$ . Note that, since g is finite-to-one, then $Y\subseteq f[\omega \times \{0\}] \in \emptyset \times \mathrm {Fin}$ . Hence $Y \in (\emptyset \times \mathrm {Fin}) \cap (\mathcal {I}_\alpha \times \emptyset )$ .

To conclude the proof, let us suppose for the sake of contradiction that $f^{-1}[Y] \in (\emptyset \times \mathrm {Fin})\cap (\mathcal {I}_\beta \times \emptyset )$ . Hence, in particular, $f^{-1}[Y] \in \mathcal {I}_\beta \times \emptyset $ , that is,

$$ \begin{align*}Z:=\{n \in\omega: f^{-1}[Y]_{(n)}\neq \emptyset\} \in \mathcal{I}_\beta. \end{align*} $$

On the other hand, we have

$$ \begin{align*} \begin{split} Z&=\{n \in \omega: \exists k \in \omega, (n,k) \in f^{-1}[Y]\}=\{n \in \omega: \exists k \in \omega, f(n,k) \in Y\}\\ &=\{n \in \omega: f(n,0) \in Y\}\supseteq \{n \in \omega: g(n) \in X\}=g^{-1}[X] \notin \mathcal{I}_\beta, \end{split} \end{align*} $$

where we used that, if $g(n) \in X$ , then $f(n,0) \in \{g(n)\}\times \omega $ and $f(n,0) \in f[\omega \times \{0\}]$ , i.e., $f(n,0) \in Y$ . This completes the proof.⊣

Proof. Let $\mathcal {I}$ be a simple density ideal. Then $\mathcal {I}$ is a density ideal and, in particular, it is nonpathological. Thanks to Remark 3.3, $\widehat {\mathcal {I}}$ is nonpathological as well. The claim follows by Lemma 3.1, Theorems 3.4–3.6.⊣

Question 3. Does there exist a tall density-like ideal which is not a generalized density ideal?

Proof. Define $S_n:=\mathrm {supp}(\mu _n)$ for each $n \in \omega $ .

Only If part. Suppose that $\boldsymbol {\mu }$ is not equi-density-like. Then there exists $\varepsilon>0$ such that for all $\delta>0$ there are $n \in \omega $ and $(F_k) \in \mathcal {F}_{\mathrm {disj}} \cap \mathcal {G}_{\mu _n,\delta }$ for which $\mu _n(\bigcup _{i \in I}F_i) \ge \varepsilon $ whenever $I\in [\omega ]^{\omega }$ . We claim that this $\varepsilon>0$ witnesses that $\varphi $ is not density-like. To this aim, fix any $\delta>0$ and let n and $(F_k)$ be as before. Define $E_k:=F_k \cap S_n$ for all $k \in \omega $ . Then $\varphi (E_k)=\mu _n(E_k)<\delta $ and $(E_k) \in \mathcal {F}_{\mathrm {disj}} \cap \mathcal {G}_{\varphi , \delta }$ . At the same time, we have

$$ \begin{align*}\textstyle \forall I \in [\omega]^{\omega}, \quad \varphi\left(\bigcup_{i \in I}E_i\right)=\mu_n\left(\bigcup_{i \in I}E_i\right)=\mu_n\left(\bigcup_{i \in I}F_i\right)\ge \varepsilon. \end{align*} $$

Therefore $\varphi $ is not density-like.

If part. Conversely, suppose that $\boldsymbol {\mu }$ is equi-density-like, and fix $\varepsilon>0$ . Then there exists a sufficiently small such that

(6) $$ \begin{align} \textstyle \forall n \in \omega, \forall (F_k) \in \mathcal{F}_{\mathrm{disj}} \cap \mathcal{G}_{\mu_n,\delta}, \exists I \in [\omega]^\omega, \mu_n\left(\bigcup_{i \in I}F_i\right)<\frac{\varepsilon}{4}. \end{align} $$

Fix $(F_k) \in \mathcal {F}_{\mathrm {disj}} \cap \mathcal {G}_{\varphi ,\delta }$ and define $(T_k) \in \mathrm {Fin}^\omega $ by $T_k:=\{n \in \omega : F_k \cap S_n \neq \emptyset \}$ for all $k \in \omega $ . At this point, we claim that there exist a sequence of infinite sets $(X_j) \in (\mathrm {Fin}^+)^\omega $ and an increasing sequence $(i_j) \in \omega ^\omega $ such that, for all $j \in \omega $ :

1. (i) $i_j=\min X_j$ ,

2. (ii) $X_{j+1}\subseteq X_j\setminus \{i_j\}$ , and

3. (iii) $\mu _t\left (F^{(j)} \cup \bigcup _{i\in X_{j+1}} F_i\right )<\frac {\varepsilon }{2}$ for each $t\in T^{(j)}$ , where $F^{(j)}:=\bigcup _{k\le j} F_{i_k}$ and $T^{(j)}:=\bigcup _{k\le j} T_{i_k}$ .

We define these sequences recursively. Start with $X_0=\omega $ and $i_0=0$ . Suppose now that $X_k$ and $i_k$ have been defined for all $k\le j \in \omega $ and satisfy (i)–(iii). If $T:=T_{i_j}\setminus T^{(j-1)}$ is empty, set $X_{j+1}:=X_j\setminus \{i_j\}$ and $i_{j+1}:=\min X_{j+1}$ ; if $T\neq \emptyset $ , since $\mu _n(F_k)\leq \varphi (F_k)<\delta $ for all $n,k\in \omega $ , we can find $X_{j+1}\subseteq X_j\setminus \{i_j\}$ such that $\mu _t\left (\bigcup _{i \in X_{j+1}}F_i\right )<\frac {\varepsilon }{4}$ for all $t \in T$ ; finally, set $i_{j+1}:=\min X_{j+1}$ . If $j\neq 0$ and $t \in T^{(j-1)}$ , it follows by the induction hypothesis that

$$ \begin{align*}\textstyle \mu_t\left(F^{(j)} \cup \bigcup_{i\in X_{j+1}} F_i\right)\le \mu_t\left(F^{(j-1)} \cup \bigcup_{i\in X_{j}} F_i\right) < \frac{\varepsilon}{2}. \end{align*} $$

On the other hand, if $j=0$ or $t \in T$ , then

$$ \begin{align*}\textstyle \mu_t\left(F^{(j)} \cup \bigcup_{i\in X_{j+1}} F_i\right)\le \mu_t(F_{i_j})+\mu_t\left(\bigcup_{i\in X_{j+1}} F_i\right) < \delta+\frac{\varepsilon}{4}< \frac{\varepsilon}{2}, \end{align*} $$

which proves the condition (iii) and completes the induction.

To complete the proof, note that if $n \notin \bigcup _k T^{(k)}$ then $\mu _n\left (\bigcup _k F^{(k)}\right )=0$ for all $n \in \omega $ . Moreover, if $n \in \bigcup _k T^{(k)}$ then

$$ \begin{align*}\textstyle \forall j \in \omega, \quad \mu_n\left(\bigcup_k F^{(k)}\right)\le \mu_n\left(F^{(j)} \cup \bigcup_{i\in X_{j+1}} F_i\right)<\frac{\varepsilon}{2}. \end{align*} $$

Thus $\varphi \left (\bigcup _k F^{(k)}\right )\le \frac {\varepsilon }{2}<\varepsilon $ , which shows that $\varphi $ is density-like.⊣

Proof. Let $\varphi _1,\ldots ,\varphi _k$ be strongly-density-like lscsms, fix $a_1,\ldots ,a_k \in [0,1]$ with $\sum _{i\le k} a_i=1$ , and define the lscsm $\varphi :=(\sum _{i\le k}a_i \varphi _i^q)^{1/q}$ . Set $c:=\frac {1}{2}\min _{i\le k}c(\varphi _i)$ , so that $2c$ is a witnessing constant for each $\varphi _i$ .

Fix $\varepsilon>0$ and a sequence $(F_j) \in \mathcal {F}_{\mathrm {disj}}\cap \mathcal {G}_{\varphi ,c\varepsilon }$ . For each $j \in \omega $ and $i \in \{1,\ldots ,k\}$ define the integer $ z_{i,j}:=\left \lfloor \varphi _i(F_j)/c\varepsilon \right \rfloor. $ Note that $z_{i,j}<a_i^{-1/q}$ , indeed

$$ \begin{align*}\textstyle \forall i=1,\ldots,k,\quad c\varepsilon>\varphi(F_j)\ge a_i^{1/q}\varphi_i(F_j) \ge a_i^{1/q}z_{i,j}c\varepsilon. \end{align*} $$

Considering that $L:=\prod _{i=1}^k (\omega \cap [0,a_i^{-1/q}))$ is finite, there exists $\ell =(\ell _1,\ldots ,\ell _k) \in L$ and an infinite set $J\subseteq \omega $ such that $z_{i,j}=\ell _i$ for all $i =1,\ldots ,k$ and $j \in J$ .

At this point, observe that $\sum _{i\le k}a_i\ell _i^q<1$ , indeed

$$ \begin{align*} \sum_{i\le k}a_i(c\varepsilon\ell_i)^q=\sum_{i\le k}a_i(c\varepsilon z_{i,\min J})^q\le \sum_{i\le k}a_i\varphi_i(F_{\min J})^q=\varphi(F_{\min J})^q < (c\varepsilon)^q. \end{align*} $$

Since each $\varphi _i$ is strongly-density-like with witnessing constant $2c$ , we have that $\varphi _i(F_j)<(z_{i,j}+1)c\varepsilon =(2c)\cdot \frac {(\ell _i+1)\varepsilon }{2}$ for all $j \in J$ . Hence there exists an infinite set $T\subseteq J$ such that $ \varphi _i(\bigcup _{t \in T}F_t)<\frac {(\ell _i+1)\,\varepsilon }{2}. $ for all $i=1,\ldots ,k$ .

Thanks to Minkowski’s inequality, we conclude that

$$ \begin{align*} \textstyle \varphi(\bigcup_{t \in T}F_t)<\frac{\varepsilon}{2}\left(\sum_{i\le k}a_i (\ell_i+1)^q\right)^{1/q}\le \frac{\varepsilon}{2}\left(\left(\sum_{i\le k}a_i\ell_i^q\right)^{1/q}+1\right)<\varepsilon, \end{align*} $$

which proves that $\varphi $ is strongly-density-like.⊣

Proof. By [Reference Farah6, Example 1.2.3(d), Theorem 1.13.3(a), and Lemma 1.13.9(Z3)], there is a sequence $\boldsymbol {\mu }=(\mu _n)$ of probability measures on $\omega $ such that $\mathcal {I}=\mathrm {Exh}(\sup _n \mu _n)$ , and $(M_n) \in \mathcal {F}_{\mathrm {disj}}$ , where $M_n:=\mathrm {supp}(\mu _n)$ for each $n \in \omega $ . Then $\widehat {\mathcal {I}}$ is isomorphic to the ideal $\mathrm {Exh}(\nu )$ on $\omega ^2$ , where $\nu $ is the lscsm defined by

$$ \begin{align*}\textstyle \forall A\subseteq \omega^2,\quad \nu(A)=\sup_{n\in \omega} \mu_n(\{k \in \omega: A_{(k)}\neq \emptyset\}), \end{align*} $$

cf. (3) in the proof of Theorem 3.2.

To conclude the proof, we show that $\nu =\psi $ , for some $\psi =\psi (\boldsymbol {\varphi },q,a,S)$ . To this aim, let $\boldsymbol {\varphi }$ be the constant sequence $(\varphi )$ , where $\varphi $ is the strongly-density-like lscsm defined by $\varphi (\emptyset )=0$ and $\varphi (S)=1$ for all nonempty $S\subseteq \omega $ . Let also q be the constant sequence $(1)$ , $S_n=M_n$ , and $a_k:=\sup _n \mu _n(\{k\})$ for all $n \in \omega $ (note that $\sum _{k\in S_n}a_k=1$ for all $n \in \omega $ ). It follows that

$$ \begin{align*} \forall A\subseteq \omega^2,\quad \psi(A)=\sup_{n \in \omega}\sum_{k \in S_n}a_k \varphi(A_{(k)})=\sup_{n \in \omega}\mu_n(\{k \in S_n: A_{(k)}\neq \emptyset\})=\nu(A). \end{align*} $$

Therefore $\widehat {\mathcal {I}}$ is a $\mathrm {DL}$ -ideal.⊣

Proof. For each $k \in \omega $ define the lscsm $\tilde {\varphi }_k$ on $\omega ^2$ by

$$ \begin{align*} \forall A\subseteq \omega^2,\quad \tilde{\varphi}_k(A)=\varphi_k(A_{(k)}). \end{align*} $$

Then each $\tilde {\varphi }_k$ is a strongly-density-like lscsm such that $c(\tilde {\varphi }_k)=c(\varphi _k)$ . Moreover, for each $n \in \omega $ , define the lscsm $\psi _n$ by

$$ \begin{align*}\forall A\subseteq \omega^2,\quad \psi_n(A)=\left(\sum_{k \in S_n}a_k\tilde{\varphi}_k(A)^{q_n}\right)^{1/q_n}. \end{align*} $$

It follows by Proposition 4.7 that each lscsm $\psi _n$ is strongly-density-like with witnessing constant $c(\psi _n)=\frac {1}{2}\min \{c(\tilde {\varphi }_k): k \in S_n\}$ . Since $\inf _n c(\varphi _n)>0$ , we have also $\inf _n c(\psi _n)>0$ , which implies that $(\psi _n)$ is equi-density-like sequence of lscsms with pairwise disjoint supports (indeed $\mathrm {supp}(\psi _n)=\bigcup _{k \in S_n}\{k\}\times \mathrm {supp}(\varphi _n)\subseteq S_n \times \omega $ ). Therefore, thanks to Theorem 4.2, $\psi =\sup _n \psi _n$ is density-like.⊣

Proof. Let $A\subseteq \omega ^2$ be an infinite set. If $A \in \mathrm {Exh}(\psi )$ then the claim is trivial. Hence, suppose hereafter that $A\in \mathrm {Exh}(\psi )^+$ , that is,

(8) $$ \begin{align} \|A\|_\psi=\inf_{F \in [\omega^2]^{<\omega}}\,\sup_{n \in \omega}\,\left(\sum_{k \in S_n}a_k \varphi_k^{q_n}(A_{(k)}\setminus F)\right)^{1/q_n}>0. \end{align} $$

Now, suppose that there exists $m \in \omega $ such that $A_{(m)}$ is infinite. Since $\mathrm {Exh}(\varphi _m)$ is tall, there exists an infinite set $B\subseteq A\cap (\{m\}\times \omega )$ such that $B_{(m)} \in \mathrm {Exh}(\varphi _m)$ . It follows by the definition of $\psi $ that $B \in \mathrm {Exh}(\psi )$ .

Otherwise $A \in (\emptyset \times \mathrm {Fin}) \cap \mathrm {Exh}(\psi )^+$ , so that $A_{(m)}$ is finite for each $m \in \omega $ . Let B be an infinite subset of A such that $|B \cap \bigcup _{m \in S_n}A_{(m)}|\le 1$ for all $n \in \omega $ (which exists, otherwise A itself would be finite, contradicting (8)). It follows that

$$ \begin{align*} \begin{split} \|B\|_\psi&\le \inf_{m\in \omega} \|B\setminus \left(\bigcup_{i\le m}S_i \times \omega\right)\|_{\psi} \le \inf_{m \in \omega}\psi\left( B\setminus \left(\bigcup_{i\le m}S_i \times \omega\right)\right)\\ &\le \inf_{m \in \omega} \sup_{n\ge m}\,\left(\sum_{k \in S_n}a_k \varphi_k^{q_n}(B_{(k)})\right)^{1/q_n} \\ &\le M \, \limsup_{n\to\infty} \max\{a_k^{1/q_n}: k \in S_n\} =0. \end{split} \end{align*} $$

Therefore $B \in \mathrm {Exh}(\psi )$ , concluding the proof.⊣

Proof. First of all, we have $\sup _n \varphi (\{n\})<\infty $ : indeed, in the opposite case, there would exist an increasing sequence $(n_k)$ in $\omega $ such that $\varphi (\{n_k\}) \ge k$ for all k and every infinite subset of $\{n_k: k \in \omega \}$ would not belong to $\mathrm {Exh}(\varphi )$ , contradicting the hypothesis that $\mathrm {Exh}(\varphi )$ is tall. Moreover, since $Q:=\sup _n q_n <\infty $ , we obtain

$$ \begin{align*}\textstyle \lim_n \max \{a_k^{1/q_n}: k \in S_n\}\le \lim_n \max \{a_k^{1/Q}: k \in S_n\} =0. \end{align*} $$

The claim follows by Proposition 4.16.⊣

Proof. Suppose for the sake of contradiction that $\mathrm {Exh}(\psi )=\mathrm {Exh}(\sup _n \mu _n)$ , where $(\mu _n)$ is a sequence of lscsms on $\omega ^2$ with finite pairwise disjoint supports and set $M_n:=\mathrm {supp}(\mu _n)$ for each n (note that $M_n \subseteq \omega ^2$ ). It follows by the standing assumptions that there exists a sequence $(F_n) \in \mathcal {F}_{\mathrm {incr}}$ such that

(9) $$ \begin{align} \forall n \in \omega,\quad \varphi_n(F_n) \ge \frac{\inf_{k \in \omega}\| \omega\|_{\varphi_k}}{2} \,\,\text{ and }\,\, X_n \cap \bigcup_{\substack{k \in \omega: \\ M_k \cap \,\bigcup_{i\le n-1}X_i \neq \emptyset}} M_k =\emptyset, \end{align} $$

where $X_n:=\{n\} \times F_n$ for each n.

Set $X:=\bigcup _n X_n$ . Then $X\notin \mathrm {Exh}(\psi )$ . Indeed

$$ \begin{align*} \begin{split} \|X\|_\psi &=\inf_{F \in \mathrm{Fin}}\sup_{n \in \omega}(\sum_{k \in S_n} a_k \varphi_k^{q_n}(X_{(k)}\setminus F))^{1/q_n}\\ &\ge \inf_{F \in \mathrm{Fin}}\sup_{n \in \omega} \min_{k \in S_n}\{\varphi_k(X_{(k)}\setminus F)\} \ge \frac{\inf_{k \in \omega}\| \omega\|_{\varphi_k}}{2}>0. \end{split} \end{align*} $$

It follows that $X \notin \mathrm {Exh}(\sup _n \mu _n)$ , i.e., there exist $\varepsilon>0$ and an increasing sequence $(n_k)$ in $\omega $ such that $\mu _{n_k}(X)>\varepsilon $ for all $k \in \omega $ . However, thanks to (9), for each k there exists a unique $m_k \in \omega $ such that $X\cap M_{n_k}\subseteq X_{m_k}$ . Therefore $\mu _{n_k}(X_{m_k})>\varepsilon $ for all $k \in \omega $ . Let M be an infinite subset of $\{m_k: k \in \omega \}$ such that $|S_n \cap M|\le 1$ for all n. Define $Y:=\bigcup _{m \in M}X_m$ and note that $Y\notin \mathrm {Exh}(\sup _n \mu _n)$ .

Then necessarily $Y\notin \mathrm {Exh}(\psi )$ . However, considering that there is at most one $k \in S_n$ such that $Y_{(k)}\neq \emptyset $ , we obtain

$$ \begin{align*} \|Y\|_\psi &\le \inf_{F \in [\omega^2]^{<\omega}}\sup_{n \in \omega} \max_{k \in S_n}\{a_k^{1/q_n}\varphi_k((Y\setminus F)_{(k)})\} \\ & \le \sup_{n \in \omega}\varphi_n(\omega) \limsup_{t\to \infty}\max_{k \in S_t}\{a_k^{1/q_t}\} = 0, \end{align*} $$

which is the wanted contradiction.⊣

Proof. Thanks to (iv), we have $\sup _{n,k}\varphi _n(\{k\})\le \sup _n \varphi _n(\omega )<\infty $ . The conclusion follows by Proposition 4.16 and Theorems 4.13 and 4.18.⊣

Proof. It follows by Corollaries 4.14, 4.17, and 4.19.⊣

Proof. Let $\mathcal {I}_d$ be the ideal of density zero sets, which is a tall ideal. Thanks to [Reference Farah6, Example 1.2.3(d), Theorem 1.13.3(a), and Lemma 1.13.9(Z3)], there exists a sequence $(\mu _n)$ of probability measures with finite pairwise disjoint supports such that $\mathcal {I}_d=\mathrm {Exh}(\varphi )$ , where $\varphi :=\sup _n \mu _n$ . In particular, $\varphi (\omega )=\|\omega \|_\varphi =1$ . The claim follows by Corollary 4.21.⊣

Question 4. How many pairwise nonisomorphic tall density-like ideals which are not generalized density ideals are there?

Proof. It is known that there exists a family $\mathscr {B}$ of $2^\omega $ subsets of $\omega $ such that

$$ \begin{align*}\forall (B,B^\prime) \in [\mathscr{B}]^2, \quad B \notin \mathrm{Fin} \,\,\,\text{ and }\,\,\, B \cap B^\prime \in \mathrm{Fin}, \end{align*} $$

see e.g., [Reference Albiac and Kalton1, Lemma 2.5.3]. Then, it is sufficient to see that $\{\,\bigcup _{n\in \omega } \{n\}\times (B\setminus n): B \in \mathscr {B}\}$ satisfies (11).⊣

Proof. Let $h: \omega ^2 \to \omega $ be the bijection defined in (1). Let also $\mathscr {M}=\{M_z: z \in \omega ^2\}$ be a partition of $\omega ^2$ into nonempty finite sets such that $M_{(n,m)}\subseteq \{n\}\times \omega $ for all $n,m\in \omega $ and

(12) $$ \begin{align} \forall n \in \omega, \quad m_{h^{-1}(n+1)}\ge (n+2)\sum_{i\le n}m_{h^{-1}(i)}, \end{align} $$

where $m_z:=|M_z|$ . Moreover, for each $(n,m) \in \omega ^2$ , let $\mu _{(n,m)}$ be the uniform probability measure given by $\mu _{(n,m)}(X)= |(\{n\}\times X)\cap M_{(n,m)}|/m_{(n,m)}$ for all $X\subseteq \omega $ .

Fix $(S_n) \in \mathcal {F}_{\mathrm {incr}}$ such that $\lim _n |S_n|=\infty $ and let $\mathscr {A}$ be a family of $2^\omega $ subsets of $\omega ^2$ which satisfies (11) (existing by Lemma 4.23). For each $A \in \mathscr {A}$ , let $\psi _A$ be the lscsm on $\omega ^2$ defined by

$$ \begin{align*} \psi_A:=\sup_{n\in \omega} \sum_{k \in S_n}\frac{1}{|S_n|}\,\varphi_{A,k}, \,\,\,\text{ where }\,\,\,\varphi_{A,k}:=\sup_{t \in A_{(k)}} \mu_{(k,t)}, \end{align*} $$

for all $k \in \omega $ . It follows that, for each $A \in \mathscr {A}$ , the lscsm $\psi _A$ is of the type (7), where $q_n=1$ for all n and $a_k=1/|S_i|$ whenever $k \in S_i$ ; hence $\lim _n a_n=0$ . In addition, for each $A \in \mathscr {A}$ and $k \in \omega $ , the ideal $\mathrm {Exh}(\varphi _{A,k})$ is tall and $\|\omega \|_{\varphi _{A,k}}=\varphi _{A,k}(\omega )=1$ . Lastly, thanks to Remark 4.5, each $\varphi _{A,k}$ is strongly-density-like with any witnessing constant $c(\varphi _{A,k})<1$ . In particular, . Therefore, by Theorem 4.20, each $\mathrm {Exh}(\psi _A)$ is a tall density-like ideal which is not a generalized density ideal. Also, by Remark 4.10, each $\mathrm {Exh}(\psi _A)$ is nonpathological.

At this point, we claim that, for all distinct $A,A^\prime \in \mathscr {A}$ , the ideals $\mathrm {Exh}(\psi _A)$ and $\mathrm {Exh}(\psi _{A^\prime })$ are not isomorphic. To this aim, fix distinct $A,A^\prime \in \mathscr {A}$ and suppose for the sake of contradiction that $\mathrm {Exh}(\psi _A)$ and $\mathrm {Exh}(\psi _{A^\prime })$ are isomorphic, witnessed by a bijection $f:\omega ^2 \to \omega ^2$ . Let $(x_n)$ be the enumeration of the infinite set $A\setminus A^\prime $ such that the sequence $(h(x_n))$ is increasing. Then, pick a sequence $(F_n) \in \mathcal {F}_{\mathrm {disj}}$ such that

$$ \begin{align*}\forall n \in \omega,\quad F_n\subseteq M_{x_n}, \,\,\, |F_n|=\left\lfloor \frac{m_{x_n}}{2}\right\rfloor, \,\,\,\text{ and }\,\,\, F_{n+1}\cap \bigcup_{\substack{z \in \omega^2: \\ h(z)<h(x_{n+1})}}f[M_z]=\emptyset. \end{align*} $$

Note that this is really possible: indeed, letting U be the latter union, it follows by (12) that

$$ \begin{align*} |U|= \sum_{h(z)<h(x_{n+1})}m_z= \sum_{i\le h(x_{n+1})-1}m_{h^{-1}(i)} \le \frac{1}{2}\,m_{x_{n+1}}. \end{align*} $$

Set $F:=\bigcup _n F_n$ . It follows by construction that for all $k \in \omega $ , hence $F \notin \mathrm {Exh}(\psi _A)$ . On the other hand, we obtain by (12) that

$$ \begin{align*} \begin{split} \forall (i,j) \in A^\prime \setminus A,\quad \mu_{(i,j)}((f^{-1}[F])_{(i)}) &=\frac{|F\cap f[M_{(i,j)}]|}{m_{(i,j)}} \\ &\le \frac{\sum_{n\in\{k:\ h(x_k)<h((i,j))\}}|F_n|}{m_{(i,j)}} \\ &\le \frac{\sum_{k\le h((i,j))-1}m_{h^{-1}(k)}}{m_{h^{-1}(h((i,j)))}} \le \frac{1}{h((i,j))+1}, \end{split} \end{align*} $$

which implies that $f^{-1}[F] \in \mathrm {Exh}(\psi _{A^\prime })$ . This contradiction concludes the proof.⊣

Proof. Let us suppose for the sake of contradiction that $\varphi $ does not satisfy condition $\mathrm {D}_{\mathrm {strong}}$ . Thanks to Proposition 2.1, we can suppose without loss of generality that there exist a sequence $(\mu _n)$ of submeasures and a sequence $(S_n) \in \mathcal {F}_{\mathrm {int}}$ of consecutive intervals of $\omega $ such that $\varphi =\sup _n\mu _n$ and $S_n=\mathrm {supp}(\mu _n)$ for all $n \in \omega $ . In particular, $\mathrm {Exh}(\varphi )=\{A\subseteq \omega : \lim _n \mu _n(A)=0\}$ and $\mu _n(\omega )\not \to 0$ .

Define $s \in \mathscr {H}$ by $s_n:=\max S_n$ for all $n \in \omega $ . Since $\varphi $ does not satisfy condition $\mathrm {D}_{\mathrm {strong}}$ , there exists $\varepsilon>0$ such that for all nonzero $m\in \omega $ there are $F^m=(F^m_n)\in \mathcal {F}_{\mathrm {incr}} \cap \mathcal {G}_{\varphi ,\frac {\varepsilon }{2^m}}$ and $k^m=(k^m_n)\in K_{s,F^m}$ for which $\varphi (\bigcup _{n} F^m_{k_n^m})\geq \varepsilon $ . Since $\varphi $ is a lscsm, for each m there exists $\ell _m \in \omega $ such that .

Set $G_m:=\bigcup _{n\leq \ell _m} F^m_{k^m_n}$ and note that $G:=\bigcup _m G_m$ does not belong to $\mathrm {Exh}(\varphi )$ . To this aim, fix a nonzero $j\in \omega $ . Since $(k^j_n)\in K_{s,F^j}$ , there are at most j many sets $F^j_{k_n^j}$ which have nonempty intersection with the set $s_j+1$ and each of them has $\varphi $ -value smaller than $\varepsilon /2^{j}$ . Thus

$$ \begin{align*} \begin{split} \varphi(G\setminus (s_j+1)) \ge \varphi(G_j\setminus (s_j+1))\ge \varphi(G_j)-\varphi(G_j \cap (s_j+1))\ge \varepsilon-j \frac{\varepsilon}{2^j}\ge \frac{\varepsilon}{2}. \end{split} \end{align*} $$

Therefore . In particular, there exists a sequence $(j_i) \in \mathscr {H}$ such that for all $i\in \omega $ . Passing eventually to a subsequence, we can assume without loss of generality that

(13) $$ \begin{align} \forall i \in \omega, \quad s_{j_i}>\max G_{i+1}. \end{align} $$

At this point, define $X:=G\cap \bigcup _i S_{j_i}$ . Then by construction , so that $X\notin \mathrm {Exh}(\varphi )$ . On the other hand, we will show that $\lim _n \mu _n(X)=0$ , reaching a contradiction. Taking into account (13), note that $S_{j_i}\cap X= S_{j_i} \cap \bigcup _{m>i}G_m$ for all i. Moreover, recall that, for all $i,m \in \omega $ , there exists at most one n such that $F^m_{k^m_n}\cap S_{j_i} \neq \emptyset $ . It follows that

$$ \begin{align*}\forall i \in \omega,\quad \varphi(S_{j_i}\cap X)=\mu_{j_i}\left(\bigcup_{m>i}G_m\right)\le \sum_{m>i}\mu_{j_i}(G_m) \le \sum_{m>i}\frac{\varepsilon}{2^m}=\frac{\varepsilon}{2^i}. \end{align*} $$

To conclude, we obtain that

$$ \begin{align*}\forall t \in \omega, \quad \varphi\left(X\setminus \bigcup_{i\le t}S_{j_i}\right)=\varphi\left(X\cap \bigcup_{i> t}S_{j_i}\right)\le \sum_{i>t}\varphi(S_{j_i} \cap X)\le \sum_{i>t}\frac{\varepsilon}{2^i}= \frac{\varepsilon}{2^t}, \end{align*} $$

which tends to $0$ as $t\to \infty $ . Hence $X \in \mathrm {Exh}(\varphi )$ , which is the wanted contradiction.⊣

Proof. Let $(\varepsilon _k)$ be a strictly decreasing sequence such that $\lim _k \varepsilon _k=0$ . Then, for each k, there are $\delta _k>0$ and a sequence $s^k=(s_n^k) \in \mathscr {H}$ such that, if $F \in \mathcal {F}_{\mathrm {incr}} \cap \mathcal {G}_{\varphi ,\delta _k}$ and $k \in K_{s^k, F}$ , then $\varphi (\bigcup _{n} F_{k_n})<\varepsilon _k$ . Without loss of generality, we can assume that $\delta _{k+1}<\delta _k<\varepsilon _k$ . Let us define $s=(s_n) \in \mathscr {H}$ as follows: $s_0:=s_0^0$ and, for each $n \in \omega $ , let $s_{n+1}$ be such that for all $m\leq n+1$ there is $\ell \in \omega $ such that $s_n\leq s^m_\ell <s_{n+1}$ . Then, set $S_0:=[0,s_0]$ and $S_{n+1}:=(s_{n},s_{n+1}]$ for all $n\in \omega $ .

Also, let $\psi $ be the lscsm defined by $\psi (\emptyset )=0$ and, for each nonempty $A\subseteq \omega $ , $\psi (A):=\delta _k$ , where k is the minimal integer such that $A\cap S_k\neq \emptyset $ . At this point, set $\nu :=\max \{\varphi ,\psi \}$ . Then $\nu $ is a lscsm such that $\mathrm {Exh}(\nu )=\mathrm {Exh}(\varphi )$ . Indeed, on the one hand, $\nu \geq \varphi $ , hence $\mathrm {Exh}(\nu )\subseteq \mathrm {Exh}(\varphi )$ . On the other hand, fix $A\in \mathrm {Exh}(\varphi )$ and $\varepsilon>0$ , hence there is $n_0\in \omega $ such that $\mu (A\setminus n)<\varepsilon $ for all $n\geq n_0$ . Also, there is $n_1\in \omega $ such that $\delta _n<\varepsilon $ for all $n\geq n_1$ . Thus, for each n such that $n\geq n_0$ and $\bigcup _{i<n_1}S_i\subseteq n$ we have $\nu (A\setminus n)<\varepsilon $ . Therefore $A\in \mathrm {Exh}(\nu )$ , which proves the opposite inclusion $\mathrm {Exh}(\varphi )\subseteq \mathrm {Exh}(\nu )$ .

Lastly, we show that $\nu $ satisfies the condition in the statement. Fix $\varepsilon>0$ and let m be the minimal integer such that $\varepsilon _m\leq \varepsilon $ . We claim that $\delta :=\delta _m$ witnesses this condition. Fix $(F_n) \in \mathcal {F}_{\mathrm {incr}} \cap \mathcal {G}_{\nu ,\delta }$ , $k \in K_{s,F}$ , and $j\in \omega $ . Then there exist $\ell ^\prime ,\ell ^{\prime \prime } \in \omega $ such that

$$ \begin{align*}\max F_{k_{2j}}\leq s_{\ell^\prime}<\min F_{k_{2j+1}}\leq\max F_{k_{2j+1}}\leq s_{\ell^{\prime\prime}}<\min F_{k_{2(j+1)}}. \end{align*} $$

Note that $m<\ell ^\prime <\ell ^{\prime \prime }$ , where the former inequality follows by the fact that $F_n \cap \bigcup _{i\le m}S_i =\emptyset $ for all n (by the definition of $\nu $ and the hypothesis $(F_n) \in \mathcal {F}_{\mathrm {incr}} \cap \mathcal {G}_{\nu ,\delta }$ ) and the latter since $s\in \mathscr {H}$ . Thus, there exists $\ell \in \omega $ such that

$$ \begin{align*}s_{\ell^\prime}\leq s^m_\ell<s_{\ell^\prime+1}\leq s_{\ell^{\prime\prime}}. \end{align*} $$

It follows that $\max F_{k_{2j}}\leq s^m_{\ell }<\min F_{k_{2(j+1)}}$ , so that $(k_{2n})\in K_{s^m, F}$ . Therefore $\varphi (\bigcup _{n} F_{k_{2n}})<\varepsilon _m\leq \varepsilon $ . It is also easy to see that $\psi (\bigcup _{n} F_{k_{2n}})\leq \delta =\delta _m<\varepsilon _m\leq \varepsilon $ . Putting all together, we conclude that $\nu (\bigcup _{n} F_{k_{2n}})<\varepsilon $ .⊣

Proof. Define $S_0:=[0,s_0]$ and $S_{n+1}:=(s_n,s_{n+1}]$ for all $n\in \omega $ . Let $\boldsymbol {\mu }=(\mu _n)$ be the sequence of lscsms defined by

$$ \begin{align*}\forall n \in \omega, \forall A\subseteq \omega,\quad \mu_n(A):=\nu(A\cap S_n). \end{align*} $$

We claim that $\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})=\mathrm {Exh}(\nu )$ , where $\varphi _{\boldsymbol {\mu }}:=\sup _n \mu _n$ .

It is clear that $\varphi _{\boldsymbol {\mu }} \le \nu $ , hence $\mathrm {Exh}(\nu )\subseteq \mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ . Conversely, fix $A\in \mathrm {Exh}(\varphi _{\boldsymbol {\mu }})$ and $\varepsilon>0$ , hence there is $\delta>0$ such that, if $F=(F_n)\in \mathcal {F}_{\mathrm {incr}} \cap \mathcal {G}_{\nu ,\delta }$ and $k\in K_{s,F}$ , then . There exists $n_0\in \omega $ such that $\mu _n(A)<\delta $ for all $n\geq n_0$ . Define $F_n:=A\cap S_{n+n_0}$ for all $n\in \omega $ . Then $\nu (F_n)=\mu _{n+n_0}(A)<\delta $ for all $n \in \omega $ . Thus $(F_n)\in \mathcal {F}_{\mathrm {incr}} \cap \mathcal {G}_{\nu ,\delta }$ and for each $k \in K_{s,F}$ we have . Note the sequences $(n)$ and $(n+1)$ belong to $K_{s,F}$ , so that and . Define $m:=\min S_{n_0}$ . Then for each $m\geq m_0$ we have

$$ \begin{align*}\nu(A\setminus m) \le \nu(A\setminus m_0)=\nu\left(\bigcup_{n \in \omega} F_n\right) \le \nu\left(\bigcup_{n\in \omega} F_{2n}\right)+\nu\left(\bigcup_{n\in \omega} F_{2n+1}\right) <\varepsilon. \end{align*} $$

We conclude that $A\in \mathrm {Exh}(\nu )$ , therefore $\mathrm {Exh}(\varphi _{\boldsymbol {\mu }})\subseteq \mathrm {Exh}(\nu )$ .⊣

Proof of Theorem 5.3. (A1) $\implies $ (A2) follows by Lemma 5.4. The implications (A2) $\implies $ (A3) $\implies $ (A5) and (A2) $\implies $ (A4) $\implies $ (A5) are obvious. Lastly, (A5) $\implies $ (A1) follows by Lemmas 5.5 and 5.6.⊣

Question 5. Does there exist a density-like ideal $\mathcal {I}$ such that $\mathcal {I}\neq \mathrm {Exh}(\varphi )$ for each strongly-density-like lscsm $\varphi $ ?