Proof. If $p = 0$ , then either the sentence $(\forall y)(\xi (y) \rightarrow \phi ^-(y))$ or $(\forall y)(\xi (y) \rightarrow \neg \phi ^-(y))$ is satisfied in $L$ (using the assumption that ${\xi }({L})$ is an inclusion-minimal definable set); hence, it holds in each $G_n$ from a certain index on. Therefore, an arbitrary choice of $r_n \in{\xi }({G_n})$ and $r \in{\xi }({L})$ meets the conclusion.

Assume that $p \geq 1$ . Let $\nu$ be the measure associated with the modelling $L$ . Define $f_L\;:\; V(L)^p \to 2^{{\xi }({L})}$ to be the function that sends $\boldsymbol{v}$ to the set $\{u \in{\xi }({L}) \;:\; L \models \phi ^-(\boldsymbol{v}, u)\}$ . Consider the pushforward measure $\mu _L$ on $2^{{\xi }({L})}$ of the $p$ -th power of $\nu$ by $f_L$ (note that for each $u \in \xi (L)$ is the set $f_L^{-1}({u})$ measurable). Viewing $2^{{\xi }({L})}$ as a lattice, we are mostly interested in the measure of the filter generated by atoms of $2^{{\xi }({L})}$ . Let $X^\uparrow = \{Y \in 2^{{\xi }({L})} \;:\; X \subseteq Y\}$ denote the filter generated by $X \in 2^{{\xi }({L})}$ . Observe that for $u \in{\xi }({L})$ , we have $\mu _L(\{u\}^\uparrow ) = \langle \phi, (L, u) \rangle$ . Suppose that $|{\xi }({L})| = t$ and define an ordering $R_L = (u_1, u_2, \dots, u_t)$ of $\xi (L)$ such that $\mu _L(R_L) = (\mu _L(\{u_i\}^\uparrow ))_{i \in [t]}$ satisfies $\mu _L(\{u_1\}^\uparrow ) \geq \mu _L(\{u_2\}^\uparrow ) \geq \dots \geq \mu _L(\{u_t\}^\uparrow )$ . Define similarly for each $n$ the function $f_n \;:\; V(G_n)^p \to 2^{{\xi }({G_n})}$ , measure $\mu _n$ (as the pushforward of the uniform measure), and the $t$ -tuple $R_n$ .

We prove that the sequence of vectors $\big (\mu _n(R_n)\big ) \subset ([0,1]^t, \|\cdot \|_\infty )$ converges to $\mu _L(R_L)$ . Then an arbitrary choice of an index $i \in [t]$ yields the sequence $(r_n)$ and vertex $r$ as the $i$ -th elements of the tuples $R_n$ and $R_L$ , respectively.

Let us argue that the convergence $\big (\mu _n(R_n)\big ) \to \mu _L(R_L)$ follows from Lemma 6. We first apply the lemma for the set $M ={\xi }({L})$ , that is, $m = t$ , and the probability distribution $\mu = \mu _L$ to show that the values of $\mu _L(R_L)$ can be computed from probabilities $\langle \psi _{k,\ell }, L \rangle, \ell \in [t]_0, k \in \left [ \binom{t}{\ell } \right ]$ for certain formulas $\psi _{k,\ell } \in \mathrm{FO}$ .

This can be done as follows. By definition, $\Pr [F_\ell ^k]$ is the probability that the intersection of $k$ independently chosen $\mu _L$ -random subsets $S_i$ of ${\xi }({L})$ has size at least $\ell$ . To compute the probability, we sample each set $S_i = f_L(\boldsymbol{v}_i) = \{u \in{\xi }({L}) \;:\; L \models \phi ^-(\boldsymbol{v}_i, u)\}$ by taking an independent $\nu$ -random $p$ -tuple $\boldsymbol{v}_i \in V(L)^p$ , and we verify that the intersection $\bigcap _{i=1}^k S_i \subseteq \xi (L)$ contains at least $\ell$ distinct elements $y_1, \dots, y_\ell$ . Therefore, $\Pr [F_\ell ^k]$ is equal to $\langle \psi _{k,\ell }, L \rangle$ for the $\mathrm{FO}$ -formulas $\psi _{k,\ell }(\boldsymbol{x}_1, \dots, \boldsymbol{x}_k) \in \mathrm{FO}_{k \cdot p}$ of the form

\begin{equation*} (\exists y_1, \dots, y_\ell ) \left ( \bigwedge _{i = 1}^\ell \xi (y_i) \land \bigwedge _{1 \leq i \lt j \leq \ell } y_i \not = y_j \land \bigwedge _{i = 1}^k \bigwedge _{j = 1}^\ell \phi ^-(\boldsymbol{x}_i, y_j) \right ), \end{equation*}

where, by definition of $\langle \cdot, \cdot \rangle$ , the independent $\nu$ -random $p$ -tuples $\boldsymbol{v}_i \in V(L)^p$ are assigned to the disjoint $p$ -tuples of variables $\boldsymbol{x}_i$ . Moreover, we have directly from the definition that the values of the multiset $A_1$ from the conclusion of Lemma 6 are exactly the values of $\mu _L(R_L)$ .

Similarly, by applying Lemma 6 for $M ={\xi }({G_n})$ and $\mu = \mu _n$ , we obtain that $\mu _n(R_n)$ depends only on $\langle \psi _{k,\ell }, G_n \rangle, \ell \in [t]_0, k \in \left [ \binom{t}{\ell } \right ]$ . Furthermore, since the dependency is continuous, and $\langle \psi _{k,\ell }, G_n \rangle \to \langle \psi _{k,\ell }, L \rangle$ for each $k, \ell \in \mathbb{N}$ by $\mathrm{FO}$ -convergence of $(G_n)$ to $L$ , we infer that the vectors $\big (\mu _n(R_n)\big )$ indeed converge to $\mu _L(R_L)$ .

Proof. Since for sentences any choice of $(r_n)$ and $r$ works, we assume that neither of $\phi _1, \dots, \phi _k$ is a sentence.

Consider an inclusion-maximal set $I \subseteq [k]$ for which there is $v \in{\xi }({L})$ such that every $i \in I$ satisfies $\langle \phi _i, (L,v) \rangle \gt 0$ . Denote $|I|$ by $k'$ . If $I = \emptyset$ , we can choose $(r_n)$ and $r$ arbitrarily; hence, assume otherwise. For $i \in I$ , set $A_i = \{\langle \phi _i, (L,u) \rangle \;:\; u \in{\xi }({L}) \} \cap (0, 1]$ . Take a vector $\boldsymbol{e} \in \mathbb{N}^{k'}$ of exponents with the property that for each distinct $\boldsymbol{a}, \boldsymbol{b} \in \textrm{X}_{i \in I} A_i$ , we have $\prod _{i \in I} a_i^{e_i} \not = \prod _{i \in I} b_i^{e_i}$ . Such a vector exists as each $A_i$ is finite and contains only positive values. The set of bad choices of rational exponents that make the values for particular $\boldsymbol{a}, \boldsymbol{b}$ coincide form a $(k'-1)$ -dimensional hyperplane in $\mathbb{Q}^{k'}$ . We can surely avoid finitely many of such hyperplanes (one for each choice of $\boldsymbol{a}$ and $\boldsymbol{b}$ ) to find a good vector of positive rational exponents and scale them to integers.

Use Lemma 3 for the formula $\psi$ of the form

\begin{equation*} \bigwedge _{i \in I} \bigwedge _{j=1}^{e_i} \phi _i(\boldsymbol{x}_{i,j}), \end{equation*}

where all the tuples $\boldsymbol{x}_{i,j}$ are pairwise disjoint, to obtain roots $(r_n)$ and $r$ . In particular, we can take the vertex $r$ such that $\langle \psi, (L,r) \rangle \gt 0$ (due to our choice of $I$ ).

We have $\lim \langle \phi _i, (G_n, r_n) \rangle = \langle \phi _i, (L, r) \rangle \gt 0$ for each $i \in I$ as

\begin{equation*} \langle \psi, (L,r) \rangle = \prod _{i \in I} \langle \phi _i, (L,r) \rangle ^{e_i}, \end{equation*}

using our selection of exponents $\boldsymbol{e}$ .

Also, it holds that $\lim \langle \phi _j, (G_n, r_n) \rangle = \langle \phi _j, (L,r) \rangle = 0$ for each $j \not \in I$ : for the formula $\chi = \bigwedge _{i \in I \cup \{j\}} \phi _i(\boldsymbol{x}_i)$ , we have $\lim \langle \chi, (G_n,r_n) \rangle = \langle \chi, (L,r) \rangle = 0$ due to the maximality of $I$ (this is for any choice of $(r_n)$ and $r$ ). We have

\begin{equation*} \langle \chi, (G_n,r_n) \rangle = \prod _{i \in I \cup \{j\}} \langle \phi _i, (G_n,r_n) \rangle \end{equation*}

and as for some $\varepsilon \gt 0$ , there is $n_0$ such that $\langle \phi _i, (G_n,r_n) \rangle \gt \varepsilon$ for each $i \in I$ and $n \geq n_0$ , the factor $\langle \phi _j, (G_n,r_n) \rangle$ must tend to $0$ .

Proof. Let $S_j$ be the set of possible limit values of $\langle \phi _j, (G_n, r_n) \rangle$ , that is, the values of $\mu _L(R_L)$ from Lemma 3. Let $T = (V,E)$ be an infinite rooted tree defined as follows: $V = \bigcup V_\ell$ , where $V_\ell$ is the set of vertices on the level $\ell$ defined as the Cartesian product of the sets $S_j$ for $j \leq \ell$ , that is, a vertex on the $\ell$ -th level is a vector with possible limit probabilities for $\phi _1, \dots, \phi _\ell$ . The only element in $V_0$ , the empty set, is the root of the tree. We put an edge between $\boldsymbol{a} \in V_\ell$ and $\boldsymbol{b} \in V_{\ell +1}$ if $a_j = b_j$ for all $j \leq \ell$ , and there are vertices $x_n \in{\xi }({G_n}), x \in{\xi }({L})$ such that

\begin{equation*} \lim \langle \phi _j, (G_n, x_n) \rangle = \langle \phi _j, (L, x) \rangle = b_j \end{equation*}

for each $j \leq \ell +1$ . Observe that if $\boldsymbol{a}\boldsymbol{b} \in E$ , then $\boldsymbol{b}$ is connected to the root (all edges of the path are witnessed by the vertices $(x_n)$ and $x$ ).

By Lemma 4, there is at least one vertex in each set $V_\ell$ connected to the root. Thus, by König’s lemma, the tree contains an infinite path $\emptyset = \boldsymbol{a}_0, \boldsymbol{a}_1, \boldsymbol{a}_2, \dots$ (all degrees are bounded as each $S_j$ is finite). The sequence $(r_n^i)$ and vertex $r^i$ are defined as the vertices $(x_n)$ and $x$ witnessing the edge from $\boldsymbol{a}_{i-1}$ to $\boldsymbol{a}_i$ .

Proof of Theorem 2. Let $(r_n^i)$ and $r^i$ for $i \in \mathbb{N}$ be the vertices from collection of extending sequences from Lemma 5 above. Let $\alpha _i = \langle \phi _i, (L,r^i) \rangle$ , and let $N_i$ to be an index satisfying

1. (i) $\langle \phi _j, (G_n, r_n^i) \rangle \in (\alpha _j - 2^{-i}, \alpha _j + 2^{-i})$ for each $n \geq N_i$ and $j \leq i$ ,

2. (ii) $N_i \gt N_j$ for each $j \lt i$ .

Set $r_n = r_n^i$ , where $i$ is the minimal positive integer satisfying that $n \lt N_{i+1}$ . For the vertex $r$ , we can set an arbitrary vertex from ${\xi }({L})$ that appears infinitely many times as $r^i$ .

It remains to verify that for an arbitrary formula $\phi _j$ , we have $\lim \langle \phi _j, (G_n, r_n) \rangle = \alpha _j = \langle \phi _j, (L, r) \rangle$ . Obviously, the second equality holds as $r = r^i$ for some $i \geq j$ . For the first equality, fix $\varepsilon \gt 0$ and set $k$ to be a positive integer satisfying $2^{-k} \lt \varepsilon$ and $k \geq \max \{j, 2\}$ . Then for $n \geq N_k$ , we have $\langle \phi _j, (G_n, r_n) \rangle = \langle \phi _j, (G_n, r_n^i) \rangle \in (\alpha _j - 2^{-i}, \alpha _j + 2^{-i})$ for some $i \geq k$ . Thus $\langle \phi _j, (G_n, r_n) \rangle \in (\alpha _j - \varepsilon, \alpha _j + \varepsilon )$ , which concludes the proof.