Fact 2.1. $Th(M^{Sh})$ has quantifier elimination and NIP.

Proof Let $M^{Sh}\prec \bar {N}$ and $(\bar {N},M^{Sh})\prec (\bar {N}',\bar {M}')$ be sufficiently saturated elementary extensions in the languages $L(M^{Sh})$ and $L(M^{Sh})\cup \{P\}$ , respectively. Let $N,M'$ , and $N'$ be the reducts of $\bar {N},\bar {M}'$ , and $\bar {N}'$ to L. Let $X\subseteq M^{\bar {x}}$ be $\emptyset $ -definable in $(M^{Sh})^{Sh}$ . Then there exists $Y\subseteq M^{\prime \bar {x}}$ definable in $\bar {M}'$ such that $X=Y\cap M^{\bar {x}}$ . We then have some $Z\subseteq M^{\prime \bar {x}\bar {y}}$ and $\bar {b}\in M^{\prime \bar {y}}$ such that Z is $\emptyset $ -definable in $\bar {M}'$ and $Y=\{\bar {a}\in M^{\prime \bar {x}}:(\bar {a},\bar {b})\in Z\}$ . By Fact 2.1, Z is defined by an $L(M^{Sh})$ -formula which defines an externally definable set of M. It follows that there exists $W\subseteq N^{\prime \bar {x}\bar {y}}$ definable in $N'$ such that $Z=W\cap M^{\prime \bar {x}\bar {y}}$ . We then have $X=\{\bar {a}\in M^{\bar {x}}:(\bar {a},\bar {b})\in W\}$ and so X is $\emptyset $ -definable in $M^{Sh}$ .

Fact 2.5. Let $A\subseteq M'$ be a small indiscernible sequence. Let $\theta (\bar {x})$ be a formula with parameters in $M'$ . Then there is a finite set $C=\{c_1, \ldots , c_k\}$ of unrealised cuts in A such that, for any tuples $\bar {b},\bar {b}^\prime $ from A, if $otp(\bar {b}/C)=otp(\bar {b}^\prime /C)$ then $M'\models \theta (\bar {b})\leftrightarrow \theta (\bar {b}^\prime )$ .

Proof By Fact 2.5, $\mathcal {C}$ is not empty. Let $C=\{c_1, \ldots , c_k\}\in \mathcal {C}$ such that k is minimal. Let $C^\prime \in \mathcal {C}$ . Suppose $C\nsubseteq C'$ . Let $c\in C\setminus C'$ and $C_1=C\setminus \{c\}$ . Since k is minimal, there are tuples $\bar {b},\bar {b}'$ from A such that $otp(\bar {b}/C_1)=otp(\bar {b}'/C_1)$ and $M'\models \theta (\bar {b})\wedge \neg \theta (\bar {b}')$ . One can deform $\bar {b}$ into $\bar {b}'$ without changing the truth value of $\theta (\bar {b})$ (by ensuring that at each stage one preserves the order type over C or the order type over $C'$ ). This is a contradiction.⊣

Note that none of the cuts in the minimum C will be at $\infty $ or $-\infty $ . If A and B are disjoint ordered sets then $A+B$ denotes $A\cup B$ equipped with the ordering which places everything in A below everything in B and agrees with the existing orderings of A and B. The following, which we take as our definition of distality, is provided by a combination of Definition 2.1 and Lemma 2.7 in [Reference Simon7].

Proof Suppose T is not distal. Then we have a small indiscernible $I+\{b\}+J$ in $M'$ , with I and J indexed by $\mathbb {Z}$ , and a small $D\subseteq M'$ such that $I+J$ is indiscernible over D but $I+\{b\}+J$ is not. It follows that there exist a formula $\phi (x,\bar {y})$ and some $a\in M'$ , with $\bar {y}$ an n-tuple of variables in the sort of $I+\{b\}+J$ , such that $M'\models \phi (a,\bar {c})$ for any strictly increasing n-tuple $\bar {c}$ from $I+J$ and $M'\models \neg \phi (a,\bar {b})$ for some strictly increasing n-tuple $\bar {b}$ from $I+\{b\}+J$ .

Let $I'$ be the set of all elements in I below $\bar {b}$ and let $J'$ be the set of all elements in J above $\bar {b}$ . Then $I'$ and $J'$ are both infinite and each is indexed by $\mathbb {N}$ , with the standard or reverse order, or by $\mathbb {Z}$ . By grouping elements together we may treat $I'$ and $J'$ as sequences of n-tuples. Then $I'+\{\bar {b}\}+J'$ is indiscernible and $I'+J'$ is indiscernible over a. We have $M'\models \phi (a,\bar {c})$ , for all $\bar {c}\in I'+J'$ , while $M'\models \neg \phi (a,\bar {b})$ .

Proof When a language other than L is intended, we shall make that clear. Let $\theta (\bar {x})$ be a formula with parameters in N. By Fact 2.5 there is a set $C_A^\theta =\{c_1, \ldots , c_k\}$ of unrealised cuts of A such that, for any two tuples $\bar {b}$ and $\bar {b}^\prime $ from A, if $otp(\bar {b}/C_A^\theta )=otp(\bar {b}^\prime /C_A^\theta )$ then $N'\models \theta (\bar {b})\leftrightarrow \theta (\bar {b}^\prime )$ . By Lemma 2.6, we may assume $C_A^\theta $ is the minimum among all possible choices (ordered by set inclusion).

For any small indiscernible sequence $B\subseteq M'$ extending A and any cut $c\in C_A^\theta $ , there must be some $c'\in C_B^\theta $ such that $c^\prime $ refines c (by which we mean that every realisation of $c^\prime $ is also a realisation of c). Otherwise $C_B^\theta $ would give rise to a finite set $C^\prime $ of unrealised cuts in A such that $C_A^\theta \nsubseteq C^\prime $ and, for any tuples $\bar {b}$ and $\bar {b}'$ from A, $otp(\bar {b}/C')=otp(\bar {b}^\prime /C')\implies N'\models \theta (\bar {b})\leftrightarrow \theta (\bar {b}^\prime )$ . This would be a contradiction.

As each $C_B^\theta $ is finite, it follows that we cannot have an infinite sequence of small indiscernible extensions $A\subseteq B_1\subseteq B_2\subseteq \cdots \subseteq M'$ such that $| C_A^\theta |<|C_{B_1}^\theta |<|C_{B_2}^\theta |<\ldots $ (since otherwise the union $\bigcup \limits _{n\in \mathbb {N}}B_n$ would be a small indiscernible sequence for which the conclusion of Fact 2.5 is false). So then, for each $\theta $ , we can find a small extension B of A such that, for any small extension $B^\prime $ of B, $|C_B^\theta |=|C_{B^\prime }^\theta |$ and in fact there is a bijection from $C_B^\theta $ to $C_{B^\prime }^\theta $ which sends each cut $c\in C_B^\theta $ to the unique $c^\prime \in C_{B^\prime }^\theta $ which refines it.

The process of extending A to such a B could be called “maximising for $\theta $ ”. Enumerate all formulas with parameters in N, add one to the enumerating indices so that only successor ordinals are used and then maximise for each formula in turn, taking unions at limit ordinals. We thereby obtain a small indiscernible sequence $A^*\subseteq M'$ extending A such that, for any formula $\theta $ with parameters in N and any small indiscernible $B\subseteq M'$ extending $A^*$ , there is a bijection from $C_{A^*}^\theta $ to $C_B^\theta $ which sends each $c\in C_{A^*}^\theta $ to the unique $c^\prime \in C_B^\theta $ which refines it. Note that “formula $\theta $ ” really means “formula $\theta (\bar {x})$ where $\bar {x}$ is a tuple of variables in the sort of A”.

We would like to consider $\bigcup \limits _\theta C_{A^*}^\theta $ and, for any small extension B of $A^*$ , the bijection taking each $c\in \bigcup \limits _\theta C_{A^*}^\theta $ to the unique $c^\prime \in \bigcup \limits _\theta C_B^\theta $ which refines it. However, we cannot be sure at this stage that such a bijection exists. The problem is that, for some $\theta _1,\theta _2$ , we might have $|C_{A^*}^{\theta _1}\cap C_{A^*}^{\theta _2}|>|C_B^{\theta _1}\cap C_B^{\theta _2}|$ . In other words, some cuts might coincide in $A^*$ but not in B. For any small $B'$ extending a small B extending $A^*$ , we must have $|C_{A^*}^{\theta _1}\cap C_{A^*}^{\theta _2}|\geq |C_B^{\theta _1}\cap C_B^{\theta _2}|\geq |C_{B'}^{\theta _1}\cap C_{B'}^{\theta _2}|$ . (To see this note, in the notation of the first inequality, that every cut in $C_B^{\theta _1}\cap C_B^{\theta _2}$ refines one in $C_{A^*}^{\theta _1}\cap C_{A^*}^{\theta _2}$ and that it would contradict the existence of our bijections if two cuts in $C_B^{\theta _1}\cap C_B^{\theta _2}$ were to refine the same cut in $C_{A^*}^{\theta _1}\cap C_{A^*}^{\theta _2}$ .) An ordinal (in this case a finite one) cannot be decreased infinitely many times. So, for each pair $\theta _1,\theta _2$ , we can extend so that $|C_B^{\theta _1}\cap C_B^{\theta _2}|$ is minimised. We can enumerate all such pairs of formulas and extend appropriately for each one in turn, taking unions at limit ordinals. This results in a small extension $A^{**}\subseteq M'$ of $A^*$ with the following property. For any small indiscernible $B\subseteq M'$ extending $A^{**}$ and any formula $\theta $ with parameters in N, let $f_B^\theta :C_{A^{**}}^\theta \rightarrow C_B^\theta $ be the bijection which maps each $c\in C_{A^{**}}^\theta $ to the unique $c^\prime \in C_B^\theta $ which refines it and define $C_B=\bigcup \limits _\theta C_B^\theta $ . Then, for each such B, the union, over all such $\theta $ , of the graphs of the functions $f_B^\theta $ is the graph of an order-preserving bijection $f_B:C_{A^{**}}\rightarrow C_B$ .

To simplify notation, let $C=C_{A^{**}}$ . Enumerate the elements of C as $(c^\alpha )_{\alpha <\kappa }$ . For each small indiscernible $B\subseteq M'$ extending $A^{**}$ and each $\beta <\kappa $ , let $c_B^\beta =f_B(c^\beta )$ . We build a chain $(B_\alpha )_{\alpha <\kappa }$ of small indiscernible sequences in $M'$ extending $A^{**}$ in the following way. Let $\beta <\kappa $ and suppose we have formed $B_\alpha $ for all $\alpha <\beta $ . Let $B_\beta '=A^{**}\cup \bigcup \limits _{\alpha <\beta } B_\alpha $ . Consider the cut $c^\beta _{B_\beta '}$ . There are four cases.

1. (1) Suppose $\{a\in B^{\prime }_\beta : a<c^\beta _{B^{\prime }_\beta }\}$ has a maximum and $\{a\in B^{\prime }_\beta : a>c^\beta _{B^{\prime }_\beta }\}$ has a minimum. Then let $B_\beta =B^{\prime }_\beta $ .

2. (2) Suppose $\{a\in B^{\prime }_\beta : a<c^\beta _{B^{\prime }_\beta }\}$ has a maximum and $\{a\in B^{\prime }_\beta : a>c^\beta _{B^{\prime }_\beta }\}$ does not have a minimum. By compactness, there exist a small indiscernible sequence B extending $B^{\prime }_\beta $ and some $v_\beta \in B$ such that $v_\beta>c^\beta _B$ and $v_\beta $ realises $c^\beta _{B^{\prime }_\beta }$ . (One uses the limit type of the indiscernible sequence $B^{\prime }_\beta $ over $NB^{\prime }_\beta $ as $c^\beta _{B^{\prime }_\beta }$ is approached from the right.) Then let $B_\beta =B^{\prime }_\beta \cup \{v_\beta \}$ .

3. (3) Suppose $\{a\in B^{\prime }_\beta : a>c^\beta _{B^{\prime }_\beta }\}$ has a minimum and $\{a\in B^{\prime }_\beta : a<c^\beta _{B^{\prime }_\beta }\}$ does not have a maximum. By compactness, there exist a small indiscernible sequence B extending $B^{\prime }_\beta $ and some $u_\beta \in B$ such that $u_\beta <c^\beta _B$ and $u_\beta $ realises $c^\beta _{B^{\prime }_\beta }$ . Then let $B_\beta =B^{\prime }_\beta \cup \{u_\beta \}$ .

4. (4) Suppose $\{a\in B^{\prime }_\beta : a<c^\beta _{B^{\prime }_\beta }\}$ does not have a maximum and $\{a\in B^{\prime }_\beta : a>c^\beta _{B^{\prime }_\beta }\}$ does not have a minimum. By compactness, there exist a small indiscernible sequence B extending $B^{\prime }_\beta $ and some $u_\beta ,v_\beta \in B$ such that $u_\beta <c^\beta _B$ , $v_\beta>c^\beta _B$ and both $u_\beta $ and $v_\beta $ realise $c^\beta _{B^{\prime }_\beta }$ . Then let $B_\beta =B^{\prime }_\beta \cup \{u_\beta ,v_\beta \}$ .

   

Let $A'=\bigcup \limits _{\alpha <\kappa }B_\alpha $ . In the remainder of this proof, if we write an indiscernible sequence as $(B,<)$ we are thinking of it as a structure and the language we are using is $\{<\}$ . The sequence $A'$ has been constructed so that, for each $\beta <\kappa $ , the cut $c^\beta _{A'}$ is definable in $(A',<)$ . Therefore, whenever we have small indiscernible sequences B and $B'$ in $M'$ , with $A'\subseteq B$ and $A'\subseteq B'$ , and $\bar {a}=(a_1, \ldots , a_n)\in B^n$ and $\bar {a}'=(a^{\prime }_1, \ldots , a^{\prime }_n)\in B^{{\prime }n}$ , if $(A',<)\prec (B,<)$ and $(A',<)\prec (B',<)$ and $qftp_{\{<\}}(\bar {a}/A')=qftp_{\{<\}}(\bar {a}'/A')$ then, for each $\beta <\kappa $ and $i\in \{1, \ldots , n\}$ , $a_i<c_B^{\beta }$ if and only if $a^{\prime }_i<c_{B'}^{\beta }$ .

We show that $A'$ has the desired property. Let $q(\bar x)$ be a complete L-type over $A'$ which is finitely realised in $A'$ . Let $\bar a=(a_1, \ldots , a_n)$ and $\bar a'=(a_1', \ldots , a_n')$ both realise $q(\bar x)$ in $M'$ . Let $(N",M",A")$ be a sufficiently saturated elementary extension of $(N',M',A')$ , in the language $L\cup \{P,Q,<\}$ , where P is a unary predicate for $M'$ , Q is a unary predicate for $A'$ , and $<$ is a binary predicate for the order relation on $A'$ . Then $q(\bar {x})$ is realised in $A"$ , say by $\bar {a}"$ . By the downward Löwenheim–Skolem theorem, there is a small $(N"',M"',A"')\prec (N",M",A")$ such that $A'\cup \{\bar {a}"\}\subseteq A"'$ and $(A',<)\prec (A"',<)$ . Then $A"'$ is indiscernible in the sense of L. We then have automorphisms $\sigma $ and $\sigma '$ of the $L\cup \{P\}$ -structure $(N",M")$ which fix $A'$ pointwise and are such that $\sigma (\bar {a}")=\bar {a}$ and $\sigma '(\bar {a}")=\bar {a}'$ . Let $B=\sigma (A"')$ and $B'=\sigma '(A"')$ . Then B and $B'$ are indiscernible (in the sense of L), $(A',<)\prec (B,<)$ and $(A',<)\prec (B',<)$ , where the orderings of B and $B'$ are determined by $\sigma $ and $\sigma '$ . We may assume $B,B'\subseteq M'$ . Since $qftp_{\{<\}}(\bar {a}/A')=qftp_{\{<\}}(\bar {a}'/A')$ we have $a_i<c_B^{\beta }$ if and only if $a^{\prime }_i<c_{B'}^{\beta }$ , for each $\beta <\kappa $ and $i\in \{1, \ldots , n\}$ . It follows that $tp(\bar {a}/N)=tp(\bar {a}'/N)$ .

Proof As in the proof of Lemma 1.3, the default language is L. Suppose, for contradiction, that T is not distal. By Lemma 2.8 there exist a model $K\models T$ , an indiscernible sequence $I+\{b\}+J$ in K, with I and J infinite and $\{b\}$ a singleton, some $a\in K$ and a formula $\phi (x,y)$ such that $I+J$ is indiscernible over a, $K\models \phi (a,d)$ for all $d\in I+J$ and $K\models \neg \phi (a,b)$ .

Let $M^{Sh}\prec \bar {N}$ and $(\bar {N},M^{Sh})\prec (\bar {N}',\bar {M}')$ be sufficiently saturated elementary extensions in the languages $L(M^{Sh})$ and $L(M^{Sh})\cup \{P\}$ , respectively. Let $N,M'$ , and $N'$ be the reducts of $\bar {N},\bar {M}'$ , and $\bar {N}'$ to L. We may assume $K\prec M^\prime $ .

By Lemma 1.3 and Remark 3.2, $I+J$ extends to a small indiscernible sequence $A'$ in $M'$ with the property that every complete type over $A'$ which is finitely realised in $A'$ implies a complete $L(M^{Sh})$ -type over $A'$ . We may assume a and b are such that $A'\cup \{b\}$ is indiscernible, with b positioned just above I, and $A'$ is indiscernible over a. (To see this, let $r(x,y)$ be the partial type over $A'$ expressing the desired properties of the pair $(a,b)$ . Any finite $r'(x,y)\subseteq r(x,y)$ involves only a finite tuple $cd$ from $A'$ , where c is bounded above by an element of I and d lies entirely above I. By indiscernibility, an automorphism of $M'$ takes c to a tuple in I and d to a tuple in J, establishing that $r'(x,y)$ can be realised. One then uses saturation.)

Now consider the structure $(\bar {N}',\bar {M}',A')$ in the language $L(M^{Sh})\cup \{P,Q\}$ . Take sufficiently saturated elementary extensions

$$\begin{align*}(\bar{N}',\bar{M}',A')\prec (\bar{N}",\bar{M}",A")\prec (\bar{N}"',\bar{M}"',A"')\prec (\bar{N}"",\bar{M}"",A"").\end{align*}$$

In all cases, let the removal of the bar correspond to taking the L-reduct. Let $p(x)=tp(a/A")$ . Let $q(y)$ be a complete type over $A"$ , where y is a single variable in the sort of $A'$ , such that q is finitely realised in $A'$ . We show that $p(x)\cup q(y)$ implies a complete type in $xy$ over $\emptyset $ .

Let $q'(y)$ be some extension of $q(y)$ to a complete type over $N""$ which is finitely realised in $A'$ . Let $(d_i)_{i\in \mathbb {Z}}$ be a Morley sequence for $q'$ over $N'$ in $A"$ . (Of course, a Morley sequence is usually indexed by $\mathbb {N}$ but, having obtained such a sequence, one can choose another one indexed by $\mathbb {Z}$ with the same EM-types.) Let $d^*$ realise q in $M"'$ . Let $(d^{\prime }_i)_{i\in \mathbb {Z}}$ be a Morley sequence in $q'$ over $N"'$ in $A""$ . We find the following picture helpful.

The sequence $(d_i)_{i\in \mathbb {Z}}+d^*+(d^{\prime }_i)_{i\in \mathbb {Z}}$ is $L(M^{Sh})$ -indiscernible. This is because every finite subsequence has a type over $A'$ which is finitely realised in $A'$ and therefore implies a complete $L(M^{Sh})$ -type over $A'$ . Also $(d_i)_{i\in \mathbb {Z}}+(d^{\prime }_i)_{i\in \mathbb {Z}}$ is L-indiscernible over $Na$ and so $L(M^{Sh})$ -indiscernible over a. By the distality of $Th(M^{Sh})$ , $(d_i)_{i\in \mathbb {Z}}+d^*+(d^{\prime }_i)_{i\in \mathbb {Z}}$ is $L(M^{Sh})$ -indiscernible over a and thus L-indiscernible over a. Since $d^*$ was an arbitrary realisation of q in $M"'$ it follows that $p(x)\cup q(y)$ implies a complete type in $xy$ over $\emptyset $ .

Since the set of all $d \in M"'$ such that $tp(d/A")$ is finitely realised in $A'$ is type-definable over $A"$ in the structure $M"'$ , a compactness argument gives us some $\bar {c}\in A^{{\prime \prime }k}$ and an L-formula $\theta (x, \bar {z})$ , with $\bar {z}$ a k-tuple of variables in the sort of y, such that $M"\models \theta (a,\bar {c})$ and $\theta (x,\bar {c})$ implies the $\phi $ -type of a over $A'$ . (To see this, suppose not. Then, for any choice of $\theta (x,\bar {c})\in tp(a/A")$ , there exist $b'\in A'\subseteq \{d\in M"':tp(d/A")\text { is finitely realised in }A'\}$ and $a'\in M"$ such that $M"\models \theta (a',\bar {c})$ and $M"\models \phi (a,b')\leftrightarrow \neg \phi (a',b')$ . Then, by compactness, there exist $a',b'\in M"'$ such that $a'\models p(x)$ , $tp(b'/A")$ is finitely realised in $A'$ and $tp(ab')\neq tp(a'b')$ which is a contradiction.) So for any finite $A\subseteq A'$ there is a $\bar {c}\in A^{{\prime }k}$ such that $M'\models \theta (a,\bar {c})$ and $\theta (x, \bar {c})$ implies the $\phi $ -type of a over A.

Recall that $A'\cup \{b\}$ is indiscernible, with b positioned just above I, and $A'$ is indiscernible over a. So $M'\models \phi (a,d)$ for all $d\in A'$ and $M'\models \neg \phi (a,b)$ . Let $A\subseteq A'$ have cardinality $k+1$ . Let $\bar {c}\in A^{{\prime }k}$ be such that $M'\models \theta (a, \bar {c})$ and $\theta (x, \bar {c})$ implies the $\phi $ -type of a over A. Let d be an element of A which does not belong to the tuple $\bar {c}=(c_1, \ldots , c_k)$ . Let f be a partial automorphism of $(A'\cup \{b\},<)$ , with domain $A \cup \{c_1, \ldots , c_k\}$ , such that $f(d)=b$ . Let $f(\bar {c})$ denote the tuple $(f(c_1), \ldots , f(c_k))$ . Then $M'\models \theta (a,f(\bar {c}))$ . It follows that $M'\models \phi (a,b)$ , since otherwise there would be some $a'\in M'$ such that $M'\models \theta (a',\bar {c})\wedge \neg \phi (a',d)$ which would be a contradiction. But $M'\models \neg \phi (a,b)$ and so we have a contradiction and the proof is finished.

Proof Let $M^{Sh}\prec \bar {N}$ and $(\bar {N},M^{Sh})\prec (\bar {N}',\bar {M}')$ be sufficiently saturated elementary extensions, the first in the language $L(M^{Sh})$ and the second in the language $L(M^{Sh})\cup \{P\}$ , where P is a new unary predicate. Let $N,M'$ , and $N'$ be the L-reducts of $\bar {N},\bar {M}'$ , and $\bar {N}'$ , respectively. Let $I+\{b\}+J$ be a small $L(M^{Sh})$ -indiscernible sequence in $\bar {M}'$ , such that I and J are both infinite without endpoints. Let $A\subseteq \bar {M}'$ be small and suppose $I+J$ is $L(M^{Sh})$ -indiscernible over A. It follows that $I+J$ is L-indiscernible over $NA$ and that $I+\{b\}+J$ is L-indiscernible over N. By distality of T, $I+\{b\}+J$ is L-indiscernible over $NA$ . Therefore $I+\{b\}+J$ is $L(M^{Sh})$ -indiscernible over A.

Fact 6.1. Let T be a complete first-order L-theory. Suppose T has NIP. Then T is distal if and only if, for every $K\models T$ , sufficiently saturated $K\prec N$ and measure $\mu $ over N, if $\mu $ is generically stable over K then its restriction to K is smooth. Moreover, if T is not distal then it has a model K such that, for every elementary extension $K\prec \hat {K}$ , there exist a sufficiently saturated $\hat {K}\prec \hat {N}$ and a measure $\mu $ over $\hat {N}$ which is generically stable over $\hat {K}$ but whose restriction to $\hat {K}$ is not smooth.

Proof Let X be a definable set of $\bar {K}'$ . Then X is a fibre of a $\emptyset $ -definable set, say $X_1$ . By Fact 2.1, $X_1$ is defined by a quantifier-free formula $R(x,z)\in L(M^{Sh})$ , without parameters. Consider the set, say $X_2$ , defined by $R(x,z)$ in $\bar {K}$ . By compactness, as is well known, $X_2$ will itself be externally definable with respect to the structure K and so there will exist an L-formula $\varphi (x,y,z)$ and parameter b from N such that $X_2$ is the set of all $(a,c)\in K$ for which $N\models \varphi (a,b,c)$ . Using compactness one can choose b so that, for every L-formula $\theta (w,y)$ , the set $\{d\in K:N\models \theta (d,b)\}$ is $\emptyset $ -definable in the structure $\bar {K}$ . We assume b has been chosen with this property. Then $X_1$ will be the set externally defined by $\varphi (x,b,z)$ in $K'$ . So, finally, we have an L-formula $\varphi (x,y,z)$ and parameters b from N and c from $K'$ such that $X=\{a\in K':N'\models \varphi (x,b,c)\}$ . Define $\mu '$ over $\bar {K}'$ such that $\mu '(X)$ is the value assigned by $\mu $ to the set defined by $\varphi (x,b,c)$ in $N'$ .

One checks that $\mu '$ is a measure over $\bar {K}'$ . Note that it is well-defined because $\varphi (x,b,c)$ is always unique up to a $\mu $ -measure zero symmetric difference, using the fact that $\mu $ is finitely realisable in K. Furthermore, $\mu '$ is definable over $\bar K$ and finitely satisfiable in $\bar K$ . For definability, we use definability of $\mu $ for the formula $\varphi (x,y,z)$ and then restrict the type-definable set to $y=b$ and project to the z-coordinate. From the way b was chosen it is clear that the resulting type-definable set is defined by a partial type using only formulas in the language of $\bar {K'}$ , with parameters in $\bar {K}$ . Finite realisability of $\mu '$ in $\bar {K}$ is immediate from the finite realisability of $\mu $ in K. So then the measure $\mu '$ over $\bar {K}'$ is generically stable over $\bar K$ . We define $\mu ^*$ to be its restriction to $\bar K$ .

For uniqueness of $\mu ^*$ (the “furthermore” statement) one uses the fact that every measure over $\bar K$ comes from a measure over $N'$ which agrees with it on K and is finitely realisable in K. Since $\mu $ is generically stable over K, it is known (by Proposition 3.3 in [Reference Hrushovski, Pillay and Simon4]) that its restriction to K has only one finitely realisable (in K) extension to $N'$ . Therefore $\mu ^*$ is unique.

Now suppose the restriction of $\mu $ to K is not smooth. Since $K\prec N$ is sufficiently saturated, this is witnessed over N and so there are two distinct extensions $\mu _1$ and $\mu _2$ to N. These extend, respectively, to measures $\mu _1^*$ and $\mu _2^*$ over $\bar {N}$ . Trivially, $\mu _1^*$ and $\mu _2^*$ are distinct. They both restrict to measures on $\bar K$ which extend the restriction of $\mu $ to K. By uniqueness of $\mu ^*$ , they both extend $\mu ^*$ . So $\mu ^*$ is not smooth.