$j_{\alpha_{i}+1,\alpha_{i+1}}$ participating in it are iterated ultrapowers
as defined over $M_{\alpha_{i}+1}$. In the proof of the lemma we will show
that the external iteration $j_{\alpha_{m}+1,\alpha}\circ\ldots\circ
j_{\alpha_{0}+1,\alpha_{1}}\circ j_{1,\alpha_{0}}$ is well defined over
$N_{m+1}$, in the sense that for every $x\in N_{m+1}$,
$\left(j_{\alpha_{i-1}+1,\alpha_{i}}\circ\ldots\circ
j_{\alpha_{0}+1,\alpha_{1}}\circ j_{0,\alpha_{0}}\right)(x)$ belongs to
$M_{\alpha_{i}+1}$. Later in this section, we will prove that such an
iteration might be an internal iteration of $N_{m+1}$, provided that the
initial nice sequence is chosen more carefully.

###### Proof.

We proceed by induction on $m$. The induction basis is given for $"m=-1"$,
namely, the case where the given nice sequence below $\alpha$ is empty. In
this case, $N_{0}=\mbox{Ult}\left(V,U\right)$, $i_{0}=j_{U}\colon V\to N_{0}$
and $k_{0}\left(i_{0}(f)(\kappa)\right)=j_{\alpha}(f)(\kappa)$, and clearly
$k_{0}=j_{0,\alpha}\colon M_{0}\to M_{\alpha}$.

Assume now that $m<\omega$ and $k_{m+1}\colon N_{m+1}\to M_{\alpha_{m+1}}$ has
been constructed (here, the embedding $k_{m+1}$ corresponds to the nice
sequence $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$ below $\alpha_{m+1}$).
Let us argue that $k_{m+2}=j_{\alpha_{m+1}+1,\alpha}\circ
k_{m+1}\restriction_{N_{m+2}}$. Indeed, given an arbitrary element
$i_{m+2}(f)\left(\kappa,\lambda_{0},\ldots,\lambda_{m},\lambda_{m+1}\right)$
of $N_{m+2}$,

$\displaystyle
k_{m+1}\left(i_{m+2}(f)\left(\kappa,\lambda_{0},\ldots,\lambda_{m},\lambda_{m+1}\right)\right)$
$\displaystyle=\ $ $\displaystyle
k_{m+1}\left(j^{N_{m+1}}_{W_{m+1}}\left(i_{m+1}(f)\right)\left(\kappa,\lambda_{0},\ldots,\lambda_{m+1}\right)\right)$
$\displaystyle=\ $ $\displaystyle
j_{U_{\mu_{\alpha_{m+1}}}}\left(j_{\alpha_{m+1}}(f)\right)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{m}},\mu_{\alpha_{m+1}}\right)$
$\displaystyle=\ $ $\displaystyle
j_{\alpha_{m+1}+1}(f)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{m}},\mu_{\alpha_{m+1}}\right)$

where we used the fact that
$W_{m+1}=i_{m+1}\left(F_{m+1}\right)\left(\kappa,\lambda_{0},\ldots,\lambda_{m}\right)$
and
$\lambda_{m+1}=i_{m+1}\left(g_{m+1}\right)\left(\kappa,\lambda_{0},\ldots,\lambda_{m}\right)$
for the computation on their values under $k_{m+1}$. Finally, apply
$j_{\alpha_{m+1}+1,\alpha}$ on both sides. ∎

If the sequence $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$ below $\alpha$ is
clear from the context, we denote $N^{*}=N_{m+1}$, $i^{*}=i_{m+1}\colon V\to
N^{*}$ and $k^{*}=k_{m+1}\colon N^{*}\to M_{\alpha}$. Note that $k^{*}$ is not
necessarily an internal iteration of $N^{*}$. Indeed, assume that
$\lambda_{0}<\mu_{\alpha_{0}}$ (this happens, e.g., if $\alpha_{0}=1$. In this
case, $\lambda_{0}=\mu_{0}$ and
$\mu_{1}=j_{U_{\mu_{0}}}\left(\mu_{0}\right)>\mu_{0}$). If $k^{*}$ was an
internal iteration of $N^{*}$, then $\lambda_{0}$ would have to be one of the
critical points participating in the iteration, since $\lambda_{0}$ is
inaccessible in $N^{*}$ and $k^{*}\left(\lambda_{0}\right)=\mu_{\alpha_{0}}$.
However, this is not possible because $\lambda_{0}$ is not measurable in
$N^{*}$.

Our goal is lemma 4.6. In the proof, it will be useful to consider a nice
sequence $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$ below $\alpha$ and its
associated iteration $N^{*}$, such that the embedding $k^{*}\colon N^{*}\to
M_{\alpha}$ is an internal iteration of $N^{*}$. This will require a more
sophisticated choice of the initial nice sequence. The example from the last
paragraph offers a lead: we would like $\lambda_{k}=\mu_{\alpha_{k}}$ to hold
for every $0\leq k\leq m$.

######  Lemma 4.3.

$k^{*}$ is an internal iteration of $N^{*}$ if and only if, for every $0\leq
k\leq m$, $\lambda_{k}=\mu_{\alpha_{k}}$.

###### Proof.

Let us assume first that $k^{*}$ is an iteration of $N^{*}$. Then
$\lambda_{k}$ is a non-measurable inaccessible in $N^{*}$, and thus
$\lambda_{k}$ cannot move by $k^{*}$. So
$\mu_{\alpha_{k}}=k^{*}\left(\lambda_{k}\right)=\lambda_{k}$.

Let us concentrate on the other direction. Assume that
$\lambda_{k}=\mu_{\alpha_{k}}$ for every $0\leq k\leq m$.
$\lambda_{0}=\mu_{\alpha_{0}}$ is measurable in $M_{U}=N_{0}$, and thus
$j_{0,\alpha_{0}}\left(\mu_{\alpha_{0}}\right)=\mu_{\alpha_{0}}$. Also,
$j_{0,\alpha_{0}}\left(W_{0}\right)=U_{\mu_{\alpha_{0}}}$. Note that–

$j_{\alpha}=j_{\alpha_{0}+1,\alpha}\circ j_{U_{\mu_{\alpha_{0}}}}\circ
j_{\alpha_{0}}=j_{\alpha_{0}+1,\alpha}\circ j^{N_{1}}_{0,\alpha_{0}}\circ
j_{W_{0}}\circ j_{U}$

where $j^{N_{1}}_{0,\alpha_{0}}$ is the iterated ultrapower consisting of the
same measures as $j_{0,\alpha_{0}}$, but acting on $N_{1}$.
$j^{N_{1}}_{0,\alpha_{0}}\colon N_{1}\to M_{\alpha_{0}+1}$ is internal to
$N_{1}$, and so is $j_{\alpha_{0}+1,\alpha}\circ j^{N_{1}}_{0,\alpha_{0}}$.

We proceed now by induction on $m$. Assume that $k_{m+1}\colon N_{m+1}\to
M_{\alpha_{m+1}}$ is an internal iteration of $N_{m+1}$ (with respect to the
nice sequence $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$ below
$\alpha_{m+1}$). Then–

$U_{\mu_{\alpha_{m+1}}}=k_{m+1}\left(W_{m+1}\right)$

and thus–

$\displaystyle j_{\alpha_{m+1}+1}=\ $ $\displaystyle
j_{U_{\mu_{\alpha_{m+1}}}}\circ k_{m+1}\circ i_{m+1}$ $\displaystyle=\ $
$\displaystyle k_{m+1}\restriction_{N_{m+2}}\circ i_{m+2}$ $\displaystyle=\ $
$\displaystyle\left(j_{\alpha_{m}+1,\alpha_{m+1}}\circ\ldots\circ
j_{\alpha_{0}+1,\alpha_{1}}\circ
j_{0,\alpha_{0}}\right)\restriction_{N_{m+2}}\circ i_{m+2}$

Where $\left(j_{\alpha_{m}+1,\alpha_{m+1}}\circ\ldots\circ
j_{\alpha_{0}+1,\alpha_{1}}\circ
j_{0,\alpha_{0}}\right)\restriction_{N_{m+2}}$ above is an internal iteration
of $N_{m+2}$, since $W_{m+1}$ is a measure over
$\lambda_{m+1}=\mu_{\alpha_{m+1}}$, and lies strictly above all the
participating critical points. Thus, the embedding $k_{m+2}$, obtained by
applying $j_{\alpha_{m+1}+1,\alpha}$ on
$\left(j_{\alpha_{m}+1,\alpha_{m+1}}\circ\ldots\circ
j_{\alpha_{0}+1,\alpha_{1}}\circ
j_{0,\alpha_{0}}\right)\restriction_{N_{m+2}}$, is an internal iteration of
$N_{m+2}$. ∎

######  Lemma 4.4.

Every nice sequence $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$ below
$\alpha$ can be completed to a nice sequence–

$\langle\alpha^{0}_{0},\ldots,\alpha^{n_{0}}_{0},\alpha^{0}_{1},\ldots,\alpha^{n_{1}}_{1},\ldots\ldots,\alpha^{0}_{m},\ldots,\alpha^{n_{m}}_{m}\rangle$

where
$\alpha^{n_{0}}_{0}=\alpha_{0},\alpha^{n_{1}}_{1}=\alpha_{1},\ldots,\alpha^{n_{m}}_{m}=\alpha_{m}$,
such that the embedding $k^{*}$ associated to the latter sequence is an
iteration of $N^{*}$.

###### Proof.

We begin with an arbitrary nice sequence
$\langle\alpha_{0},\ldots,\alpha_{m}\rangle$, and complete it to a nice
sequence–

$\langle\alpha^{0}_{0},\ldots,\alpha^{n_{0}}_{0},\alpha^{0}_{1},\ldots,\alpha^{n_{1}}_{1},\ldots\ldots,\alpha^{0}_{m},\ldots,\alpha^{n_{m}}_{m}\rangle$

where
$\alpha^{n_{0}}_{0}=\alpha_{0},\alpha^{n_{1}}_{1}=\alpha_{1},\ldots,\alpha^{n_{m}}_{m}=\alpha_{m}$.

We first extend the sequence below $\alpha_{0}$, namely define
$\alpha^{0}_{0},\ldots,\alpha^{n_{0}}_{0}$.

Denote $N^{0}_{0}=\mbox{Ult}\left(V,U\right)$ and $i^{0}_{0}=j_{U}\colon V\to
N^{0}_{0}$. Let $\lambda^{0}_{0}=i^{0}_{0}\left(g_{0}\right)(\kappa)$. Let
$\alpha^{0}_{0}\leq\alpha_{0}$ be the first such that
$\lambda^{0}_{0}\leq\mu_{\alpha^{0}_{0}}$.
$\left(\mbox{cf}\left(\lambda^{0}_{0}\right)\right)^{V}>\kappa$, so actually
$\lambda^{0}_{0}=\mu_{\alpha^{0}_{0}}$. If $\alpha^{0}_{0}=\alpha_{0}$, we set
$n_{0}=0$ and we are done extending the sequence below $\alpha_{0}$. Assume
otherwise.

Work in $N^{0}_{0}$ and define there
$W^{0}_{0}=i^{0}_{0}\left(\mathcal{U}\right)\left(\lambda^{0}_{0}\right)$. Let
$N^{1}_{0}=\mbox{Ult}\left(N^{0}_{0},W^{0}_{0}\right)$ and
$i^{1}_{0}=j^{N^{0}_{0}}_{W^{0}_{0}}\circ i^{0}_{0}\colon V\to N^{1}_{0}$.
Define $k^{1}_{0}\colon N^{1}_{0}\to M_{\alpha^{0}_{0}+1}$ to be such that for
every $f\in V$,

$k^{1}_{0}\left(i^{1}_{0}\left(f\right)\left(\kappa,\lambda^{0}_{0}\right)\right)=j_{\alpha^{0}_{0}+1}\left(f\right)\left(\kappa,\mu_{\alpha^{0}_{0}}\right)$

by lemma 4.3, $k^{1}_{0}$ is an iterated ultrapower of $N^{1}_{0}$. The
measures participating in this iteration lie on measurables below
$\alpha^{0}_{0}$ (actually, $k^{1}_{0}=j^{N^{1}_{0}}_{1,\alpha^{0}_{0}}$). In
$N^{1}_{0}$, let $\lambda^{1}_{0}=j_{W^{0}_{0}}\left(\lambda^{0}_{0}\right)$,
and note that $\lambda^{1}_{0}$ is measurable in $N^{1}_{0}$ above
$\lambda^{1}_{0}$. Thus, $\lambda^{1}_{0}$ does not participate in the
iteration $k^{1}_{0}$, namely
$k^{1}_{0}\left(\lambda^{1}_{0}\right)=\lambda^{1}_{0}$. So $\lambda^{1}_{0}$
is a measurable cardinal in $M_{\alpha^{0}_{0}+1}$, and
$\left(\mbox{cf}\left(\lambda^{1}_{0}\right)\right)^{V}>\kappa$. Thus, there
exists an index $\alpha^{1}_{0}$, such that
$\mu_{\alpha^{1}_{0}}=\lambda^{1}_{0}$ and
$\alpha^{0}_{0}<\alpha^{1}_{0}\leq\alpha_{0}$. If $\alpha^{1}_{0}=\alpha_{0}$,
we finish extending the sequence below $\alpha_{0}$ and set $n_{0}=1$. Assume
otherwise. Define in $N^{1}_{0}$ the measure
$W^{1}_{0}=i^{1}_{0}(\mathcal{U})\left(\lambda^{1}_{0}\right)$. Let
$N^{2}_{0}=\mbox{Ult}\left(N^{1}_{0},W^{1}_{0}\right)$ and
$i^{2}_{0}=j^{N^{1}_{0}}_{W^{1}_{0}}\circ i^{1}_{0}\colon V\to N^{2}_{0}$.
Define $k^{2}_{0}\colon N^{2}_{0}\to M_{\alpha^{1}_{0}+1}$ in the natural way,
namely, for every $f\in V$,

$k^{2}_{0}\left(i^{2}_{0}(f)\left(\kappa,\lambda^{0}_{0},\lambda^{1}_{0}\right)\right)=j_{\alpha^{1}_{0}+1}(f)\left(\kappa,\lambda^{0}_{0},\lambda^{1}_{0}\right)$

and by 4.3, $k^{2}_{0}$ is an iterated ultrapower of $N^{2}$ with measurables
below $\mu_{\alpha^{1}_{0}}$. Denote
$\lambda^{2}_{0}=j^{N^{1}_{0}}_{W^{1}_{0}}\left(\lambda^{1}_{0}\right)>\lambda^{1}_{0}$.
Arguing as before, $\lambda^{2}_{0}$ is measurable in $M_{\alpha^{1}_{0}+1}$
with cofinality above $\kappa$, and thus, there exists $\alpha^{2}_{0}$ such
that $\lambda^{2}_{0}=\mu_{\alpha^{2}_{0}}$ and
$\alpha^{0}_{0}<\alpha^{1}_{0}<\alpha^{2}_{0}\leq\alpha_{0}$.

Continue in this fashion, and construct an increasing sequence
$\alpha^{0}_{0}<\alpha^{1}_{0}<\ldots\leq\alpha_{0}$. We argue that the
construction stops after finitely many steps. Assume otherwise, and let
$\langle\alpha^{n}_{0}\colon n<\omega\rangle$ be a strictly increasing
sequence of ordinals below $\alpha_{0}$, such that for every $n<\omega$,

$\mu_{\alpha_{0}}>\mu_{\alpha^{n+1}_{0}}=\lambda^{n+1}_{0}=j_{W^{n}_{0}}\left({\lambda^{n}_{0}}\right)>{\lambda^{n}_{0}}=\mu_{\alpha^{n}_{0}}$

and–

$\mu_{\alpha^{n+1}_{0}}=\lambda^{n+1}_{0}={k^{n+1}_{0}}\left(\lambda^{n+1}_{0}\right)={k^{n+1}_{0}}\left(i^{n+1}_{0}\left(g_{0}\right)(\kappa)\right)=j_{\alpha^{n}_{0}+1}\left(g_{0}\right)(\kappa)=j_{U_{\mu_{\alpha^{n}_{0}}}}\left(\mu_{\alpha^{n}_{0}}\right)$

let $\alpha^{*}_{0}=\sup\\{\alpha^{n}_{0}\colon n<\omega\\}\leq\alpha_{0}$.
Note that–

$j_{\alpha^{*}_{0}}(g_{0})(\kappa)=j_{\alpha^{0}_{0},\alpha^{*}_{0}}\left(\mu_{\alpha^{0}_{0}}\right)=\sup\\{\mu_{\alpha^{n}_{0}}\colon
n<\omega\\}<\mu_{\alpha^{*}_{0}}$

and thus–

$\mu_{\alpha_{0}}=j_{\alpha_{0}}\left(g_{0}\right)(\kappa)=j_{\alpha^{*}_{0},\alpha_{0}}\left(j_{\alpha^{*}_{0}}\left(g_{0}\right)(\kappa)\right)=j_{\alpha^{*}_{0}}\left(g_{0}\right)(\kappa)$

which contradicts the fact that
$j_{\alpha^{*}_{0}}\left(g_{0}\right)(\kappa)<\mu_{\alpha^{*}_{0}}\leq\mu_{\alpha_{0}}$.

Thus, there exists $n_{0}<\omega$ and a sequence
$\alpha^{0}_{0}<\alpha^{1}_{0}<\ldots<\alpha^{n_{0}}_{0}=\alpha_{0}$ such that
for every $n<n_{0}$,

$\mu_{\alpha^{n+1}_{0}}=j_{U_{\mu_{\alpha^{n}_{0}}}}\left(\mu_{\alpha^{n}_{0}}\right)=j_{\alpha^{n+1}_{0}}\left(g_{0}\right)(\kappa)$

where the last equality follows by induction, since–

$j_{\alpha^{n+1}_{0}}(g_{0})(\kappa)=j_{\alpha^{n}_{0},\alpha^{n+1}_{0}}\left(j_{\alpha^{n}_{0}}(g_{0})(\kappa)\right)=j_{\alpha^{n}_{0},\alpha^{n+1}_{0}}\left(\mu_{\alpha^{n}_{0}}\right)=j_{U_{\alpha^{n}_{0}}}\left(\mu_{\alpha^{n}_{0}}\right)$

let us justify the last equality in the above equation. If $\mu_{\alpha_{0}}$
is not a limit of measurables, then $\alpha^{n+1}_{0}=\alpha^{n}_{0}+1$ and
the equation is clear. Otherwise, $\mu_{\alpha^{n}_{0}}$ is a limit of
measurables. Therefore
$\mu_{\alpha^{n+1}_{0}}=j_{U_{\alpha^{n}_{0}}}\left(\mu_{\alpha^{n}_{0}}\right)$
is a limit of measurables, and each factor in
$j_{\alpha^{n}_{0}+1,\alpha^{n+1}_{0}}$ is an ultrapower embedding with one of
them. Thus, each such factor maps $\mu_{\alpha^{n+1}_{0}}$ to itself.

This finishes the completion of the initial nice sequence below $\alpha_{0}$.
Let $N^{*}_{0}$ be the iterated ultrapower associated to the nice sequence
$\langle\alpha^{0}_{0},\ldots\alpha^{n}_{0}\rangle$, with a corresponding
embedding $i^{*}_{0}\colon V\to N^{*}_{0}$. Let $k^{*}_{0}\colon N^{*}_{0}\to
M_{\alpha_{0}+1}$ be defined as follows: for every $f\in V$,

$k^{*}_{0}\left(i^{*}_{0}(f)\left(\kappa,\mu_{\alpha^{0}_{0}},\ldots,\mu_{\alpha^{n_{0}}_{0}}\right)\right)=j_{\alpha_{1}}(f)\left(\kappa,\mu_{\alpha^{0}_{0}},\ldots,\mu_{\alpha^{n_{0}}_{0}}\right)$

By lemma 4.3, the embedding $k^{*}_{0}$ is an iterated ultrapower of
$N^{*}_{0}$, and $j_{\alpha_{0}+1}=k^{*}_{0}\circ i^{*}_{0}$. All the
ultrapowers in $k^{*}_{0}$ are taken on measurables below $\alpha_{0}$.

Now work over $N^{*}_{0}$, define
$\lambda^{0}_{1}=i^{*}_{0}\left(g_{1}\right)\left(\kappa,\mu_{\alpha_{0}}\right)$.
$\lambda^{0}_{1}>\mu_{\alpha_{0}}$ is measurable in $N^{*}_{0}$ and thus is
not moved by $k^{*}_{0}$. Also, it has cofinality above $\kappa$ in $V$. Let
$\alpha^{0}_{1}\leq\alpha_{1}$ be such that
$\lambda^{0}_{1}=\mu_{\alpha^{0}_{1}}$. If $\lambda^{0}_{1}=\mu_{\alpha_{1}}$,
we set $n_{1}=0$ and move on to extend the sequence below $\alpha_{2}$. Assume
otherwise. Let
$W^{0}_{1}=i^{*}_{0}\left(\mathcal{U}\right)\left(\lambda^{0}_{1}\right)$. Let
$N^{0}_{1}=\mbox{Ult}\left(N^{*}_{0},W^{0}_{1}\right)$, and
$i^{0}_{1}=j^{N^{*}_{0}}_{W^{0}_{1}}\circ i^{*}_{0}$. Let $k^{0}_{1}\colon
N^{1}_{0}\to M_{\alpha^{0}_{1}+1}$ be the natural embedding, and continue the
construction as above. It will stop after finitely many steps.

By repeating the same argument for $\alpha_{2},\ldots,\alpha_{m}$, we generate
the desired completion of $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$. ∎

######  Remark 4.5.

For every $0\leq i\leq m$, $\mu_{\alpha_{i}}$ appears in the Prikry sequence
of $\mu^{*}_{\alpha_{i}}=k_{\alpha_{i}}\left(\mu_{\alpha_{i}}\right)$ in
$M\left[H\right]$. Note that in the above proof, the completion below
$\mu_{\alpha_{i}}$, namely the sequence
$\langle\alpha^{0}_{k}\ldots,\alpha^{n_{i}}_{i}\rangle$, is a subsequence of
the Prikry sequence of $\mu^{*}_{\alpha_{i}}$ below $\mu_{\alpha_{i}}$. In
lemma 4.8 we will prove that this subsequence is actually a segment in this
Prikry sequence.

######  Lemma 4.6.

Assume that $\alpha<\kappa^{*}$. Denote
$\bar{\mu}=\sup\\{\mu_{\alpha^{\prime}}\colon\alpha^{\prime}<\alpha\\}$. Let
$\lambda>\bar{\mu}$ be an inaccessible cardinal in $M_{\alpha}$. Then
$\left(\mbox{cf}(\lambda)\right)^{V}>\kappa$.

###### Proof.

Let us first consider the case where there is no $\beta<\alpha$ and
$\lambda^{\prime}<\lambda$ such that
$j_{\beta,\alpha}\left(\lambda^{\prime}\right)=\lambda$. Let
$\langle\alpha_{0},\ldots,\alpha_{m}\rangle$ be a nice sequence below $\alpha$
such that
$\lambda=j_{\alpha}(h)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{m}}\right)$
for some function $h\in V$. We can assume that the sequence in complete as in
lemma 4.4, and so $k^{*}\colon N^{*}\to M_{\alpha}$ is an internal iterated
ultrapower. Denote–

$\lambda^{*}=i^{*}(h)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{m}}\right)$

and note that $k^{*}\left(\lambda^{*}\right)=\lambda$. It suffices to prove
that $\lambda^{*}=\lambda$, since every inaccessible above $\kappa$ in a
finite iteration of $V$ has cofinality $>\kappa$ in $V$. Assume that
$\lambda^{*}<\lambda$. Because $\lambda^{*}$ is inaccessible in $N^{*}$,
$\lambda^{*}$ is one of the measurables participating in the iteration $k$,
namely $\lambda^{*}=\mu_{\beta}$ for some $\beta<\alpha$. Since
$\lambda^{*}>\mu_{\alpha_{m}}$, $\beta>\alpha_{m}$. Then–

$\displaystyle\lambda=$ $\displaystyle\ k^{*}\left(\lambda^{*}\right)$
$\displaystyle=$ $\displaystyle\left(j_{\beta,\alpha}\circ
j_{\alpha_{m}+1,\beta}\circ j_{\alpha_{m-1}+1,\alpha_{m}}\circ\ldots\circ
j_{\alpha_{0}+1,\alpha_{1}}\circ
j_{1,\alpha_{0}}\right)\left(\lambda^{*}\right)$ $\displaystyle=$
$\displaystyle\ j_{\beta,\alpha}\left(\lambda^{*}\right)$

where we used the fact that $\lambda^{*}=\mu_{\beta}$ is inaccessible in
$N^{*}$ above $\mu_{\alpha_{m}}$, and thus is fixed by ultrapowers below
$\mu_{\alpha_{m}}$ and by $j_{\alpha_{m}+1,\beta}$. It follows that there
exists $\beta<\alpha$ and $\lambda^{*}<\lambda$ such that
$j_{\beta,\alpha}\left(\lambda^{*}\right)=\lambda$, which is a contradiction.

Let us now take care of the case where, for some $\beta<\alpha$ and
$\lambda_{0}<\lambda$, $j_{\beta,\alpha}\left(\lambda_{0}\right)=\lambda$. Let
$\beta<\alpha$ be the least such that such $\lambda_{0}$ exists. Since
$\lambda_{0}$ is inaccessible in $M_{\beta}$ and
$\lambda_{0}<j_{\beta,\alpha}\left(\lambda_{0}\right)$, $\lambda_{0}$ is one
of the measurables participating in the iteration $j_{\beta,\alpha}$. Thus,
$\lambda_{0}=\mu_{\gamma_{0}}$, for some $\gamma_{0}<\alpha$.

Denote $\lambda_{1}=j_{U_{\mu_{\gamma_{0}}}}\left(\mu_{\gamma_{0}}\right)$.
This is an inaccessible cardinal in $M_{\gamma_{0}+1}$. Let us argue that
$\left(\mbox{cf}\left(\lambda_{1}\right)\right)^{V}>\kappa$.

Pick a complete nice sequence $\langle\alpha_{0},\ldots,\alpha_{m}\rangle$
below $\gamma_{0}+1$ such that, for some function $h\in V$,

$\lambda_{1}=j_{\gamma_{0}+1}(h)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{m}}\right)$

we can assume that $\alpha_{m}=\gamma_{0}$ (else, add it. The sequence will
remain complete since there is no $\lambda^{\prime}<\lambda_{0}$ and
$\gamma^{\prime}\leq\gamma_{0}$ such that
$j_{\gamma^{\prime},\gamma_{0}+1}\left(\lambda^{\prime}\right)=\lambda_{0}$).
Let $N^{*}$ be the associated finite iteration, with an embedding $i^{*}\colon
V\to N^{*}$. let $k^{*}\colon N^{*}\to M_{\gamma_{0}+1}$ be the corresponding
iterated ultrapower such that $k^{*}\circ i^{*}=j_{\gamma_{0}+1}$. Denote
$\lambda^{*}_{1}=i^{*}(h)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{m}}\right)$.
Then $k^{*}\left(\lambda^{*}_{1}\right)=\lambda_{1}$. Let us argue that
$\lambda^{*}_{1}=\lambda_{1}$. Assume that $\lambda^{*}_{1}<\lambda_{1}$. Then
$\lambda^{*}_{1}$, which is measurable in $N^{*}$, is one of the measurables
participating in $k^{*}$. Note that–

$\lambda_{1}=k^{*}\left(\lambda^{*}_{1}\right)=j_{\alpha_{m}+1,\gamma_{0}+1}\circ
j_{\alpha_{m-1}+1,\alpha_{m}}\circ
j_{\alpha_{m-2}+1,\alpha_{m-1}}\circ\ldots\circ
j_{1,\alpha_{0}}\left(\lambda^{*}_{1}\right)$

but $\alpha_{m}=\gamma_{0}$, so $j_{\alpha_{m}+1,\gamma_{0}+1}$ is the
identity. So $\lambda^{*}_{1}<\mu_{\alpha_{m}}=\mu_{\gamma_{0}}$.
$\mu_{\gamma_{0}}$ is already a non-measurable inaccessible in $N^{*}$ (since
we started from a complete nice sequence which includes it), and thus
$k^{*}\left(\mu_{\gamma_{0}}\right)=\mu_{\gamma_{0}}$. Namely
$j_{U_{\gamma_{0}}}\left(\mu_{\gamma_{0}}\right)=\lambda_{1}=k^{*}\left(\lambda^{*}_{1}\right)<\mu_{\gamma_{0}}$,
a contradiction.

Thus $\left(\mbox{cf}(\lambda_{1})\right)^{V}>\kappa$. If
$\lambda_{1}=\lambda$, we are done. Else, $\lambda_{1}<\lambda$ is
inaccessible in $M_{\gamma_{0}+1}$, and is mapped via
$j_{\gamma_{0}+1,\alpha}$ to $\lambda$. Hence, arguing as before,
$\lambda_{1}\leq\bar{\mu}$ is one of the measurables participating in the
iterated ultrapower $j_{\gamma_{0}+1,\alpha}$. Therefore, there exists
$\gamma_{1}\in\left(\gamma_{0},\alpha\right)$ such that
$\lambda_{1}=\mu_{\alpha_{\gamma_{1}}}$. Denote
$\lambda_{2}=j_{U_{\mu_{\gamma_{1}}}}\left(U_{\mu_{\gamma_{1}}}\right)>\lambda_{1}$.
As above, $\left(\mbox{cf}\left(\lambda_{2}\right)\right)^{V}>\kappa$. If
$\lambda_{2}=\lambda$, we are done. Assume otherwise, and continue in this
fashion.

Let us argue that the process stops after finitely many steps. Assume
otherwise. Then we have constructed an $\omega$-sequence of ordinals below
$\alpha$, $\langle\gamma_{n}\colon n<\omega\rangle$, and an increasing
sequence–

$\lambda_{0}=\mu_{\gamma_{0}}<\lambda_{1}=\mu_{\gamma_{1}}<\lambda_{2}=\mu_{\gamma_{2}}<\ldots<\lambda$

such that, for every $n<\omega$,
$\lambda_{n+1}=\mu_{\gamma_{n+1}}=j_{U_{\mu_{\gamma_{n}}}}\left(\mu_{\gamma_{n}}\right)$.
Denote $\gamma^{*}=\sup\\{\gamma_{n}\colon n<\omega\\}$ (possibly
$\gamma^{*}=\alpha$). Let $\lambda^{*}=\sup\\{\lambda_{n}\colon n<\omega\\}$.
Then–

$j_{\gamma^{*},\alpha}\left(\lambda^{*}\right)=\lambda$

however, $j_{\gamma^{*},\alpha}\left(\lambda^{*}\right)=\lambda^{*}$: if
$\gamma^{*}=\alpha$ this is clear. Else, note that $\mu_{\gamma^{*}}$ is
chosen strictly above $\sup\\{\mu_{\xi}\colon\xi<\gamma^{*}\\}=\lambda^{*}$.
Therefore, the critical point of $j_{\gamma^{*},\alpha}$ is above
$\lambda^{*}$, and
$j_{\gamma^{*},\alpha}\left(\lambda^{*}\right)=\lambda^{*}$.

It follows that $\lambda^{*}=\lambda$. But $\lambda^{*}\leq\bar{\mu}$
(equality may hold if $\gamma^{*}=\alpha$), contradicting the fact that
$\lambda>\bar{\mu}$. ∎

We now return to our context, and assume that $\langle
M_{\alpha}\colon\alpha\leq\kappa^{*}\rangle$ is the iteration described in the
previous section, with the same notations. We can first simplify the
definition of the critical points $\mu_{\alpha}$:

######  Corollary 4.7.

Assume that $\alpha<\kappa^{*}$. Let
$\bar{\mu}=\sup\\{\mu_{\alpha^{\prime}}\colon\alpha^{\prime}<\alpha\\}$.
If $\alpha$ is successor, $\mu_{\alpha}$ is the first measurable above
$\bar{\mu}$ in $M_{\alpha}$.
If $\alpha$ is limit and
$\left(\mbox{cf}\left(\alpha\right)\right)^{V}\leq\kappa$, then $\mu_{\alpha}$
is the first measurable above $\bar{\mu}$ in $M_{\alpha}$.
If $\alpha$ is limit and
$\left(\mbox{cf}\left(\alpha\right)\right)^{V}>\kappa$, then $\mu_{\alpha}$ is
the first measurable in $M_{\alpha}$ which is greater or equal to $\bar{\mu}$.

###### Proof.

If $\bar{\mu}$ is measurable in $M_{\alpha}$ and
$\left(\mbox{cf}\left(\alpha\right)\right)^{V}>\kappa$, then
$\left(\mbox{cf}\left(\bar{\mu}\right)\right)^{V}>\kappa$ and thus
$\mu_{\alpha}=\bar{\mu}$ by the definition. Else, $\mu_{\alpha}$ is chosen to
be the least measurable in $M_{\alpha}$ above $\bar{\mu}$ with cofinality
above $\kappa$ in $V$, which is, by the previous lemma, the least measurable
above $\bar{\mu}$ in $M_{\alpha}$. ∎

######  Lemma 4.8.

Assume that $\alpha<\kappa^{*}$ and $\lambda$ appears after $\mu_{\alpha}$ in
the Prikry sequence of $\mu^{*}=k_{\alpha}\left(\mu_{\alpha}\right)$. Then
$\lambda=j_{U_{\mu_{\alpha}}}\left(\mu_{\alpha}\right)$.

###### Proof.

Since $j_{U_{\mu_{\alpha}}}\left(\mu_{\alpha}\right)$ is measurable in
$M_{\alpha+1}$ above
$\bar{\mu}_{\alpha+1}=\sup\\{\mu_{\alpha^{\prime}}\colon\alpha^{\prime}\leq\alpha\\}$,
it follows, by lemma 4.6, that–

$\left(\mbox{cf}\left(j_{U_{\mu_{\alpha}}}\left(\mu_{\alpha}\right)\right)\right)^{V}>\kappa$

Thus there exists $\beta>\alpha$ such that
$j_{U_{\mu_{\alpha}}}\left(\mu_{\alpha}\right)=\mu_{\beta}$, and appears in
the Prikry sequence of
$k_{\beta}\left(\mu_{\beta}\right)=j_{\beta,\kappa^{*}}\left(j_{\alpha,\beta}\left(\mu_{\alpha}\right)\right)=\mu^{*}$.

Let us prove now that
$j_{U_{\mu_{\alpha}}}\left(\mu_{\alpha}\right)=\mu_{\beta}$ is the immediate
successor of $\mu_{\alpha}$ in the Prikry sequence of $\mu^{*}$.

Assume, for contradiction, that
$\mu_{\alpha}<\lambda<j_{U_{\mu_{\alpha}}}\left(\mu_{\alpha}\right)$, and
$\lambda$ appears after $\mu_{\alpha}$ in the Prikry sequence of $\mu^{*}$.
Assume that
$\lambda=j_{\alpha+1}(g)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{k}},\mu_{\alpha}\right)$,
for some $g\in V$ and $\alpha_{0}<\ldots<\alpha_{k}<\alpha$. Assume also that
$h\in V$ is a function such that
$\mu_{\alpha}=j_{\alpha}(h)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{k}}\right)$
for the same $\alpha_{0}<\ldots<\alpha_{k}<\alpha$ (this can always be
arranged by changing the sequence
$\langle\alpha_{0},\ldots,\alpha_{k}\rangle$). Then–

$j_{\alpha+1}(g)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{k}},\mu_{\alpha}\right)<j_{\alpha+1}(h)\left(\kappa,\mu_{\alpha_{0}},\ldots,\mu_{\alpha_{k}}\right)$

so we may assume that for every $\xi,\nu_{0},\ldots,\nu_{k},\eta$, below
$\kappa$,
$g\left(\xi,\nu_{0},\ldots,\nu_{k},\eta\right)<h\left(\xi,\nu_{0},\ldots,\nu_{k}\right)$.
Assume also that $\mu_{\alpha}$ is the $n$-th element in the Prikry sequence
of $\mu^{*}$. In $V\left[G\right]$, let $\lambda(\xi)$ be the
$\left(n+1\right)$-th element in the Prikry sequence of
$h\left(\xi,\vec{\mu}(\xi)\right)$, so that
$\left[\xi\mapsto\lambda(\xi)\right]_{W}=\lambda$.

Assume that the sequence
$\langle\alpha_{0},\ldots,\alpha_{k}\rangle\subseteq\alpha$ is nice (else, add
more coordinates). Now apply the Multivariable Fusion Lemma. For every
$\langle\xi,\vec{\nu}\rangle$, let–

$\displaystyle e\left(\xi,\vec{\nu}\right)=\\{$ $\displaystyle r\in
P\setminus\nu_{k}+1\colon\mbox{there exists a bounded subset }A\subseteq
h\left(\xi,\vec{\nu}\right)\mbox{ such that}$ $\displaystyle r\mbox{ forces
that the }\left(n+1\right)\mbox{-th element in the Prikry sequence of
}h\left(\xi,\vec{\nu}\right)$ $\displaystyle\mbox{belongs either to }A\mbox{
or to the club of closure points of the function }$ $\displaystyle\eta\mapsto
g\left(\xi,\vec{\nu},\eta\right)\\}$

We argue that $e\left(\xi,\vec{\nu}\right)$ is $\leq^{*}$ dense open above any
condition which forces that
$\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{$\sim$}}{{\vec{\mu}}}}(\xi)=\vec{\nu}$.
Let $p\in P\setminus\nu_{k}+1$ be such a condition. Denote for simplicity
$h=h\left(\xi,\vec{\nu}\right)$. Direct extend $p\restriction_{h}$ such that
it decides the length of $t^{p}_{h}$; if the length is $\geq\left(n+1\right)$,
direct extend $p\restriction_{h}$ further, so that it forces that there exists
a bounded subset $A\subseteq h$ such that the $\left(n+1\right)$-th element in
the Prikry sequence of $h$ belongs to it. Finally, shrink
$\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{$\sim$}}{{A}}}^{p}_{h}$ by
intersecting with the club of closure points of the function which maps each
$\eta<h$ to $g\left(\xi,\vec{\nu},\eta\right)$. The condition obtained this
way indeed belongs to $e\left(\xi,\vec{\nu}\right)$.

Now, fix $p\in G$ and a $C$-tree $T$ such that for every
$\langle\xi,\vec{\nu}\rangle\in T$ which is admissible for $p$,

$\displaystyle\left(p^{\frown}\langle\xi,\vec{\nu}\rangle\right)\restriction_{\nu_{k}+1}\Vdash\mbox{there
exists a bounded subset }A\subseteq h\left(\xi,\vec{\nu}\right)\mbox{ such
that}$ $\displaystyle
p^{\frown}\langle\xi,\vec{\nu}\rangle\setminus\left(\nu_{k}+1\right)\mbox{
forces that the }\left(n+1\right)\mbox{-th element in the Prikry sequence}$
$\displaystyle\mbox{of }h\left(\xi,\vec{\nu}\right)\mbox{ belongs either to
}A\mbox{ or to the club of closure points of the function }$
$\displaystyle\eta\mapsto g\left(\xi,\vec{\nu},\eta\right)\\}$

Let
$\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{$\sim$}}{{A}}}\left(\xi,\vec{\nu}\right)$
be a $P_{\nu_{k}+1}$-name for the set $A$ above, and set–

$A^{*}\left(\xi,\vec{\nu}\right)=\\{\gamma<h\left(\xi,\vec{\nu}\right)\colon\exists
r\geq p^{\frown}\langle\xi,\vec{\nu}\rangle\restriction_{\nu_{k}+1},\
r\Vdash\gamma\in\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{$\sim$}}{{A}}}\left(\xi,\vec{\nu}\right)\\}$

It follows that for a set of $\xi$-s in $W$, $\lambda(\xi)$ either belongs to
$A^{*}\left(\xi,\vec{\mu}(\xi)\right)$ of to the club of closure points of
$\eta\mapsto g\left(\xi,\vec{\mu}(\xi),\eta\right)$.

However, it cannot hold that for a set of $\xi$-s in $W$, $\lambda(\xi)\in
A^{*}\left(\xi,\vec{\mu}(\xi)\right)$. Indeed assume otherwise. Denote–

$A^{*}=j_{\alpha}\left(\langle\xi,\vec{\nu}\rangle\mapsto
A^{*}\left(\xi,\vec{\nu}\right)\right)\left(\kappa,\mu\right)$

then $A^{*}$ is bounded in $\mu_{\alpha}$, and, under the above assumption,
$\lambda\in k_{\alpha}\left(A^{*}\right)=A^{*}\subseteq\mu_{\alpha}$, which is
a contradiction.

Thus, in $M\left[H\right]$, $\lambda$ is a closure point of $\eta\mapsto
j_{W}(g)\left(\kappa,\vec{\mu},\eta\right)$. Recall that
$\mu_{\alpha}<\lambda$, and thus
$j_{W}(g)\left(\kappa,\vec{\mu},\mu_{\alpha}\right)<\lambda=j_{\alpha}(g)\left(\kappa,\vec{\mu},\mu_{\alpha}\right)$,
which is a contradiction. ∎

######  Corollary 4.9.

Let $\alpha<\kappa^{*}$ and denote
$\mu^{*}=k_{\alpha}\left(\mu_{\alpha}\right)$. Then the Prikry sequence of
$\mu^{*}$ in $M\left[H\right]$ has a final segment of the form–

$\langle\mu_{\alpha_{0}},\mu_{\alpha_{1}},\mu_{\alpha_{2}},\ldots,\mu_{\alpha_{n}},\ldots\rangle$

where $\alpha_{0}=\alpha$, and for every $n<\omega$,
$\mu_{\alpha_{n+1}}=j_{U_{\mu_{\alpha_{n}}}}\left(\mu_{\alpha_{n}}\right)$.
Furthermore, the above sequence belongs to $V$, namely
$\left(\mbox{cf}\left(\mu^{*}\right)\right)^{V}=\omega$.

###### Proof.

The first part follows immediately from the previous lemma. Let us concentrate
on the second part. Assume that there is no $\beta<\alpha_{0}$ and
$\mu<\mu_{\alpha_{0}}$ such that
$j_{\beta,\alpha}\left(\mu\right)=\mu_{\alpha_{0}}$ (if there is, replace
$\mu_{\alpha_{0}}$ with the least such $\mu$). Let
$\beta_{0},\ldots,\beta_{k}$ be a complete nice sequence such that
$\mu_{\alpha_{0}}=j_{\alpha_{0}}(h)\left(\beta_{0},\ldots,\beta_{k}\right)$
for some $h\in V$. It follows that the sequence
$\langle\beta_{0},\ldots,\beta_{k},\alpha_{0},\alpha_{1},\ldots,\alpha_{n}\rangle$
is complete, for every $n<\omega$. Then, for every $n<\omega$, a finite
iteration $\langle N_{i}\colon i\leq n+1\rangle$ can be defined as in the
beginning of this section. If $f\in V$ is a function such that
$U_{\mu_{\alpha_{0}}}=j_{\alpha_{0}}(f)\left(\kappa,\mu_{\beta_{0}},\ldots,\mu_{\beta_{k}}\right)$,
then the sequence $\langle N_{i}\colon i<\omega\rangle$ is definable in $V$,
since each step above the first $k$-many steps in the iteration, uses a
measure represented by $f$. Because each sequence
$\langle\beta_{0},\ldots,\beta_{k},\alpha_{0},\ldots,\alpha_{n}\rangle$ is
complete, the sequence
$\langle\mu_{\alpha_{0}},\mu_{\alpha_{1}},\ldots,\mu_{\alpha_{n}},\ldots\rangle$
is a final segment of the sequence of critical points in the iteration
$\langle N_{i}\colon i<\omega\rangle$, and thus belongs to $V$. ∎

######  Remark 4.10.

We would like to emphasize the point that the characterization of Prikry
sequences given in the previous corollary is given only up to some finite
initial segment. Let us denote $\mu=\mu_{0}$, which is the first measurable
above $\kappa$ in $M_{U}$, and $\mu^{*}=k_{0}\left(\mu_{0}\right)$ which is
the first measurable above $\kappa$ in $M$. We argue that the Prikry sequence
of $\mu^{*}$ in $M\left[H\right]$ may have any prescribed finite initial
segment $t\in\left[\mu\right]^{<\omega}$. We use those notations only in the
following claim:

######  Claim 4.11.

For every finite, increasing sequence $t\in\left[\mu\right]^{<\omega}$, there
exists a condition $p\in P_{\kappa}$ which forces that $t$ is an initial
segment of the Prikry sequence of $\mu^{*}$ in $M\left[H\right]$.

###### Proof.

Assume that $\xi\mapsto t(\xi)$ is a function in $V$ such that
$\left[\xi\mapsto t(\xi)\right]_{U}=t$. For each $\xi<\kappa$, let $s(\xi)$ be
the first measurable strictly above $\xi$. We can assume that for every
$\xi<\kappa$, $\max\left(t(\xi)\right)<s(\xi)$.

Note that the set $\\{s(\xi)\colon\xi<\kappa\\}\cap\lambda$ is nonstationary
in any inaccessible $\lambda\leq\kappa$: This is clear if $\lambda$ is not a
limit of measurables. If it is, $\\{s(\xi)\colon\xi<\kappa\\}$ is disjoint to
the club of limit points of $\Delta=\\{\alpha<\kappa\colon\alpha\mbox{ is
measurable}\\}$ below $\lambda$.

Now, let us define a condition $p\in P_{\kappa}$, with
$\mbox{supp}(p)=\\{s(\xi)\colon\xi<\kappa\\}$. We first choose a set $X\in U$
on which the function $\xi\mapsto s(\xi)$ with domain $X$ is injective. Note
that by normality of $U$, every function is either one-to-one or constant
modulo $U$, so such a set $X\in U$ exists.

Set, for a given $\xi\in X$, $p(s(\xi))=\langle
t(\xi),s\left(\xi\right)\rangle$. This is forced by any condition in
$P\restriction_{s\left(\xi\right)}$ to be a legitimate element of
$\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{$\sim$}}{{Q}}}_{s\left(\xi\right)}$.
The condition $p\in P_{\kappa}$ defined in this way forces that the Prikry
sequence of $\mu^{*}$ starts with $t$: Indeed, in $V\left[G\right]$,

$\\{\xi<\kappa\colon t(\xi)\mbox{ is an initial segment of the Prikry sequence
of }s(\xi)\\}\supseteq X\in W$

thus, in $M\left[H\right]$, $\left[\xi\mapsto t(\xi)\right]_{W}$ is an initial
segment of the Prikry sequence of the measurable cardinal $\left[\xi\mapsto
s(\xi)\right]_{W}$. But by lemma 2.3,

$\left[\xi\mapsto t(\xi)\right]_{W}=k\left(\left[\xi\mapsto
t(\xi)\right]_{U}\right)=k(t)=t$

and clearly–

$\left[\xi\mapsto s(\xi)\right]_{W}=\mu^{*}$

so in $M\left[H\right]$, $t$ is an initial segment of the Prikry sequence
added to $\mu^{*}$.∎

Let us prove now that for every measurable $\mu^{*}$ above $\kappa$ in $M$,
$\mu^{*}$ has the form $k_{\alpha}\left(\mu_{\alpha}\right)$ for some
$\alpha<\kappa^{*}$. In particular, in the light of corollary 4.9,
$\left(\mbox{cf}(\mu^{*})\right)^{V}=\omega$.

######  Lemma 4.12.

Assume that $\mu^{*}\in\left(\kappa,\kappa^{*}\right)$ is measurable in $M$.
Then $\mu^{*}=k_{\alpha}\left(\mu_{\alpha}\right)$ for some
$\alpha<\kappa^{*}$.

###### Proof.

Let $\beta<\kappa^{*}$ be the first such that, for some $\mu\leq\mu^{*}$,
$\mu^{*}=k_{\beta}\left(\mu\right)$. Then $\mu$ is measurable in $M_{\beta}$.
$\beta$ is either $0$ or a successor by its minimality. Assume first that
$\beta=\alpha+1$. $\mu=\mu_{\alpha}$ cannot hold since $\mu_{\alpha}$ is not
measurable in $M_{\alpha+1}$. If $\mu<\mu_{\alpha}$ then
$j_{\alpha,\beta}\left(\mu\right)=\mu$, contradicting the minimality. Thus
assume that
$\mu>\mu_{\alpha}=\bar{\mu}_{\beta}=\sup\\{\mu_{\beta^{\prime}}\colon\beta^{\prime}<\beta\\}$.
Recall that $\mu$ is measurable in $M_{\beta}$. By lemma 4.6,
$\left(\mbox{cf}\left(\mu\right)\right)^{V}>\kappa$. Therefore, for some
$\gamma\in\left[\beta,\kappa^{*}\right)$, $\mu=\mu_{\gamma}$. Hence
$k_{\gamma}\left(\mu_{\gamma}\right)=\mu^{*}$.

If $\beta=0$ then $\mu$ is measurable in $M_{0}$ above $\kappa$ and below
$\kappa^{*}$, and clearly $\left(\mbox{cf}(\mu)\right)^{V}>\kappa$. So, again,
there exists $\gamma<\kappa^{*}$ such that $\mu=\mu_{\gamma}$, and
$k_{\gamma}\left(\mu_{\gamma}\right)=\mu^{*}$. ∎

######  Corollary 4.13.

Assume that $\alpha<\kappa^{*}$ is limit, and denote
$\bar{\mu}=\sup\\{\mu_{\alpha^{\prime}}\colon\alpha^{\prime}<\alpha\\}$.
Assume that $\bar{\mu}$ is measurable in $M_{\alpha}$. Then
$\left(\mbox{cf}(\alpha)\right)^{V}$ is either $\omega$ or $\kappa^{+}$. In
the former case, $\bar{\mu}$ is measurable in $M$. In the latter case,
$\bar{\mu}=\mu_{\alpha}$ is a non-measurable inaccessible cardinal in $M$.
Moreover:

  1. 1.

If $\bar{\mu}$ is not measurable in $M_{\alpha}$ or
$\left(\mbox{cf}\left(\alpha\right)\right)^{V}>\kappa$, $\mu_{\alpha}$ is the
first measurable $\geq$ $\bar{\mu}$ in $M_{\alpha}$ (this includes the case
where $\alpha$ is successor, since, in this case, $\mu_{\alpha-1}$ is not
measurable in $M_{\alpha}$).

  2. 2.

Else, $\bar{\mu}$ is measurable in $M_{\alpha}$ and
$\left(\mbox{cf}(\alpha)\right)^{V}=\omega$, and then
$\mu_{\alpha}=\bar{\mu}$.

###### Proof.

Assume that $\bar{\mu}$ is measurable in $M_{\alpha}$. If
$\left(\mbox{cf}(\alpha)\right)^{V}\leq\kappa$, then $\bar{\mu}<\mu_{\alpha}$,
so $\bar{\mu}=k_{\alpha}\left(\bar{\mu}\right)$ is measurable in $M$. By the
previous lemma, $\bar{\mu}=k_{\gamma}\left(\mu_{\gamma}\right)$ for some
$\gamma<\kappa^{*}$. By corollary 4.9,
$\left(\mbox{cf}\left(\bar{\mu}\right)\right)^{V}=\omega$. Hence
$\left(\mbox{cf}(\alpha)\right)^{V}=\omega$. ∎

######  Remark 4.14.

Recall that $j_{W}\left(\mathcal{U}\right)\setminus\kappa\in M$ is sufficient
for the definability of $j_{W}\restriction_{V}$ over $V$. Let us argue that it
is not necessary.

For every measurable $\eta<\kappa$, let $\langle s^{n}(\eta)\colon
n<\omega\rangle$ be the increasing enumeration of the first $\omega$-many
measurables above $\eta$ which carry at least $\eta$-many normal measures of
Mitchell order $0$. For each such $\eta$ and $n<\omega$, let
$\vec{F}\left(s^{n}(\eta)\right)$ be an enumeration for all the normal
measures of order $0$ on $s^{n}(\eta)$. Fix an unbounded nonstationary subset
$X\subseteq\Delta$ such that for every $\eta\in X$ and $n<\omega$,
$s^{n}(\eta)\notin X$. Let $P$ be the forcing notion which uses, at stage
$s^{n}(\eta)$ where $\eta\in X$ and $n<\omega$, the measure which extends
$\left(\vec{F}\left(s^{n}\left(\eta\right)\right)\right)\left(\eta_{n}\right)$.
Here, $\eta_{n}<\eta$ is the $n$-th element in the Prikry sequence of $\eta$
in $M\left[H\right]$. For every other measurable, use the measure chosen first
with respect to a prescribed well order of $V_{\kappa}$.

Pick a generic set $G\subseteq P$ such that $G$ contains a condition $p$ such
that $X\subseteq\mbox{supp}(p)$, but for every $\xi\in X$,
$p\restriction_{\xi}\Vdash t^{p}_{\xi}=\langle\rangle$.

Then $j_{W}\left(\mathcal{U}\right)\setminus\kappa\notin M$, since the
measures used in $j_{W}(P)$ on $M$-measurables above $\kappa$ code the Prikry
sequences of all the measurables in $j_{W}(X)\setminus\kappa$.

However, $j_{W}\restriction_{V}$ is definable in $V$: Assume that
$\alpha<\kappa^{*}$. If there is no $\eta\in j_{\alpha}(X)$ and $n<\omega$
such that $\mu_{\alpha}=s^{n}(\eta)$, $U_{\mu_{\alpha}}$ is the first measure
on $\alpha$ with respect to the image under $j_{\alpha}$ of the prescribed
well order on $V_{\kappa}$. Otherwise, assume that $\eta\in j_{\alpha}(X)$,
$n<\omega$ and $\mu_{\alpha}=s^{n}(\eta)$. Denote
$\eta^{*}=k_{\alpha}\left(\eta\right)$, so that
$k_{\alpha}\left(\mu_{\alpha}\right)=s^{n}\left(\eta^{*}\right)$. Let
$\beta=\beta_{0}<\kappa^{*}$ be the least such that
$k_{\beta}\left(\mu_{\beta}\right)=\eta^{*}$. We argue that the Prikry
sequence of $\eta^{*}$ in $M\left[H\right]$ is the sequence of critical points
taken by iteration $U_{\mu_{\beta}}$ $\omega$-many times over $M_{\beta}$.
This will follow once we prove that $\mu_{\beta}$ is the first element in the
Prikry sequence of $\eta^{*}$, and this is true since $\eta^{*}\in j_{W}(X)$
and there exists a condition $p\in G$ which forces that
$t^{p}_{\xi}=\langle\rangle$ for every $\xi\in X$. Thus, we can assume that
$\langle\mu_{\beta_{0}},\mu_{\beta_{1}},\ldots,\mu_{\beta_{n}},\ldots\rangle$
is the Prikry sequence of $\eta^{*}$ in $M\left[H\right]$.

Recall that
$k_{\alpha}\left(U_{\mu_{\alpha}}\right)=j_{W}\left(\mathcal{U}\right)\left(k_{\alpha}\left(\mu_{\alpha}\right)\right)$;
by the definition of the forcing,
$j_{W}\left(\mathcal{U}\right)\left(k_{\alpha}\left(\mu_{\alpha}\right)\right)$
is the Prikry forcing taken with the measure–

$\left(j_{W}\left(\vec{F}\right)\left(k_{\alpha}\left(\mu_{\alpha}\right)\right)\right)\left(\mu_{\beta_{n}}\right)$

thus, $U_{\mu_{\alpha}}$ can be computed in $V$ as follows: first, calculate
over $M_{\beta_{0}}$ (which is already definable in $V$ by induction) the
sequence $\langle\mu_{\beta_{n}}\colon n<\omega\rangle$, which are the
critical points in the iteration of length $\omega$ with $U_{\mu_{\beta_{0}}}$
over $M_{\beta_{0}}$ (here, $U_{\mu_{\beta}}$ is the least measure with
respect to the image under $j_{\beta}$ of the prescribed well order on
$V_{\kappa}$); then, compute
$U_{\mu_{\alpha}}=\left(j_{\alpha}\left(\vec{F}\right)\left(\mu_{\alpha}\right)\right)\left(\mu_{\beta_{n}}\right)$.

## References

  * [1] Omer Ben-Neria and Spencer Unger.  Homogeneous changes in cofinalities with applications to hod.  Journal of Mathematical Logic, 17(02):1750007, 2017.
  * [2] Moti Gitik and Eyal Kaplan.  On restrictions of ultrafilters from generic extensions to ground models.
  * [3] Joel David Hamkins.  Gap forcing.  Israel Journal of Mathematics, 125(1):237–252, 2001.
  * [4] Ralf Schindler.  Iterates of the core model.  Journal of Symbolic Logic, pages 241–251, 2006.