Proposition 15.19.22.65049DB7855E Let $g_ j$ be smooth Riemannian metrics on $X$ such that $g_ j\to \pi ^*g_0$ uniformly on every compact subset of $X\setminus E$. Suppose each $s_ a$ has a nested neighborhood basis $V_{a,r}$, with $U_{a,r}=\pi ^{-1}(V_{a,r})$, satisfying
\[ \lim _{r\downarrow 0}\operatorname {diam}_{\widehat d_0}(\iota (V_{a,r}))=0, \qquad \lim _{r\downarrow 0}\limsup _{j\to \infty } \operatorname {diam}^{\mathrm{int}}(U_{a,r},g_ j)=0. \]
Then, for the global length distances $d_ j$ of $g_ j$,
\[ \lim _{j\to \infty }\sup _{x,x'\in X} |d_ j(x,x')-\widehat d_0(G(x),G(x'))|=0. \]
Consequently $(X,d_ j)$ converges to $(\widehat Y,\widehat d_0)$ in Gromov–Hausdorff distance.
Proof.
If $S$ is empty, uniform metric comparison on compact $X$ proves the distance convergence. Assume $N\geq 1$. Fix $\tau >0$. Choose pairwise disjoint neighborhoods $V_ a$ from the specified bases with
\[ \operatorname {diam}_{\widehat d_0}(\iota (V_ a))<\tau , \qquad \operatorname {diam}^{\mathrm{int}}(U_ a,g_ j)<\tau \]
for all sufficiently large $j$, where $U_ a=\pi ^{-1}(V_ a)$. Pick $q_ a\in V_ a\setminus S$, and then choose open neighborhoods $W_ a$ of $s_ a$ with $\overline W_ a\subset V_ a$ and $q_ a\notin \overline W_ a$. The regular cores
\[ K_ Y=Y\setminus \bigcup _ a W_ a, \qquad K_ X=\pi ^{-1}(K_ Y) \]
are compact and identified by $\pi $. Uniform convergence and positivity of $g_0$ on this core give numbers $A_ j\geq 1$, with $A_ j\to 1$, such that lengths of curves contained in the cores compare in both directions by the factor $A_ j$.
Consider a piecewise smooth path in $Y^\circ $ joining two points of $K_ Y$. The preimages of the closed sets $\overline W_ a$ are compact subsets of its parameter interval. At its first visit to any such set, delete the portion from that visit through its last visit to the same set. Resume with the remaining tail. That set is never visited again; thus there are at most $N$ deleted portions. Each retained segment lies in $K_ Y$, and each deleted portion has endpoints in a single $\overline W_ a\subset V_ a$. The lifts of those endpoints can be joined in $U_ a$ with $g_ j$-length less than $\tau $, up to an arbitrarily small error. Applying this procedure to almost minimizing paths, comparing retained lengths, and letting the errors tend to zero gives, for $x,x'\in K_ X$,
\[ d_ j(x,x')\leq A_ j\widehat d_0(G(x),G(x'))+N\tau . \]
Conversely start with an almost minimizing $g_ j$-path in $X$ and apply the same deletion procedure to the closed sets $\pi ^{-1}(\overline W_ a)$. Project its retained segments, which avoid $E$, to $K_ Y$. The two images of each deleted portion have completion distance less than $\tau $. The triangle inequality in $\widehat Y$ therefore gives
\[ \widehat d_0(G(x),G(x'))\leq A_ j d_ j(x,x')+N\tau , \qquad x,x'\in K_ X. \]
For any $x\notin K_ X$ with $\pi (x)\in W_ a$, replace $x$ by $p=\pi ^{-1}(q_ a)\in U_ a\cap K_ X$. Then $d_ j(x,p)<\tau $ and $\widehat d_0(G(x),G(p))<\tau $. Leave points of $K_ X$ fixed. Replacing both endpoints extends the two inequalities to all $x,x'\in X$, with the same error
\[ C_ j=(N+2+2A_ j)\tau . \]
If $D=\operatorname {diam}(\widehat Y,\widehat d_0)$, the first inequality gives $\operatorname {diam}(X,d_ j)\leq A_ jD+C_ j$. The positive and negative distance errors are consequently bounded by
\[ (A_ j-1)D+C_ j, \qquad (A_ j-1)(A_ jD+C_ j)+C_ j, \]
respectively. Their upper limits are at most $(N+4)\tau $. Letting $\tau \downarrow 0$ proves uniform convergence.
The graph of $G$ is a correspondence between $X$ and $\widehat Y$, and its distortion $b_ j$ for the distances $d_ j$ and $\widehat d_0$ is the supremum in the statement. By lemma 15.19.20,
\[ d_{\mathrm{GH}}((X,d_ j),(\widehat Y,\widehat d_0))\leq b_ j/2\longrightarrow 0. \]
$\square$
Pome
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