Lemma 19.7.9806DFD35CCE5

Proof. We may assume that $t_{\min }<t_{\max }$ as otherwise there is nothing to prove.

For each $t\in [t_{\min },t_{\max }]$, we write $K_ t=\{ y\in \mathbb {R}^{n-1}:(t,y)\in K\} $. Let $t_ j\to t$ be a convergent sequence in $[t_{\min },t_{\max }]$, we want to show that $K_{t_ j}$ converges to $K_ t$ with respect to the Hausdorff metric. Recall that this amounts to the following two assertions:

1. 4164F65EB6F1

   For each convergent sequence $y_ j\in K_{t_ j}$ with limit $y$, we have $y\in K_ t$;

2. 5031A2661BB4

   Given any $y\in K_ t$, up to replacing $t_ j$ by a subsequence, we can find $y_ j\in K_{t_ j}$ converging to $y$.

The first assertion is obvious. Let us prove the second. Take $y\in K_ t$. Up to replacing $t_ j$ by a subsequence and taking the symmetry into account, we may assume that $t_ j> t$ for all $j$. In particular, $t<t_{\max }$.

We can find a point $z=(z^1,z')\in K$ such that $z^1>t$ (for example, there is always such a point with $z^1=t_{\max }$). Replacing $t_ j$ by a subsequence, we may assume that $t_ j\in (t,z^1)$ for all $j$. Then it suffices to take

This completes the proof. $\square$