Proposition 21.7.2EDD78BC61BD7

Proof. Let $n>0$. If $d_{\mathrm H}(K,L)\leq \varepsilon $, then every $p\in K$ has a point $q\in L$ with $|p-q|\leq \varepsilon $, and for $|x|=1$ we get $\langle p,x\rangle \leq \langle q,x\rangle +\varepsilon \leq h_ L(x)+\varepsilon $; hence $h_ K\leq h_ L+\varepsilon $ on the unit sphere, and by symmetry $|h_ K-h_ L|\leq \varepsilon $ there. Conversely, suppose that $|h_ K-h_ L|\leq \varepsilon $ on the unit sphere. The support function of the convex body $L+\overline{B(0,\varepsilon )}$ is $h_ L(x)+\varepsilon |x|$. If some $p\in K$ were outside $L+\overline{B(0,\varepsilon )}$, lemma 21.2.3(1) would give a unit vector $x$ with $\langle p,x\rangle >h_ L(x)+\varepsilon \geq h_ K(x)$, which is impossible. Hence $K\subset L+\overline{B(0,\varepsilon )}$, and by symmetry $d_{\mathrm H}(K,L)\leq \varepsilon $. This proves the formula. It shows that $d_{\mathrm H}$ satisfies the triangle inequality, and by lemma 21.2.3(1) a convex body is determined by its support function; thus $d_{\mathrm H}$ is a metric.

Let $\varepsilon _ j=d_{\mathrm H}(K_ j,K)\to 0$. For large $j$ all $K_ j$ lie in the bounded set $K+\overline{B(0,1)}$. A point $p$ outside $K$ has positive distance from $K$, so $p\notin K_ j$ once $\varepsilon _ j<\operatorname {dist}(p,K)$. If $p$ is an interior point of $K$, choose $r>0$ with $\overline{B(p,r)}\subset K$. Then, for $|x|=1$,

If $\varepsilon _ j<r$ and $p\notin K_ j$, strict separation would give a unit vector $x$ with $\langle p,x\rangle >h_{K_ j}(x)$, contradicting this inequality; hence $p\in K_ j$ for large $j$. Thus $\mathbf1_{K_ j}\to \mathbf1_ K$ outside the boundary of $K$. If $K$ has nonempty interior, its boundary is a null set by lemma 21.2.3(4); if $K$ has empty interior, it lies in a proper affine subspace and is itself a null set, and $\mathbf1_{K_ j}\to 0=\mathbf1_ K$ almost everywhere by the outside-point argument. In either case dominated convergence, with the dominating function $\mathbf1_{K+\overline{B(0,1)}}$, gives $\operatorname {vol}(K_ j)\to \operatorname {vol}(K)$.

The outer parallel bodies $K^\varepsilon $ have finite volume and decrease to the compact set $K$ as $\varepsilon \downarrow 0$, so continuity of measure from above gives the first limit. The inner sets $K_\varepsilon $ increase, as $\varepsilon $ decreases, to the interior of $K$, which differs from $K$ by a null set, as we saw in the previous paragraph; monotone convergence gives the second limit. In dimension zero both $K^\varepsilon $ and $K_\varepsilon $ equal the single point for every $\varepsilon $. $\square$