Theorem 4.1.10872A6887585B

Proof. (1) Since $p$ is upper semicontinuous (proposition 3.4.11), $\varlimsup _{z\to x}p(z)\leq p(x)$; in particular $p$ is continuous at $x$ if $p(x)=-\infty $. Assume that $p(x)$ is finite; we must show that $\varliminf _{z\to x}p(z)\geq p(x)$. For $z\in \mathbb R^ d$ choose a point $z'\in K$ closest to $z$. Then $|z'-x|\leq |z'-z|+|z-x|\leq 2|z-x|$, so $z'\to x$ as $z\to x$, and $|z'-y|\leq |z'-z|+|z-y|\leq 2|z-y|$ for every $y\in K$. Consequently, for $y\in K$,

Fix $\delta >0$ and split $p=p'_\delta +p''_\delta $, where $p'_\delta (w)=\int _{B(x,\delta )}E_ d(w-y)\, \mathrm d\mu (y)$ is the near part and $p''_\delta $ the far part. The far part is continuous at $x$, since $E_ d(w-y)\to E_ d(x-y)$ uniformly for $y\in K$ with $|y-x|\geq \delta $ as $w\to x$, by the uniform continuity of $E_ d$ on compact subsets of $\mathbb R^ d\setminus \{ 0\} $. The near part, restricted to $K$, is continuous at $x$, being the difference of $p|_ K$ and $p''_\delta |_ K$. Since $\mu $ is carried by $K$, we may integrate the inequalities above with respect to $\mu $ over $B(x,\delta )$; letting $z\to x$, we obtain for $d\geq 3$

and for $d=2$ the bound $\varliminf _{z\to x}p(z)\geq p(x)-(2\pi )^{-1}\mu (B(x,\delta ))\log 2$. Since $p(x)$ is finite, the function $E_ d(x-\cdot )$ is $\mu $-integrable and $\mu (\{ x\} )=0$. By dominated convergence and continuity of $\mu $ from above, $p'_\delta (x)\to 0$ and $\mu (B(x,\delta ))\to 0$ as $\delta \downarrow 0$, which proves $\varliminf _{z\to x}p(z)\geq p(x)$.

(2) Let $M\geq 0$ be an upper bound of $E_ d$ on the compact set $K-K$. If $\nu $ is the restriction of $\mu $ to a Borel set, then $p=E_ d*\nu +E_ d*(\mu -\nu )$ and $E_ d*(\mu -\nu )\leq M\mu (K)$ on $K$, so $E_ d*\nu \geq p-M\mu (K)$ on $K$; hence $E_ d*\nu $ is finite $\nu $-almost everywhere.

We construct inductively pairwise disjoint compact sets $K_1,K_2,\ldots \subset K$ such that the residual measures $\nu _ j=\mu |_{K\setminus (K_1\cup \cdots \cup K_ j)}$ satisfy $\nu _ j(\mathbb R^ d)<2^{-j}$, and such that each $E_ d*(\mu |_{K_ j})$ is finite and continuous on $\mathbb R^ d$. Put $\nu _0=\mu $, and suppose $K_1,\ldots ,K_{j-1}$ have been chosen. Apply lemma 4.1.9 to $\nu =\nu _{j-1}$ and $E_ d*\nu $: there is a compact set $L$ with $\nu (\mathbb R^ d\setminus L)<2^{-j-1}$ on which $E_ d*\nu $ is finite and continuous. Since $\nu $ is carried by $K$ and does not charge $K_1\cup \cdots \cup K_{j-1}$, inner regularity gives a compact set $K_ j\subset (L\cap K)\setminus (K_1\cup \cdots \cup K_{j-1})$ with $\nu (\mathbb R^ d\setminus K_ j)<2^{-j}$. Put $\mu _ j=\nu |_{K_ j}=\mu |_{K_ j}$; then $\nu _ j=\nu -\mu _ j$ has mass less than $2^{-j}$.

It remains to show that $E_ d*\mu _ j$ is finite and continuous. On $K_ j$ the two functions $E_ d*\mu _ j$ and $E_ d*(\nu -\mu _ j)$ are upper semicontinuous and less than $+\infty $, and their sum $E_ d*\nu $ is finite and continuous; hence both are finite on $K_ j$. Moreover, for $x\in K_ j$ and $K_ j\ni w\to x$,

so the restriction of $E_ d*\mu _ j$ to $K_ j$ is continuous. The support of $\mu _ j$ is contained in $K_ j$, so part (1), applied to $\mu _ j$ and the compact set $K_ j$, shows that $E_ d*\mu _ j$ is continuous at every point of $K_ j$. At a point $x\notin K_ j$, the function $E_ d*\mu _ j$ is finite and continuous, because $E_ d(w-y)\to E_ d(x-y)$ uniformly in $y\in K_ j$ as $w\to x$, as for the far part in (1). Thus $E_ d*\mu _ j$ is finite and continuous on $\mathbb R^ d$. Finally, $\mu -\sum _{i\leq j}\mu _ i=\nu _ j$ has mass less than $2^{-j}$, so $\mu =\sum _ j\mu _ j$. $\square$