Proposition 4.2.13A13D01405A7F

Proof. Enlarging $A$ or $s$ shrinks the class of admissible functions, which proves the two monotonicity assertions. For $c>0$, multiplication by $c$ is a bijection between the admissible functions for $s$ and those for $cs$; for $c=0$, both sides vanish (with $0\cdot \infty =0$), since the zero function is admissible. If $h_1$ and $h_2$ are admissible for $s$ and $t$ respectively, then $h_1+h_2$ is admissible for $s+t$, so $R_{s+t}^ A\leq R_ s^ A+R_ t^ A$. Monotonicity and homogeneity are preserved by lower semicontinuous regularization. For subadditivity, recall that each reduction agrees with its regularization outside a polar set, hence almost everywhere. Therefore $\widehat R_{s+t}^ A\leq \widehat R_ s^ A+\widehat R_ t^ A$ almost everywhere. Both sides are superharmonic, so the inequality holds everywhere by the ball-average formula above.

Now let $U$ be bounded and $A$ polar. Then $A$ has compact closure, so proposition 4.1.11 provides a finite positive measure $\nu $ of compact support with $E_ d*\nu =-\infty $ on $A$. The kernel $E_ d$ is bounded above on the bounded set of differences $x-y$ with $x\in U$ and $y\in \operatorname{Supp}\nu $, so adding a constant to $-E_ d*\nu $ gives a nonnegative superharmonic function $p$ on $U$ with $p=+\infty $ on $A$; by proposition 3.4.11, $p$ is locally integrable. For every $\varepsilon >0$, the function $\varepsilon p$ is admissible, even at points of $A$ where $s=+\infty $. Hence $R_ s^ A\leq \varepsilon p$ for every $\varepsilon >0$, and so $R_ s^ A=0$ on the set $\{ p<\infty \} $. This set has full Lebesgue measure, because $p$ is locally integrable, and is therefore dense in $U$. Consequently $\widehat R_ s^ A=0$. $\square$