Theorem 17.5.20F21AE677873A

Proof. We derive a recursive inequality for the capacities of the sublevel sets of $u-\phi $ and iterate it along levels that converge to a finite depth. For $s>0$ put

The sets $E_ s$ decrease as $s$ increases, so $g$ is nonincreasing.

We first compare the measures on $E_ s$. Both $u$ and $\phi -s$ belong to $\mathcal E(X,\theta ,\phi )$, so the comparison principle of proposition 16.25.4, applied to $u$ and $\phi -s$, gives $\mu _\phi (E_ s)\leq \mu _ u(E_ s)$. By (17.5.21),

and hence $(1-a)\mu _ u(E_ s)\leq \nu (E_ s)$. Since $u$ has full mass relative to $\phi $, proposition 16.25.7 gives $t^ n\operatorname {Cap}_{\theta ,\phi }(E_{s+t})\leq \mu _ u(E_ s)$ for $0<t\leq 1$. Combining this with the last inequality and the capacity bound in (17.5.21), we obtain $t^ ng(s+t)^ n\leq B^ ng(s)^{n(1+\delta )}$, that is,

By the second assertion of proposition 16.25.7, $g(s)\to 0$ as $s\to \infty $. Choose $s_0>0$ with $2Bg(s_0)^\delta \leq 1$, and put

We claim that $g(C)=0$. If $g(s_0)=0$, then $C=s_0$ and there is nothing to prove. Otherwise, as long as $g(s_ j)>0$, define inductively

Since $g$ is nonincreasing, $t_ j\leq 2Bg(s_0)^\delta \leq 1$, and (17.5.23) with $s=s_ j$ and $t=t_ j$ gives

Hence $g(s_ j)\leq 2^{-j}g(s_0)$ and $t_ j\leq 2^{-j\delta }\cdot 2Bg(s_0)^\delta $, so that

If the construction stops because $g(s_ j)=0$ for some $j$, then $g(C)\leq g(s_ j)=0$. Otherwise $g(C)\leq g(s_ j)\leq 2^{-j}g(s_0)$ for every $j$, and again $g(C)=0$. This proves the claim.

Thus $\operatorname {Cap}_{\theta ,\phi }(E_ C)=0$. The capacity bound in (17.5.21) gives $\nu (E_ C)=0$. Since $\phi $ itself is a competitor for $\operatorname {Cap}_{\theta ,\phi }$ (definition 16.25.5), we also have $\mu _\phi (E_ C)\leq \operatorname {Cap}_{\theta ,\phi }(E_ C)=0$. The measure inequality in (17.5.21) therefore gives $\mu _ u(\{ u<\phi -C\} )=0$. As $\phi -C\in \mathcal E(X,\theta ,\phi )$, the domination principle, theorem 16.22.10, yields $u\geq \phi -C$ everywhere.

Now suppose that $I<\infty $. On $E_ s$ the weight $\phi -u$ exceeds $s$, so Markov's inequality gives $\nu (E_ s)\leq I/s$. Taking $t=1$ in the sublevel estimate and using the measure inequality on $E_ s$, we find

Choose $s=1+I(2B)^{n/\delta }/(1-a)$ and $s_0=s+1$. Since $(1-a)s>I(2B)^{n/\delta }$, we get $g(s_0)^ n\leq (2B)^{-n/\delta }$, that is, $2Bg(s_0)^\delta \leq 1$. Thus $s_0$ satisfies the condition imposed on it above, and the constant in (17.5.24) is at most $s_0+(1-2^{-\delta })^{-1}$. Since $(2B)^{n/\delta }=\bigl (2(A/(1-a))^{1/n}\bigr )^{n/\delta }$, this is the right side of (17.5.22). The lower bound $\phi -C\leq u$ remains true when $C$ is increased, which proves (17.5.22). $\square$