Finite-measure growth increment lemma

For every integer $m\ge 1$ the superlevel set $\{r\ge r_0:u(r)\ge m\}$ is closed in $[r_0,\mathrm{\infty })$, and it is nonempty for $m\le {\lim }_{r\to \mathrm{\infty }}u(r)=\mathrm{\infty }$; let $s_m$ be its least element, and put $m_0:=\lceil u(r_0)\rceil$. Then $s_m\le s_{m+1}$.

(Failure set) Put $\phi (t):=t^{-1-\epsilon }$ and $F:=\{r\ge r_0:u(r+\phi (u(r)))\ge u(r)+1\}$. The function $r\mapsto u(r+\phi (u(r)))-u(r)-1$ is continuous on $[r_0,\mathrm{\infty })$ because $u$ and $\phi$ are continuous; hence $F$ is closed in $[r_0,\mathrm{\infty })$.

(Application to $T$) Let $f$ be a nonconstant meromorphic function. By [F4] the function $T(\cdot ,f)$ is continuous and nondecreasing in $r$ (the constant $C_{\mathrm{\infty }}(f)$ is additive). It is also unbounded: $T$ has a limit because it is nondecreasing, and if that limit were finite then $T(r,f)=O(\log r)$ for large $r$; [F5] would then make $f$ rational of some degree $d\ge 1$, and the same item gives $T(r,f)=d\log r+O(1)\to \mathrm{\infty }$, a contradiction, while constant $f$ is excluded. Increasing $r_0$ so that $T(r_0,f)\ge 1$, the lemma applies to $u=T(\cdot ,f)$.

(Measurability and finite measure of the cover) Every bounded closed interval $[a,b]$ is a countable intersection of half-open intervals $(a-1\mathrm{/}n,b]$, hence Lebesgue measurable by [F1]; the same intervals cover it with measures tending to $b-a$, so $\lambda ([a,b])\le b-a$. Set $E:=[r_0,\mathrm{\infty })\cap ([r_0,s_{m_0}]\cup {\bigcup }_{m\ge m_0}[s_{m+1}-m^{-1-\epsilon },s_{m+1}])\mathrm{.}$ This set is measurable, is contained in $[r_0,\mathrm{\infty })$, and by [F2] $\lambda (E)\le s_{m_0}-r_0+{\sum }_{m\ge m_0}{m}^{-1-\epsilon }<\mathrm{\infty }$ by [F3] and $\epsilon >0$.

If $m>u(r_0)$ then $s_m>r_0$ and $u(s_m)=m$: by definition $u(s_m)\ge m$, and if $u(s_m)>m$ then by continuity $u>m$ on a left neighbourhood of $s_m$ inside $[r_0,\mathrm{\infty })$, contradicting minimality. If $m=u(r_0)$ the same holds with $s_m=r_0$.

(Cover of the failure set) Recall $m_0=\lceil u(r_0)\rceil$ from step 1.1 and let $r\in F$ with $r\ge s_{m_0}$. Write $m:=\lfloor u(r)\rfloor \ge m_0$. Then $u(r)\in [m,m+1)$, so $r\ge s_m$ and $r<s_{m+1}$. Moreover $u(r)\ge m$, so by step 2.2, $u(r+\phi (u(r)))\ge u(r)+1\ge m+1=u(s_{m+1})$; monotonicity of $u$ then forces $r+\phi (u(r))\ge s_{m+1}$, i.e. $r\ge s_{m+1}-\phi (u(r))\ge s_{m+1}-\phi (m)$. Hence $F\cap [s_{m_0},\mathrm{\infty })\subseteq {\bigcup }_{m\ge m_0}[s_{m+1}-m^{-1-\epsilon },s_{m+1}]\cup [r_0,s_{m_0}]\mathrm{.}$

(Conclusion off $E$) If $r\ge r_0$ and $r\mathrm{\notin }E$, then $r\mathrm{\notin }[r_0,s_{m_0}]$, so $r>s_{m_0}$; since $r$ lies in no interval $[s_{m+1}-m^{-1-\epsilon },s_{m+1}]$ with $m\ge m_0$ either, step 3.1 gives $r\mathrm{\notin }F$: unwinding the definition of $F$, $u(r+\phi (u(r)))<u(r)+1$, as required.

The argument selects nothing: each $s_m$ is the least element of a nonempty closed set, and the covering intervals are defined from the $s_m$ and the given constants. Countable Choice is used only through the published Lebesgue measure interface of [F1].