Nevanlinna deficiency and ramification defect relations

(Per-target bound) Since $N_1(r,a;f)\le N(r,a;f)$ and $T(r,f)>0$ for all large $r$, $\epsilon (a,f)={\mathrm{lim\hspace{0.17em}inf}}_r\frac{N_1(r,a;f)}{T(r,f)}\le {\mathrm{lim\hspace{0.17em}inf}}_r\frac{N(r,a;f)}{T(r,f)}\le {\mathrm{lim\hspace{0.17em}sup}}_r\frac{N(r,a;f)}{T(r,f)}=1-\delta (a,f)$, so $\delta (a,f)+\epsilon (a,f)\le 1$, and both indices are nonnegative by [F1].

(Second Main Theorem bound for finite target sets) Let $A$ be a finite set of distinct sphere targets with $\mathrm{\mid }A\mathrm{\mid }\ge 3$. By [F2], ${\sum }_{a\in A}m(r,a;f)+N_1(r,f)\le 2T(r,f)+S(r,f)$ outside a set $E$ of finite linear measure; since ${\sum }_{a\in A}{N}_1(r,a;f)\le N_1(r,f)$ by [F2], also ${\sum }_{a\in A}(m(r,a;f)+N_1(r,a;f))\le 2T(r,f)+S(r,f)$ outside $E$.

(Target sets of at most two points) If $\mathrm{\mid }A\mathrm{\mid }\le 2$ then ${\sum }_{a\in A}(\delta (a,f)+\epsilon (a,f))\le \mathrm{\mid }A\mathrm{\mid }\le 2$ by step 1.1, so the asserted bound holds for every finite target set of at most two points.

(Growth separation) If $f$ is transcendental, [F3] gives $T(r,f)\mathrm{/}\log r\to \mathrm{\infty }$ and $T(r,f)\to \mathrm{\infty }$, so the general error bound in [F2] satisfies $S(r,f)\mathrm{/}T(r,f)\to 0$ off $E$. If $f$ is rational of degree $d\ge 1$, [F2] supplies the stronger error $O_f(1)$ at every large radius, while [F3] gives $T(r,f)=d\log r+O(1)\to \mathrm{\infty }$; thus this error divided by $T$ also tends to zero.

(Defect bound for finite target sets) For $\mathrm{\mid }A\mathrm{\mid }\ge 3$, dividing step 1.2 by $T(r,f)$ and using step 2.2 gives ${\sum }_{a\in A}(\frac{m(r,a;f)}{T(r,f)}+\frac{N_1(r,a;f)}{T(r,f)})\le 2+o(1)$ along $r\mathrm{\notin }E$; since $E$ has finite measure its complement is unbounded, so the lower limit of the left side is at most $2$. As a finite sum of lower limits is at most the lower limit of the sum, ${\sum }_{a\in A}(\delta (a,f)+\epsilon (a,f))\le 2$; with step 2.1 this covers every finite target set $A$.

(Supremum over finite sets) By [F1] the total deficiency sum is the supremum of the finite subsums, each of which is at most $2$ by step 3.1, so ${\sum }_a(\delta (a,f)+\epsilon (a,f))\le 2$; since $\delta \ge 0$, also ${\sum }_a\delta (a,f)\le 2$.

(The positive set is at most countable) For every integer $k\ge 1$ put $F_k=\{a\in \widehat{\mathbb{C}}:\delta (a,f)+\epsilon (a,f)>1\mathrm{/}k\}$; were $\mathrm{\mid }{F}_k\mathrm{\mid }\ge 2k+1$, then the finite subsum over any $2k+1$ points of $F_k$ would exceed $\frac{2k+1}{k}=2+\frac{1}{k}>2$, contradicting step 4.1, so each $F_k$ is finite and hence at most countable. Since $\delta +\epsilon \ge 0$, the set where either index is positive equals ${\bigcup }_{k\ge 1}{F}_k$, a countable union of at most countable sets, which is at most countable by [F4].