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The growth order of a meromorphic function isThe order of the integrated count is defined analogously using ${\log }^{+}N(R;a)$, with an identically zero count assigned order zero. Constants have function order zero. These are growth orders, not local zero or pole orders. For nonconstant $f$, the first main theorem gives $\rho (N(\cdot ;a))\le \rho (f)$ for every target.

Use the truncated Nevanlinna second main theorem in the following standard form: for distinct fixed targets $a_1,\dots ,a_q$,outside an exceptional set of radii of finite linear measure. For finite order the error is $O(\log R)$; replacing each truncated count by its full count preserves the inequality.

Suppose $0<\rho (f)<\infty$ and three targets have counting orders below $\rho (f)$. Choose $\mu$ strictly between the maximum of their three orders and $\rho (f)$. Each full count is at most $R^{\mu }$ eventually. The theorem with $q=3$ then gives $T_f(R)\le C{R}^{\mu }$ for sufficiently large radii outside the exceptional set. Its tail measure is eventually less than one, so every interval $[R,R+1]$ contains a nonexceptional radius $R^{\prime }$. Monotonicity of the characteristic yields $T_f(R)\le T_f(R^{\prime })\le C(R+1{)}^{\mu }$ for all large $R$. This contradicts its order. If $\rho (f)=0$, all the counting orders are already zero. Hence the counting-order exceptions for a finite-order meromorphic function satisfyThe exponential has order one but omits both zero and infinity, so those two counts are zero; this realizes two exceptions.

For a Möbius map $U(w)=(aw+b)/(cw+d)$ with $ad-bc\ne 0$, translation and multiplication by a nonzero constant change $T$ by $O(1)$, as follows directly from the proximity inequalities and unchanged pole multiplicities. The first main theorem also gives $T(1/h)=T(h)+O(1)$. If $c=0$, $U$ is affine. If $c\ne 0$, write $U(w)=a/c+K/(cw+d)$ with $K\ne 0$ and use these three elementary operations. Thus $T_{U\circ f}=T_f+O(1)$, proving

For the sum, a pole of $f_1+f_2$ has order no larger than the sum of the two pole orders, andTherefore $T_{f_1+f_2}\le T_{f_1}+T_{f_2}+O(1)$ andIf $\rho (f_1)>\rho (f_2)$, apply the same inequality to $f_1=(f_1+f_2)-f_2$. It forces $\rho (f_1)\le \max (\rho (f_1+f_2),\rho (f_2))$, hence the sum's order equals $\rho (f_1)$. The reversed case is identical. This reasoning also handles an infinite larger order. For equal orders cancellation can be strict: $f_1=e^z$, $f_2=-e^z$ both have order one, but their sum is zero of order zero.