Let $p\in \mathbb{C} \lbrack z],~p=p\left( re^{it}\right) ,n=\deg p$ and we want to compute de limit $$ \lim_{r\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}\frac{p^{\prime}\left( re^{it}\right) re^{it}}{p\left( re^{it}\right) }\operatorname*{dt}% =\lim_{r\rightarrow\infty}F\left( r\right) =\lim_{r\rightarrow\infty}% \frac{1}{2\pi}\int_{0}^{2\pi}f\left( r,t\right) \operatorname*{dt}. $$
Since $x\mapsto xp^{\prime}\left( x\right) /p\left( x\right) $ is continous on $ \mathbb{C} $ and $$ \lim_{n\rightarrow\infty}\frac{p^{\prime}\left( re^{it}\right) re^{it}% }{p\left( re^{it}\right) }=n, $$ hence $x\mapsto xp^{\prime}\left( x\right) /p\left( x\right) $ is bounded.
Then exists $K>0$ such that $$ \left\vert \frac{p^{\prime}\left( re^{it}\right) re^{it}}{p\left( re^{it}\right) }\right\vert <K,\forall\left( r,t\right) \in \mathbb{R} _{+}\times\lbrack0,2\pi]. $$ So, the conditions of Dominated convergence theorem are verified for $f_{n}=f_{r},$ then $$ \lim_{r\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}\frac{p^{\prime}\left( re^{it}\right) re^{it}}{p\left( re^{it}\right) }\operatorname*{dt}=\frac {1}{2\pi}\int_{0}^{2\pi}\lim_{r\rightarrow\infty}\frac{p^{\prime}\left( re^{it}\right) re^{it}}{p\left( re^{it}\right) }\operatorname*{dt}=n. $$ Just a little problem I have to ask you: here $r\in% \mathbb{R} _{+},$ so $f_{r}$ is a kind of generalised sequence and in Lebesgue theorem $f_{n}$ is defined with $n\in% \mathbb{N} $; it is correct to apply this theorem in that case?