Existence of the limit of an integral

Question

This is a question related to the problem I asked in Show that $|f'(z)|>\frac{n_k}{10k}$ for $1-\frac{1}{n_k}<|z|<1-\frac{1}{2n_k}$.. Suppose that \begin{equation} f(z)=\sum_{k=1}^n \frac{z^{n_k}}{k}, \end{equation} where $\{n_k\}$ is a sequence of positive integers with $n_k>e^{n_{k-1}}$ for every $k \ge 2$. Does the limit \begin{equation} \lim_{R \to 1} \int_0^R f'(re^{it}) dr \end{equation} exist for almost all $t$ on $[-\pi,\pi]$? My idea is that \begin{equation} f(re^{it})=\sum_{k=1}^\infty \frac{n_k r^{n_k-1}}{k}e^{i(n_k-1)t} \end{equation} implies \begin{equation} \int_0^R f'(re^{it}) dr=\sum_{k=1}^\infty \frac{n_k}{k} \Big[\int_0^R r^{n_k-1} dr\Big] e^{i(n_k-1)t}=e^{-it}f(Re^{it}), \end{equation} so it seems that the limit exist $\textit{for all}$ $t$. However, I feel that the question is not so easy. What mistake(s) have I made?

Answer 1

I think I have got a solution. In fact, since $\sum \frac{1}{k^2}<\infty$, $f \in H^2$ and thus the nontangential limits $f^\ast(e^{it})$ exists a.e. on $T$. Hence we are done.