Let $f$ be a holomorphic function function on the open upper half of the complex plane ($\{\lambda = a + ib \in \mathbb{C} : b > 0 \}$) that takes values in the closed upper half plane (such a function is called a Herglotz function). Define the set $M = \{\lambda \in \mathbb{R} : \limsup_{\varepsilon \searrow 0} \text{Im}\, f(\lambda + i \epsilon) = \infty\}$.
Is it true that $M$ is closed?
One fact that I know to be true is that the (finite or infinite) limit $\lim_{\varepsilon \searrow 0} \text{Im}\,f(\lambda + i \varepsilon)$ exists for almost all $\lambda \in \mathbb{R}$ with respect to Lebesgue measure. So far I have tried to prove the above by naive argument: if $M \ni \lambda_k$ converge to $\lambda$, look at: $$ \text{Im}\,f(\lambda + i \varepsilon) \ge - |\text{Im}\,f(\lambda + i \varepsilon) - \text{Im}\,f(\lambda_k + i \varepsilon)|+ \text{Im}\,f(\lambda_k + i \varepsilon). $$ Using this lower bound I want to find a sequence of epsilon's $\varepsilon_j \searrow 0$ so that $\sup_{\varepsilon \in (0, \varepsilon_j)} \text{Im}\,f(\lambda + i \varepsilon) \ge j$. Here is where I am stuck (I can't get the $\lambda_k$ and $\varepsilon$ to "play nicely" with each other on the right hand side). Hint or solutions are greatly appreciated.