Let $F$ be holomorphic in the strip $S := \{ z \in \mathbb{C} : 0 < \mathrm{Re} z < 1\}$ and continuous on its closure $\overline{S}$. Further define
$\displaystyle h(z) := \frac{1}{\pi i} \log \left( i \frac{1 + z}{1 - z}\right)$
for $\vert z \vert \leqslant 1$ and $z \neq \pm 1$ ($\log$ denotes the branch of the logarithm, where $\log 1 = 0$). Why exactly is $\log \vert F \circ h\vert$ upper semicontinuous for $\vert z \vert \leqslant 1$ and $z \neq \pm 1$?
It is, because we can choose a continuous branch of the logarithm and since the composition of a upper semicontinuous function with a continuous function is upper semicontinuous, we are done.