Let $\mathbb T$ be the unit circle and suppose that $f\in L^1(\mathbb T)$ is realvalued. Then its Poisson integral $F=P[f]$ is realvalued, too. Let $$Osz[f](e^{i\theta}):=\limsup_{z\to e^{i\theta}\atop z\in S_\alpha(\theta)} F(z) \liminf_{z\to e^{i\theta}\atop z\in S_\alpha(\theta)} F(z)$$ be the oscillation of $F$ in the cone ${S_\alpha(\theta):=\{z\in \mathbb D: \arg(1e^{i\theta}z)<\alpha\}}$, $0<\alpha<\pi/2$. By The HardyLittlewood maximality theorem, $Osz[f]$ is welldefined and finite a.e. (for details, one may see the book "Bounded analytic functions" by J. B.Garnett). Why $Osz[f]$ is measurable? Note that, in general, the supremum over an uncountable family of measurable functions is not measurable, in general.
2 Answers
$F(z)$ is harmonic and therefore continuous inside the unit disk. This means that the maximum for points in $S_\alpha(\theta)$ with absolute value between $11/n$ and $11/(n+1)$ is a continuous function of $\theta$. Call this $G_n(\theta)$. Similarly, define $g_n(\theta)$ to be the minimum. The limits superior and inferior that you want are the limits superior and inferior of $G_n$ and $g_n$ respectively, which are measurable.
Some additional details (responding to a request for clarification from the OP):
Set $C_n(\theta)=S_\alpha(\theta)\cap A_n\cap\{z:ze^{i\theta}< 1/\sqrt n\}$, where $A_n=\{z:11/n\lez\le 11/(n+1)\}$. (The point of the $1/\sqrt n$ is to ensure that you don’t get point from the opposite side of the disk).
Notice that $F$ is uniformly continuous on $A_n$. By definition, $G_n(\theta)=\max_{z\in C_n(\theta)}F(z)$ and $g_n(\theta)=\min_{z\in C_n(\theta)}F(z)$. Let $\epsilon>0$ and choose $\delta$ sufficiently small that if $z$ and $z’$ are within $\delta$, then $F(z)F(z’)<\epsilon$.
Now suppose $G_n(\theta)=F(z)$ with $z\in C_n(\theta)$. Then if $\theta\theta’<\delta$, there exists a point $z’$ of $C_n(\theta’)$ within $\delta$ of $z$. Hence $G_n(\theta’)\ge G_n(\theta)\epsilon$. By symmetry, $G_n(\theta)\ge G_n(\theta’)\epsilon$ also, so we have shown if $\theta\theta’<\delta$, then $G_n(\theta)G_n(\theta’)<\epsilon$. Since $\epsilon$ was chosen arbitrarily, we have shown that $G_n$ is uniformly continuous.

$\begingroup$ Could you please give us some hints why this maximum function is continuous? (there may be whole arcs on where the maximum is taken). $\endgroup$– raySep 27, 2019 at 18:31

$\begingroup$ I am not yet convinced. In fact, since the argument of $z$ may be far away from $\theta$, I do not yet see why this $z'$ in the other cone and close to $z$ should exist. The whole must somehow use the relation $\arg z\theta< C (1z)$ between the argument of $z$ and $\theta$ (so also with $\theta'$) which comes from the hypothesis that $z$ belongs to a cone with endpoint $e^{i\theta}$. $\endgroup$– raySep 28, 2019 at 11:12

1
May be one could proceed as follows:
It obviously suffices to prove that
$$V:=\{\theta\in\mathbb R: \hspace{3mm} \limsup_{z\to e^{i\theta},~z\in S_\alpha(\theta)} F(z)>\eta\}$$ and
$$\{\theta\in\mathbb R:\hspace{3mm} \liminf_{z\to e^{i\theta}~ z\in S_\alpha(\theta)} F(z)<\eta\}$$
are open sets in $\mathbb R$ for each $\eta\in \mathbb R$. Note that
there is a constant $k_\alpha>1$ such that
for every $z=re^{it}\in S_\alpha(\theta)$, we have $\mu:=t\theta<k_\alpha(1r)$. Hence,
if $\theta\in V$ and $F(re^{it})>\eta$ for some $r\geq r_0$, we choose $\theta'$ so close to $\theta$ that
$$t\theta'\leq t\theta+\theta\theta'<\mu+\delta<k_\alpha(1r).$$
Hence $re^{it}\in S_\alpha(\theta')$ and so
$$\limsup_{z\to e^{i\theta},~z\in S_\alpha(\theta')} F(z)\geq F(re^{it})>\eta.$$
We conclude that ${\theta'\in V}$.