Measurability of the angular limit function

Question

Let $\mathbb T$ be the unit circle and suppose that $f\in L^1(\mathbb T)$ is real-valued. Then its Poisson integral $F=P[f]$ is real-valued, 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 ${S_\alpha(\theta):=\{z\in \mathbb D: |\arg(1-e^{-i\theta}z)|<\alpha\}}$, $0<\alpha<\pi/2$. By The Hardy-Littlewood maximality theorem, $Osz[f]$ is well-defined 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.

Answer 1

$F(z)$ is harmonic and therefore continuous inside the unit disk. This means that the maximum for points in $S_\alpha(\theta)$ with absolute value between $1-1/n$ and $1-1/(n+1)$ 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 $C_n(\theta)=S_\alpha(\theta)\cap A_n\cap\{z:|z-e^{i\theta}|< 1/\sqrt n\}$, where $A_n=\{z:1-1/n\le|z|\le 1-1/(n+1)\}$. (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.

Answer 2

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 $\mu:=|t-\theta|<k_\alpha(1-|r|)$. Hence, if $\theta\in V$ and $F(re^{it})>\eta$ for some $r\geq r_0$, we choose $\theta'$ so close to $\theta$ that $$|t-\theta'|\leq |t-\theta|+|\theta-\theta'|<\mu+\delta<k_\alpha(1-|r|).$$ 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 ${\theta'\in V}$.