# Measurability of the angular limit function

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.

$$F(z)$$ 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.

• Could you please give us some hints why this maximum function is continuous? (there may be whole arcs on where the maximum is taken). – ray Sep 27 '19 at 18:31
• 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 $|\arg z-\theta|< C (1-|z|)$ between the argument of $z$ and $\theta$ (so also with $\theta'$) which comes from the hypothesis that $z$ belongs to a cone with endpoint $e^{i\theta}$. – ray Sep 28 '19 at 11:12
• Take $z’=e^{i(\theta’-\theta)z}$ – Anthony Quas Sep 28 '19 at 18:20

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 $$\mu:=|t-\theta|. Hence, if $$\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 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 $${\theta'\in V}$$.