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$F(z)$ 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 $C_n(\theta)=S_\alpha(\theta)\cap A_n$, where $A_n=\{z:1-1/n\le|z|\le 1-1/(n+1)\}$. 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.

$F(z)$ 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.

$F(z)$ 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 $C_n(\theta)=S_\alpha(\theta)\cap A_n$, where $A_n=\{z:1-1/n\le|z|\le 1-1/(n+1)\}$. Notice that $F$ is uniformly continuous on $A_n$. By definition, $G_n(\theta)=\max_{z\in C_n(\theta)}F(z)$ and $g_n(\theta)=\min_{z\in C_n(\theta)}F(z)$. Suppose that

$F(z)$ 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 $C_n(\theta)=S_\alpha(\theta)\cap A_n$, where $A_n=\{z:1-1/n\le|z|\le 1-1/(n+1)\}$. 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.

$F(z)$ 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.

$F(z)$ 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 $C_n(\theta)=S_\alpha(\theta)\cap A_n$, where $A_n=\{z:1-1/n\le|z|\le 1-1/(n+1)\}$. Notice that $F$ is uniformly continuous on $A_n$. By definition, $G_n(\theta)=\max_{z\in C_n(\theta)}F(z)$ and $g_n(\theta)=\min_{z\in C_n(\theta)}F(z)$. Suppose that