Let $v \in \mathbb{S}^{n-1}$ be a unit vector. Given $r>0$, Let $B_r(x)$ be the ball of radius $r$ around $x \in \mathbb{R}^n$, and define
$$B_r^+(x,v)=\{ y \in B_r(x) \, | \, \langle y-x,v\rangle >0\},$$ and $$B_r^-(x,v)=\{ y \in B_r(x) \, | \, \langle y-x,v\rangle <0 \}. $$
Now, let $\Omega \subseteq \mathbb{R}^n$ be an open subset, and let $u \in L^1_{\text{loc}}(\Omega)$.
Let $a \neq b \in \mathbb{R}$, and suppose that
$$ \lim_{r \to 0} \frac{1}{\mu(B_r^+(x,v))}\int_{B_r^+(x,v)} |u(y)-a|dy=0, $$ and $$ \lim_{r \to 0} \frac{1}{\mu(B_r^-(x,v))}\int_{B_r^-(x,v)} |u(y)-b|dy=0. $$
Question: How to prove that the triple $(a,b,v)$ is uniquely determined up to a permutation of $a$ and $b$, and a change of sign of $v$?
(I stumbled on this fact in a geometric measure theory book- it seems rather plausible, but I don't see an immediate way to prove it).
Up to translation and rescaling, this is a claim about a sequence of functions in $B(0,1)$. To wit, define $u_{r} : B(0,1) \to \mathbb{R}$ by $u_{r}(\xi) = u(x + r \xi)$ to find that \begin{equation*} \lim_{r \to 0^{+}} \int_{B^{+}_{1}(0,v)} |u_{r}(\xi) - a| \, d \xi = 0, \quad \lim_{r \to 0^{-}} \int_{B^{-}_{1}(0,v)} |u_{r}(\xi) - b| \, d \xi = 0. \end{equation*} Now that the domain is no longer moving, this is a question about $L^{1}$-convergence of functions in $B(0,1)$. In fact, notice that \begin{equation*} \lim_{r \to 0^{+}} \int_{B_{1}(0)} |u_{r}(\xi) - a \chi_{B_{1}^{+}(0,v)} - b \chi_{B_{1}^{+}(0,v)}| \, d\xi = 0. \end{equation*} Since limits in $L^{1}(B(0,1))$ are unique and $a \chi_{B_{1}^{+}(0,v)} + b \chi_{B_{1}^{+}(0,v)}$ uniquely determines $a$, $b$, and $v$ (up to the permutation $(a,b,v) \mapsto (b,a,-v)$), we are done.