Let $D$ be a domain and $u\colon D\rightarrow[-\infty,+\infty)$ be a locally integrable, locally bounded above function. Let us suppose that $u$ satisfies the local sub-mean inequality, that is for every $z\in D$ there exists an $\rho>0$, such that $$u(z)\leq \frac{1}{2\pi}\int_{0}^{2\pi}u(z+re^{i\theta})\,d\theta\qquad(0\leq r<\rho).$$ We assume that u is not necessarily upper semi continuous, so $u$ is not subharmonic.
However, is it true that $$\lim_{r\to 0}\frac{1}{2\pi}\int_{0}^{2\pi}u(z+re^{i\theta})\,d\theta= u^*(z)\,,$$ where $$u^*(z)\doteqdot\lim_{s\to 0}\left(\sup_{|w-z|<s}u(w)\right)$$ is the upper semi regularization of $u$?