Consider an upper semicontinuous function $\phi: \Omega \to (-\infty, \infty]$, in the sense that $\phi = \phi^*$, where $\phi^*$ denotes the upper semicontinuous regularization $$ \phi^*(z) = \varlimsup_{\zeta \to z} \phi(\zeta) $$ and $\Omega$ is some domain in $\mathbb{C}^n$, let's say strictly pseudoconvex if it helps.

Forming the usual Perron-Bremermann envelope of $\phi$, i.e. $$ U(z) = \sup\{ u(z) : u \in \mathcal{PSH}(\Omega), u \le \phi \}, $$ it is clear (assuming that $U$ is locally bounded) that $U^* \le \phi^* = \phi$, so $U^*$ is a member in the defining family for $U$, and thus $U \ge U^*$ and consequently, $U = U^*$. In particular, $U$ is already upper semi-continuous, and thus plurisubharmonic without having to take the upper semicontinuous regularization.

Now, let us consider the following variation, where we take the envelope only using upper bounded functions: $$ V(z) = \sup\{ u(z) : u \in \mathcal{PSH}(\Omega), u \le \phi, \sup_\Omega u < +\infty \}. $$ Note that in general, $V < U$ (take for example $\phi$ as the Poisson kernel with a pole at $z=1$ on the unit disc: then $U = \phi$, but $V = 0$).

**Question:** Is it true that $V$ is upper semicontinuous and thus plurisubharmonic? The problem being, of course, that $V$ is in general not upper bounded, so it doesn't follow immediately that $V^*$ is a member of the defining family for $V$.