 Solution

Before the Loewner swallowing time $T_z$, put $Z_t=g_t(z)-U_t=X_t+i{Y}_t$ and ${\theta }_t=\arg Z_t\in (0,\pi )$. The Itô formula applied to the holomorphic logarithm in the complex upper half-plane givesTaking imaginary parts yields the SLE angle processFor $\kappa =4$ the drift vanishes, sois a continuous local martingale. Since $0<{\theta }_t<\pi$, every localized stopped version is a bounded true martingale. After resolving the lifetime issue below, the dominated convergence theorem applied to conditional expectations removes localization and gives a true bounded martingale.

For completeness, there is no finite lifetime ambiguity for a fixed interior $z$ in this parameter range. The preceding Bessel process and Conformal Markov property of SLE argument gives a simple trace staying in the interior, so no point is swallowed in a disconnected pocket. If $T_z$ were finite, the trace would have to reach $z$, forcing the Loewner conformal radius to tend to zero by the Koebe quarter theorem. ButThe Dambis-Dubins-Schwarz theorem says that a bounded continuous local martingale cannot have infinite quadratic variation before a finite terminal time: that would require a Brownian motion to stay in a bounded interval for all clock times. Therefore ${{\rm Y}}_t$ cannot tend to zero at a finite $T_z$, a contradiction. Thus $T_z=\infty$ almost surely for each fixed $z$. This proves the SLE4 angle martingale assertion: the angle is a bounded continuous martingale. In particular $\mathbb{E}[{\theta }_t]=\arg z$. The upper-half-plane branch of the argument is used throughout.