= Solution
Before the <Loewner swallowing time> $T_z$, put $Z_t=g_t(z)-U_t=X_t+iY_t$ and $\theta_t=\arg Z_t\in(0,\pi)$. The <Itô formula> applied to the <holomorphic logarithm> in the <complex upper half-plane> gives
$$
d\log Z_t=\frac{2-\kappa/2}{Z_t^2}\,dt-\frac{\sqrt\kappa}{Z_t}\,dB_t.
$$
Taking imaginary parts yields the <SLE angle process>
$$
d\theta_t=(\kappa-4)\frac{X_tY_t}{|Z_t|^4}\,dt+\sqrt\kappa\frac{Y_t}{|Z_t|^2}\,dB_t.
$$
For $\kappa=4$ the drift vanishes, so
$$
\boxed{\theta_t=\arg z+2\int_0^t\frac{Y_s}{|Z_s|^2}\,dB_s}
$$
is 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>. But
$$
\log\frac{\Upsilon_t}{\Upsilon_0}=-4\int_0^t\frac{Y_s^2}{|Z_s|^4}\,ds=-[\theta]_t.
$$
The <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 $\Upsilon_t$ 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: \b[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.
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