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For parameter four, the SLE4 angle martingale is bounded, so the Continuous-time martingale convergence theorem gives an almost sure limit. The imaginary part of the Chordal Loewner equation givesFor a point off the full simple Loewner trace the flow exists at every finite time. If the limiting angle lay strictly between $0$ and $\pi$, the last derivative would eventually be bounded above by a strictly negative constant. That would make $Y_t^2$ negative, a contradiction. Hence the terminal angle is either zero or $\pi$.

The simple transient chordal Loewner trace from $0$ to infinity splits the complex upper half-plane into a left component $D^{-}$, adjacent to the negative real boundary, and a right component $D^{+}$, adjacent to the positive real boundary. The Dirichlet boundary values of the SLE angle process identify which terminal value occurs. Indeed, run an independent planar Brownian motion from $z$ until it first hits the full Loewner trace or the real axis. This time is finite. The Brownian path up to that time is bounded, and Transience of chordal SLE ensures that any Loewner trace point it meets belongs to a finite initial segment. Thus its exit side for the truncated domains eventually agrees with its exit side for the full component. In $D^{-}$ every such exit is through the left bank or negative real boundary; in $D^{+}$ every exit is through the right bank or positive real boundary. Dominated convergence in the harmonic-measure representation therefore givesThis is a random side indicator, not the deterministic value $\arg z$. Uniform integrability also gives $\mathbb{E}[F(z)]=\arg z$, so the SLE4 left-passage probability is

Use the Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-U_t)$ with $U_t=\sqrt{\kappa }{B}_t$. For a fixed $z\in \mathbb{H}$, let $T_z$ be its Loewner swallowing time, and write $Z_t=X_t+i{Y}_t=g_t(z)-U_t$ and $J_t=|g_t^{\prime }(z)|$ before that time. The Loewner conformal radius is ${{\rm Y}}_t=Y_t/J_t$, half of the conformal radius of $D_t=\mathbb{H}\setminus K_t$ at $z$. The Koebe quarter theorem bounds the conformal radius above by four times the distance to the boundary. For the reverse comparison, if $\varphi :\mathbb{D}\to D_t$ maps zero to $z$, apply the Schwarz lemma to ${\varphi }^{-1}(z+dw)$ on the unit disc, where $d=\mathrm{dist}(z,\partial D_t)$; it gives $d\le |{\varphi }^{\prime }(0)|$. ThusWe use the basic trace theorems that, for $\kappa >8$, the Loewner trace is continuous and Transience of chordal SLE gives $|\gamma (t)|\to \infty$. These facts make its image relatively closed in $\mathbb{H}$; they do not assume the space-filling conclusion. The continuity and transience statements are available in Rohde and Schramm's basic trace theorems.

Take $\rho =\kappa -8>0$ in the supplied SLE interior-point martingale. The derivative exponent vanishes, leavingA nonnegative local martingale is a supermartingale. Applying the optional stopping theorem after localization at its first hit of a level $R$ gives the maximal inequality for a nonnegative supermartingaleThus $C={\sup }_{t<T_z}{M}_t<\infty$ almost surely.

Suppose the Loewner trace avoids some open ball $B(z,r)$ whose closure is inside $\mathbb{H}$. Before $T_z$, the whole open ball is in $D_t$: a connected open ball disjoint from the trace cannot be partly in the unbounded connected component. Hence ${{\rm Y}}_t\ge r/2$. The bound on $M$ forces $S_t\ge c>0$ throughout this interval, for a positive random constant $c$.

The Chordal Loewner equation and its derivative yieldThe first identity forces $T_z<\infty$, since otherwise $Y_t^2\le Y_0^2-4c^{2}t$ becomes negative. At a finite maximal lifetime, $Y_t\to 0$; otherwise the continuous Loewner driving function and the ordinary differential equation continue past that time. Integrating the last identity, using $S_t\ge c$, givescontradicting ${{\rm Y}}_t\ge r/2$. Thus every fixed rational ball inside $\mathbb{H}$ is hit almost surely. A countable intersection makes the trace a dense subset almost surely, and its relatively closed image, established by continuity and transience, then contains all of $\mathbb{H}$:This proves space-filling SLE above parameter eight. It rules out unvisited open regions, rather than inferring visits merely from membership in the filled compact H-hulls. The choice $\rho =\kappa -8$ does not address the critical value eight.

The Scaling invariance of SLE says that $r^{-1}\gamma (r^{2}t)$ has the same law as $\gamma (t)$. Therefore its first time reaching height one is $r^{-2}{\sigma }_r$, andfor every $r>0$.

To identify $p$, it suffices to look at small heights. The trace must have a point of positive imaginary part: otherwise it lies entirely on the real axis and its hull in $\mathbb{H}$ is empty, contradicting $\mathrm{hcap}(A_t)=2t>0$. By continuity from $\gamma (0)=0$, any positive height attained by the trace forces it to pass through every smaller positive height.

Consequently the increasing union of events $\{{\sigma }_{1/n}<\infty \}$ has probability one. Each event has probability $p$, so continuity of probability gives $p=1$. This proves that SLE reaches every positive height:Equivalently, a positive random height supremum whose law is invariant under every dilation must be infinite almost surely. This argument uses neither Transience of chordal SLE nor any assertion about where the trace goes as time tends to infinity.

Fix $R>0$. Part i, with $ϵ$ arbitrarily small and $t$ sufficiently large, shows thatThe hulls increase with time. Once this half-disc lies in $K_t$, the stated future-avoidance property givesIntersecting these probability-one events over positive integer $R$ provesalmost surely. This is Transience of chordal SLE.

For a chordal SLE trace $\gamma$, the Scaling invariance of SLE makes $\mathbb{P}({\sup }_t\mathrm{Im}\gamma (t)>r)$ independent of $r>0$. The trace must enter the upper half-plane because its positive-time hull has positive half-plane capacity. Letting $r\downarrow 0$ therefore makes this probability one, and continuity implies that every positive height is reached. This argument does not assume Transience of chordal SLE.

For $\kappa >8$, the Loewner trace visits every point of the complex upper half-plane almost surely. Choose $\rho =\kappa -8$ in the SLE interior-point martingale to obtain $M_t={{\rm Y}}_t^{(\kappa -8)/8}{S}_t^{-(\kappa -8)/\kappa }$. The maximal inequality for a nonnegative supermartingale bounds its running supremum. If a fixed interior ball were avoided, its center would have ${{\rm Y}}_t\ge c_1>0$ before swallowing, forcing $S_t\ge c_2>0$. But the Chordal Loewner equation gives $d{Y}_t^2/dt=-4S_t^2$, forcing a finite lifetime with $Y_t\to 0$, and $d\log {{\rm Y}}_t/d\log Y_t=2S_t^2$ then forces ${{\rm Y}}_t\to 0$, a contradiction. Countably many rational balls give a dense trace; continuity and Transience of chordal SLE make its image relatively closed, so it is all of the complex upper half-plane. This argument does not cover the critical parameter eight.