= Solution
The <Scaling invariance of SLE> says that $r^{-1}\gamma(r^2t)$ has the same law as $\gamma(t)$. Therefore its first time reaching height one is $r^{-2}\sigma_r$, and
$$
\mathbb P(\sigma_r<\infty)=\mathbb P(\sigma_1<\infty)=:p
$$
for 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 $\operatorname{hcap}(A_t)=2t>0$. 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>:
$$
\boxed{\mathbb P(\sigma_r<\infty)=1\qquad(r>0).}
$$
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.
Back to article page
