= Solution
A regular <hypersurface> $f=0$, with $df\ne0$, is a <null hypersurface> if its normal <covector> has zero squared <norm>:
$$
g^{ab}(\partial_af)(\partial_bf)=0\quad\text{on }f=0.
$$
Its normal <vector> is then also tangent to the <hypersurface>; the induced <metric tensor> is degenerate.
For the <Kerr metric> the normal to a constant-$r$ surface has squared <norm>
$$
g^{rr}=\frac{\Delta}{\Sigma},\qquad
\Delta=(r-r_+)(r-r_-),\qquad
r_\pm=M\pm\sqrt{M^2-a^2}.
$$
For $0<a<M$ the two roots are positive and distinct. To verify that they are genuine regular <null hypersurfaces>, rather than trusting singular <Boyer-Lindquist coordinates>, introduce <ingoing Kerr coordinates>
$$
dv=dt+\frac{r^2+a^2}{\Delta}\,dr,\qquad
d\psi=d\phi+\frac a\Delta\,dr.
$$
Substitution gives
$$
\begin{aligned}
ds^2={}&-\left(1-\frac{2Mr}{\Sigma}\right)dv^2
+2\,dv\,dr+\Sigma\,d\theta^2
-2a\sin^2\theta\,dr\,d\psi\\
&-\frac{4Mar\sin^2\theta}{\Sigma}\,dv\,d\psi
+\sin^2\theta\left(r^2+a^2+
\frac{2Ma^2r\sin^2\theta}{\Sigma}\right)d\psi^2.
\end{aligned}
$$
There are no $\Delta^{-1}$ terms. At either root $\Sigma>0$, and the <determinant> is $-\Sigma^2\sin^2\theta$ away from the angular coordinate poles. Thus the <metric tensor> extends smoothly across each root. The raised normal on either surface is
$$
\left.(d r)^\sharp\right|_{r_\pm}
=\frac{r_\pm^2+a^2}{\Sigma}
\left(\partial_v+\Omega_\pm\partial_\psi\right),
\qquad
\Omega_\pm=\frac{a}{r_\pm^2+a^2}.
$$
This is a nonzero null <Killing vector field> normal to the surface. Both surfaces are <Killing horizons>, with <horizon angular velocities> $\Omega_\pm$.
\b[The event-horizon terminology in the paper needs qualification.] In the usual asymptotically flat <Kerr black hole> extension, relative to a chosen exterior's <future null infinity>, the future branch of $r=r_+$ is the <event horizon>, the boundary $\partial J^-(\mathscr I^+)$. In the intervening black-hole region $\Delta<0$, the coordinate $r$ has timelike gradient and future-directed curves move toward smaller $r$; they cannot return across the outer boundary to that exterior. The inner surface $r=r_-$ is a <Cauchy horizon>, not a second <event horizon> for the same asymptotic end. Nullness is a local property; identifying an <event horizon> additionally requires this global <causal structure>.<General Relativity by David Tong>[Sections 6.3.2 and 6.2.4: regular extension across the two Kerr roots, global diagram and the inner Cauchy horizon]{ref}
The allowed endpoint $a=0$ provides an especially concrete counterexample to the unrestricted description of both roots. The <Kerr metric> then becomes <Schwarzschild spacetime>, $r_+=2M$ is its regular <event horizon>, but $r_-=0$ is a <curvature singularity>:
$$
R_{abcd}R^{abcd}=\frac{48M^2}{r^6}\longrightarrow\infty.
$$
It is not a regular inner <null hypersurface>. Thus the two-horizon calculation above assumes $0<a<M$.
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