 Solution

Write $A=r^2+a^2$, $s=\sin \theta$, $\Sigma =A-a^2{s}^2$, and $\Delta =A-2Mr$. Use ingoing Kerr coordinates, so $dt=dv-Adr/\Delta$ and $d\varphi =d\chi -adr/\Delta$. To see the cancellations without expanding every term, write the Kerr metric in the equivalent formThe combinations becomeThe first square contributes $-\Sigma d{r}^2/\Delta$, cancelling the explicit radial term. Expanding the remaining terms givesThere is no denominator $\Delta$ in this Lorentzian metric. At the outer horizon $r_{+}=M+\sqrt{M^2-a^2}$, $\Sigma >0$ and the components are smooth. The determinant is $-{\Sigma }^2{\sin }^2\theta$, so away from the usual polar-coordinate degeneracy the metric is nondegenerate and extends across $r_{+}$. The axis can be covered by regular angular charts. Thus the Boyer-Lindquist coordinates are singular there, while the ingoing Kerr coordinates are regular at the future horizon.

For physical nonextremality the invariant parameter condition is $M>|a|$, $M>0$. The printed $M>a$ is sufficient when the rotation orientation has been chosen so that $a\ge 0$; without that convention it needs the absolute value.