KL is invariant under a measurable bijection

If $P\ll ̸Q$, choose E_0 with $Q(E_0)=0<P(E_0)$. Since ${\varphi }^{-1}$ is measurable, $\varphi (E_0)=({\varphi }^{-1}{)}^{-1}(E_0)$ is measurable; then ${\varphi }_{\mathrm{\#}}Q(\varphi (E_0))=0<{\varphi }_{\mathrm{\#}}P(\varphi (E_0))$. Both divergences are $+\mathrm{\infty }$ by [F1].

If $P\ll Q$, let $r=dP\mathrm{/}dQ$. [given] The function $s=r\circ {\varphi }^{-1}$ is measurable. For every B in F, ${\int }_{B}sd({\varphi }_{\mathrm{\#}}Q)={\int }_{{\varphi }^{-1}(B)}rdQ=P({\varphi }^{-1}(B))={\varphi }_{\mathrm{\#}}P(B),$ so s is a Radon–Nikodym density of ${\varphi }_{\mathrm{\#}}P$ with respect to ${\varphi }_{\mathrm{\#}}Q$. By [F2], the push-forward integral identity applies separately to the nonnegative functions $(\ln s{)}^{+}$ and $(\ln s{)}^{-}$; the negative-part integrals are finite by [F1]. Therefore $D({\varphi }_{\mathrm{\#}}P\mathrm{\parallel }{\varphi }_{\mathrm{\#}}Q)=\int (\ln s{)}^{+}d({\varphi }_{\mathrm{\#}}P)-\int (\ln s{)}^{-}d({\varphi }_{\mathrm{\#}}P)=\int (\ln r{)}^{+}dP-\int (\ln r{)}^{-}dP=D(P\mathrm{\parallel }Q),$ including the case where the positive-part integral is $+\mathrm{\infty }$.