if P and Q are probability measures over a set X, and P is absolutely continuous with respect to Q,then set $$ D_{\mathrm{KL}}(P\|Q) = \int_X \ln\frac{{\rm d}P}{{\rm d}Q} \, {\rm d}P, $$
set $$D_{\mathrm{TV}}(P\|Q) =2\sup\{|P(E)-Q(E)|\big| E\in \mathcal{F}\}$$ where $\mathcal{F}$ is a $\sigma-$algebra of $X$
the Pinsker theorem said: $$D_{TV}\le\sqrt{\frac{D_{KL}}{2}}$$
but the reverse is not true:
for $\forall\varepsilon>0$,there exist $P,Q$ such that $$D_{TV}<\varepsilon,D_{KL}=\infty$$
Could you show me such an example?
Choose $P,Q$ such that $D_{TV} < \epsilon$ but the support of $P$ - support of $Q$ has positive measure under $P$. Then $D(P||Q) = \infty$.
You can do this with the counting measure and $P$ being $0$ with probabilty $1-\epsilon$ and $1$ with probability $\epsilon$, $Q$ being $0$ with probability $1-\epsilon$ and $2$ with probability $\epsilon$. The total variation distance is $\epsilon$, but the KL divergence is infinite.