If \(X\sim P\) then for any distribution \(Q\) it is unlikely that \(Q\) ascribes much greater density to \(X\)'s outcome than \(P\) does. In fact if \(P,Q\) have PDFs \(f_P, f_Q\), then:

\begin{align}
\mathbb{P}(f_P(X)\leq c f_Q(X)) &= \int \mathbf{1}_{\{x:f_P(x)\leq cf_Q(x) \}} f_P(x)dx \\
&\leq \int c f_Q(x) dx \\
&= c.
\end{align}

This carries over to relative entropy:

Noting \(D_{KL}(P\|Q)=\mathbb{E}[Z]\) for \(Z=\ln \frac{f_P(X)}{f_Q(X)}\), then for any \(z,\)

\begin{align}
\mathbb{P}(Z\leq z) &= \mathbb{P}\left(\ln \frac{f_P(X)}{f_Q(X)} \leq z \right) \\
&= \mathbb{P}(f_P(X)\leq e^z f_Q(X)) \\
&\leq e^z.
\end{align}

This is actually just an interesting instance of the Chernoff bound. The same thing can be done when \(P,Q\) aren't over \(\mathbb{R}\) or don't have CDFs, or even with other types of divergences.