Let $\mu\lll\nu$ be $\sigma$-finite Borel measures, which are not finite, on a topological space $X$. Under what conditions is $0<\operatorname{ess-supp}(\frac{d\mu}{d\nu}I_K)<\infty$ for every compact subset $\emptyset\subset K\subseteq X$.
In other words when is $\frac{d\mu}{d\nu} \in L^{\infty}_{\nu,\mathrm{loc}}(X)$?
Let $$f:=\frac{d\mu}{d\nu}.$$ Then $$f\in L^{\infty}_{\nu,loc}(X)\iff\text{$\forall$ compact $K\subseteq X$ $\exists$ $c_K\in(0,\infty)$ $\forall$ Borel $A$ we have $\mu(A\cap K)\le c_K\nu(A\cap K)$.}$$
Indeed, for the $\Rightarrow$ implication, take any compact $K\subseteq X$. Then $\exists$ $c_K\in(0,\infty)$ such that $f\le c_K$ $\nu$-a.e. on $K$. So, for any Borel $A$ we have $$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\le c_K\nu(A\cap K),$$ as desired.
Vice versa, for the $\Leftarrow$ implication, take any compact $K\subseteq X$ and suppose that $\mu(A\cap K)\le c_K\nu(A\cap K)$ for some $c_K\in(0,\infty)$ and all Borel $A$. Let now $A:=f^{-1}((c_K,\infty))$, so that $f>c_K$ on $A$. Then $$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\ge c_K\nu(A\cap K),$$ and the latter inequality is strict (and hence contradicts condition $\mu(A\cap K)\le c_K\nu(A\cap K)$) if $\nu(A\cap K)>0$. So, $\nu(A\cap K)=0$, that is, $f\le c_K$ $\nu$-a.e. on $K$, as desired.
Similarly, for any compact $K\subseteq X$, $$\operatorname{esssup}_Kf>0\iff \text{ $\exists$ $b_K\in(0,\infty)$ $\exists$ Borel $A$ such that $\mu(A\cap K)\ge b_K\nu(A\cap K)$.}$$
