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 $\frac{d\mu}{d\nu} \in L^{\infty}\_{\nu,\mathrm{loc}}(X)$?



When is the Radon-Nikodym derivative locally essentially bounded
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. 




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Similarly, for any compact $K\subseteq X$, 
$$\operatorname{esssup}\_Kf>0\iff \text{ $\exists$ $b\_K\in(0,\infty)$ $\exists$ Borel $A$ such that $\mu(A\cap K)\ge b\_K\nu(A\cap K)$.}$$


