# When is the Radon-Nikodym derivative locally essentially bounded

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

• My google ninja's skill got me the following: math.stackexchange.com/questions/3119941/… is this helpful? – Alan Apr 30 at 18:26
• @Alan It is pretty interesting, but I don't want to assume any such geometric structure. Btw, I like the ninja skills terminology I'll have to borrow it :) – AnnieLeKatsu Apr 30 at 18:29
• I got this terminology from someone else in physicsforums, can't remember from whom though. :-D – Alan Apr 30 at 18:33
• What type of conditions do you need? – Fedor Petrov Apr 30 at 19:04
• in fact $\| \frac{d\mu}{d\nu}\|_{\infty,K}=\sup_{H\subset K} \frac{\mu(H)}{\nu(H)}$ – Pietro Majer Apr 30 at 19:33

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