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Let $(E,d)$ be a metric space, $\mu$ be a nonnegative Radon$^1$ measure on $\mathcal B(E)$ and $\nu$ be a finite (signed) Radon measure on $\mathcal B(E)$.

I'm searching for a Radon-Nikodým-like theorem yielding a sufficient condition to ensure that there is a locally $\mu$-integrable $f:E\to\mathbb R$ with $$\int g\:{\rm d}\nu=\int fg\:{\rm d}\mu\;\;\;\text{for all }g\in C_c(E)\tag1.$$

Context: Given $p\in[1,\infty)$ and $q\in(1,\infty]$ with $p^{-1}+q^{-1}=1$, I would like to prove that $L^p(\mu)'$ is isometrically isomorphic to $L^q(\mu)$ by the following argumentation: If $\mathcal M(E)$ denotes the space of finite (signed) measures on $\mathcal B(E)$, we know that $$\langle f,\nu\rangle:=\int f\:{\rm d}\nu\;\;\;\text{for }(f,\nu)\in C_b(E)\times\mathcal M(E)$$ is a duality pairing. Given a nonnegative Radon measure $\mu$ on $\mathcal B(E)$ and $\varphi\in L^p(\mu)'$, we easily see that $$Lf:=\langle f,\varphi\rangle\;\;\;\text{for }f\in C_b(E)$$ is a linear functional which is continuous with respect to the topology $\sigma(C_b(E),\mathcal M(E))$. So, there is a $\nu\in\mathcal M(E)$ with $$Lf=\langle f,\nu\rangle\;\;\;\text{for all }f\in C_b(E).$$ Now I would like to find a condition ensuring that $\nu$ has a density with respect to $\mu$.

Remark: Please don't stick to close to the stated assumptions. I'm willing to impose further assumptions if they lead to a positive result.

$^1$ i.e. for all $B\in\mathcal B(E)$ and $\varepsilon>0$, there is a compact $K\subseteq E$ with $\mu(B\setminus K)<\varepsilon$.

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    $\begingroup$ Something is wrong in your footnote defining Radon: B is never used... $\endgroup$
    – Matthew Daws
    Commented Dec 1, 2020 at 20:44
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    $\begingroup$ In (1) do you mean $C_c(E)$ or $C_b(E)$? $\endgroup$
    – Nate Eldredge
    Commented Dec 1, 2020 at 21:00
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    $\begingroup$ If I understand the intent of your definition, then a Radon measure is $\sigma$-finite. Since the intended fact about duality of $\mathrm L^p$ spaces is true for all $\sigma$-finite measures, what's the point of imposing additional hypotheses to give a new proof of it? $\endgroup$
    – LSpice
    Commented Dec 1, 2020 at 21:07
  • $\begingroup$ Do you want to use the lattice structure of $L^p(\mu)'$? Then $\varphi$ writes as a difference of positive functionals, $\varphi=\varphi_+-\varphi_-$, and it's sufficient to consider the case of $\varphi\ge0$. Once you got $\nu$ (not completely clear whence), it seems easy that $\nu<<\mu$, so that RN applies. For any $\phi\ge0$, any closed $\mu$-null set $F$ and $\epsilon>0$ there is a continuous $f$ with $\chi_F\le f$ and $\|f\|_p<\epsilon$. So $\nu(F)\le \|\varphi\|_{L^p(\mu)'}\epsilon$, whence $\nu(F)=0$. By the regularity, $(\mu(S)=0\implies \nu(S)=0)$ holds true for any $S\in B(E)$. $\endgroup$
    – Pietro Majer
    Commented Dec 1, 2020 at 23:36
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    $\begingroup$ Then it seems to me that you should assume $E$ is locally compact; otherwise $C_c(E)$ may be too small to be useful. For instance, in an infinite-dimensional Banach space $E$, we have $C_c(E) = 0$, so the conclusion is trivially true for every $\mu, \nu$, taking $f$ to be any function you want. $\endgroup$
    – Nate Eldredge
    Commented Dec 2, 2020 at 14:27

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