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Let $\mathcal{S}$ be the set of all finite signed measures on the measurable space $(\mathcal{X}, \mathcal{A})$. Does there exist a dominating measure $\mu$ on $(\mathcal{X}, \mathcal{A})$ for which every $\gamma \in \mathcal{S}$ has a (not necessarily unique-a.s.) density $d\gamma/d\mu$?

This seems to be implied in the answer to another question. Would anyone mind expounding a bit more or at least pointing me to a more detailed explanation?

On the other hand, I'm a little worried that there are some additional conditions needed on the space $(\mathcal{X}, \mathcal{A})$, as seems to be the case with other facts in this vein, e.g. Radon-Nikodym Thoreom, and also here but perhaps that post was only using the extra conditions to show the existence of a finite dominating measure.

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It's more complicated than that. In general, there is no universally dominating measure (on countable spaces, the counting measure dominates all finite measures, so sometimes there is).

In this comment:

Take a maximal family $\mathcal{A}$ of mutually singular probability measures. (Use Zorn's Lemma.) The space of measures is isometrically the $l_1$-sum of $L^1(\nu)$ as $\nu$ ranges over the family $\mathcal{A}$.

GEdgar, the author of the answer, explains the construction.

Since you used $\mathcal{A}$ for the $\sigma$-algebra, let us call the maximal family of mutually singular probability measures $\mathcal{P}$.

Now if $\mu$ is a finite (signed) measure on $\mathcal{A}$, for every $\nu \in \mathcal{P}$ you have the Lebesgue decomposition

$$\mu = \mu_{\nu} + \mu_{\nu}^{\perp}$$

of $\mu$ with respect to $\nu$, where $\mu_{\nu}$ is absolutely continuous with respect to $\nu$ and $\mu_{\nu}^{\perp}$ and $\nu$ are mutually singular. Since the members of $\mathcal{P}$ are mutually singular, $\mu_{\nu_1}$ and $\mu_{\nu_2}$ are mutually singular for $\nu_1\neq \nu_2$.

Then you have

$$\mu = \sum_{\nu \in \mathcal{P}} \mu_{\nu},$$

for otherwise $\mu$ would have a nonzero part that is singular with respect to all $\nu \in \mathcal{P}$, and from that you could construct a probability measure (the positive or negative part of the remainder would be nonzero, and after scaling gives a probability measure) that is singular with respect to all $\nu\in \mathcal{P}$, contradicting the maximality of $\mathcal{P}$.

And identifying $\mu_{\nu}$ with its density with respect to $\nu$ we have the map sending each measure into the space

$$\bigoplus_{\nu \in \mathcal{P}}{}^{l_1} L^1(\nu),\tag{1}$$

the $l_1$-direct sum of all the $L^1(\nu)$. One then verifies that this map is a bijective isometry.

We can identify the space $(1)$ with $L^1(\mu)$ for some measure $\mu$, but that can in general not be a measure on $\mathcal{A}$. The obvious construction is to let $\mathcal{Y} = \mathcal{X}\times \mathcal{P}$, and take the $\sigma$-algebra

$$\mathcal{B} = \bigl\{ M \subset \mathcal{Y} : \bigl(\forall \nu \in \mathcal{P}\bigr)\bigl( \{ x \in \mathcal{X} : (x,\nu) \in M\} \in \mathcal{A}\bigr)\bigr\}$$

and set

$$\mu(M) = \sum_{\nu \in \mathcal{P}} \nu\bigl(\{ x\in \mathcal{X} : (x,\nu) \in M\}\bigr)$$

for $M \in \mathcal{B}$. Sometimes, smaller constructions are possible.

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