# Existence of a dominating measure that has densities w.r.t. all finite signed measures

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.

## 1 Answer

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.