A general notion of the support of a measure? Let $(\Omega, \mathcal A, \mu)$ be a finite measure space. If $\Omega$ is finite and $\mathcal A = 2^\Omega$, then the support of $\mu$ is $s(\mu) = \{\omega: \mu\{\omega\} > 0\}$.
As far as I know, there isn't a very broad generalization of $s(\mu)$ to uncountable $\Omega$. The most common approach seems to involve assuming that $\Omega$ comes equipped with a topology $\mathcal T$ that generates $\mathcal A$. One then defines $$s(\mu) = \{\omega: \omega \in U \in \mathcal T \implies \mu(U) > 0\}.$$
That is, $\omega$ is in the support of $\mu$ if every open set containing $\omega$ has positive $\mu$-measure. This definition coincides with the one for the finite case when $\mathcal T$ is the discrete topology.
It seems to me that it would be nice to have a general notion of support that doesn't rely on $\Omega$ having a topology. Here is an idea I had.
Say that a measure $\mu$ on $(\Omega, \mathcal A)$ is discrete if it is of the form $\mu = \sum_{i=1}^n \alpha_i \delta_{\omega_i}$ with $\alpha_i$ positive and $\sum_{i=1}^n\alpha_i = 1$, and where $\delta_{\omega_i}$ is point mass at $\omega_i$.
The general definition of support that I will propose is motivated by the following interesting fact:

Every finitely additive finite measure on $(\Omega, \mathcal A)$ is the pointwise limit of a net of discrete measures. That is, for all $\mu$ there is a net $(\mu_d)$ of discrete measures such that $\mu(A) = \lim_d \mu_d(A)$ for all $A \in \mathcal A$.

Let $\mathcal P$ be the set of all finitely additive measures on $(\Omega, \mathcal A)$ equipped with the topology of pointwise convergence.

Now say that $\omega$ is in $s(\mu)$ if and only if every open subset of $\mathcal P$ containing $\mu$ contains a discrete measure $\nu = \sum_{i=1}^n \alpha_i \delta_{\omega_i}$ such that $\omega_i = \omega$ for some $i$.

It's clear that this definition of support generalizes the one given when $\Omega$ is finite. And while it does utilize some topology (on $\mathcal P$), it doesn't assume that $\Omega$ is a topological space, and can therefore be applied to any measure space whatsoever (even a finitely additive one).

This is open-ended, but my question is basically: Is this a good definition of support? Has it been studied before? Does anyone have any interesting observations or comments about the definition?


One potentially interesting observation is that on any space there will always be finitely additive measures $\mu$ with full support, i.e. $s(\mu) = \Omega$.
Proof. Suppose not. Then for every $\mu \in \mathcal P$ there is an open subset $N_\mu$ of $\mathcal P$ and $\omega_\mu \in \Omega$ such that no discrete measure $\nu = \sum_{i=1}^n \alpha_i \delta_{\omega_i}$ in $N_\mu$ is such that $\omega_i = \omega_\mu$ for some $i$. The collection $\{N_\mu: \mu \in \mathcal P\}$ is an open cover of $\mathcal P$, and thus for some $n$ and $\mu_1,...,\mu_n$ the collection $\{N_{\mu_i}: 1 \leq i \leq n\}$ covers $\mathcal P$ because $\mathcal P$ is compact ($\mathcal P$ is a closed subset of $[0,1]^\mathcal A$ with the product topology.) But now for any positive $\alpha_i$, $1 \leq i \leq n$, summing to $1$ the discrete measure $\sum_{i=1}^n \alpha_i \delta_{\omega_{\mu_i}}$ is not in any of the $N_{\mu_i}$, which is a contradiction.
This raises the question:

Under what conditions can one guarantee that there is a countably additive measure with full support?

 A: If I understand your definition correctly, I think every measure (including the zero measure) has full support.  Here is why:  given any measure
$\mu$, and any point $\omega $ on $\Omega $, let's prove that $\omega $ lies in the support of $\mu $.  For this pick any open subset $V$ of
$\mathcal P$ containing $\mu $.
By definition of the topology of pointwise
convergence, there is some $\varepsilon >0$,  and  measurable sets $A_1, A_2, \ldots , A_n$, such that
$$
  U_{\varepsilon ;A_1, A_2, \ldots , A_n}:= \{\nu \in  \mathcal P: |\nu (A_i)-\mu (A_i)|<\varepsilon ,  \text{ for } i=1, \ldots , n\}\subseteq  V.
  $$
We will now prove that there exists a discrete measure $\nu $ in $V$ such that $\nu (\omega )\neq 0$.
Using your claim, pick  some discrete measure $\nu $ in $U_{\varepsilon /2;A_1, A_2, \ldots , A_n}$.
If $\nu (\omega )\neq 0$, we are done.  Otherwise,
let
$$
  \nu '= \nu +(\varepsilon /2)\delta _\omega .
  $$
It is then clear that $\nu '(\omega )=\varepsilon /2\neq 0$, and for every $i$ we have
$$
  |\nu '(A_i)-\mu (A_i)| \leq  |\nu '(A_i)-\nu (A_i)| +   |\nu (A_i)-\mu (A_i)| < (\varepsilon /2)\delta _w(A_i) + \varepsilon /2 \leq \varepsilon ,
  $$
This proves that $\nu '$ belongs to $V$.
