Topology and Measures I apologize if this question is a bit vague; I'm just wondering if there is a concept like what I'm talking about, or if I'm just lost. I'll start with just some thoughts. I looked a bit, and I don't think I found what I was looking for. Admittedly, my topology is pretty much limited to a bit of point-set topology.
In measure theory, we talk about a measure $\mu$ being "continuous" with respect to $\nu$ if the measures are related appropriately. It's my understanding that the idea of continuity is inextricable from some notion of topology.
Moreover, in the context of Lebesgue measure $\lambda$, in the context of $[0, 1]$ with a typical $\sigma$-algebra, say $\mathcal{B}$, $\lambda$'s behavior seems to be related to the topology of $[0, 1]$, e.g. if I have a dense subset of $[0, 1]$, say $\mathbb{Q}$, then I can define the measure of a measurable set by a sequence of better and better coverings of the set by intervals with rational endpoints. And this seems to hold for any measure that is continuous with respect to $\lambda$.
Further, say I defined $\mu(F) = \int_{F}g \, \mathrm{d}\lambda$, where $g$ is an appropriate non-negative integrable function. Then if I set $M(x) = \mu([a, x])$, then $M$ is continuous, but if I set $\mu$ to be something not $\lambda$-continuous, i.e. I don't define it by $g$, but $\mu$ gives positive measure to some singleton, then $M$ ceases to be continuous in $[0, 1]$.
So my question: Is there a link between the "continuity" of a measure and the topology of the measure space? If so, then is there some way to define a topology on a measure space by the measure? For example, is there some way to say some set is "dense" with respect to the measure, in the way that we can approximate any measurable set's Lebesgue measure in $[0, 1]$ by covering it with intervals with rational endpoints? Any help is appreciated. Thanks!
 A: I would try to challenge a statement made in another answer:

For a completely general pair of measures, there is definitely nothing topological that can be said.

Recall that $\mu \ll \nu$ iff for all $\epsilon > 0$ there exists $\delta>0$ such that $\mu(F)\leq\epsilon$ whenever $\nu(F) \leq \delta$. Now, let us endow the $\sigma$-algebra $\mathcal F$ with the $L^1(\nu)$ pseudometric given by $d_\nu(A,B):=\nu(A\triangle B)$ where $A\triangle B = (A\setminus B)\cup(B\setminus A)$ denotes the symmetric difference between two sets. Note that $d_\nu$ is a pseudometric only since $d_\nu(A,B)  =0 $ does not imply $A = B$.
Note that $|\mu(A) - \mu(B)|\leq \mu(A\triangle B)$ so that $\mu$ is a continuous function (on $\mathcal F)$ with respect to the pseudometric $d_\nu$, and hence the corresponding topology. I think, it is a rather good justification to say that $\mu$ is continuous w.r.t. $\nu$.
If you are more comfortable in thinking about L-spaces of functions, just consider $\mu$ as a functional on functions endowed with $L^1(\nu)$ distance.
