Consider the set of all probability measures on a Polish space $X$ (equipped with the Borel $\sigma$field $\mathcal{B}(X)$). I am wondering if there exist conditions under which a subset of measures that are abolutely continuous w.r.t. a given (fixed) measure, say $\mu$, is dense (w.r.t. the weak topology). When $X=\mathbb{R}^d$ and $\mu$ is atomless, this follows from the convolution argument.
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1$\begingroup$ I think this works if and only if $\mu$ has topological support equal to all of $X$. $\endgroup$– Christian RemlingCommented Mar 9 at 0:15

$\begingroup$ @d.k.o. Well, the usual proof just works in that case, no? $\endgroup$– Martin HairerCommented Mar 9 at 9:02

$\begingroup$ @MartinHairer What do you mean exactly by the usual proof? $\endgroup$– d.k.o.Commented Mar 9 at 9:26

1$\begingroup$ @d.k.o. I meant the proof that every measure can be approximated by an atomic measure supported on some fixed countable dense set. Just replace the atoms by the restriction of $\mu$ to small balls, see for example the construction at the bottom of p.16 of my SPDE notes: hairer.org/notes/SPDEs_Course.pdf. $\endgroup$– Martin HairerCommented Mar 9 at 13:18

$\begingroup$ @d.k.o. You're right, thanks! $\endgroup$– Martin HairerCommented Mar 9 at 18:55
