I have the following question: Suppose $\mathcal{M}^a(P)$ to be the set of all probability measures $Q$ dominated by some probability measure $P$, i.e. $Q \ll P$ for all $Q \in \mathcal{M}^a(P)$. I am trying to figure out whether the closure of the set of Radon-Nikodym densities $Z = \{ \frac{dQ}{dP}; Q \in \mathcal{M}^a(P)\}$ is compact in the weak* topology of $L^1(P)$ and - if this isn't true in general - which assumption need to be made.
Until now I've read a lot in Bogachev's measure theory book. In theorem 4.7.25 we get under some assumptions the existence of a dominating measure and in that case the closure of $Z$ is, according to the theorem, weak*-closed. I tried to figure out whether this holds true in general for densities w.r.t. a dominating measure or depends on the one constructed in the theorem. But unfortunately it wasn't apparent to me.
Thanks for helping!
edit: Ok I think I have made some progress. I studied the proof mentioned above and what I got is the following:
- For every nonnegative measure $\nu$, the space $L^1(\nu)$ is embedded as a closed linear subspace in $\mathcal{M}(X, \mathcal{A})$ (the space of all bounded countably additive measures on $\mathcal{A}$) if we identify $f \in L^1(\nu)$ with the measure $f \cdot \nu$. With this identification, the topology $\sigma(\mathcal{M}, \mathcal{M^\star})$ induces on $L^1(\nu)$ the topology $\sigma(L^1 , L^\infty)$.
- Then the proof proceeds with the construction of a dominating measure $\nu$ for a set of measures $M \subset \mathcal{M}$. For $\mu_1, \dots, \mu_n \in M$ we can then use the RN-density and write $\mu_n = f_n \cdot \nu$ where $f_n \in L^1(\nu)$.
- And now comes the point where I'm not sure about the justification: The proof then says It is clear by point 1. that the sequence $\{f_n\}$ has compact closure in the weak topology $\sigma(L^1(\nu), L^\infty(\nu))$. - Is this a direct consequence of the embedding? And if so, does this mean that this works for every dominating measure, meaning that that for a dominating measure the set of RN-densities is always weak compact?
edit:Preliminaries:
4. $\mathcal{M} = \mathcal{M}(X,\mathcal{A})$ is a Banach space and one can therefore consider here the usual weak topology of a Banach space, i.e. we recall that a sequence of vectors $x_n$ in a normed space $E$ is called weakly convergent to a vector $x$ if $l(x_n) \rightarrow l(x)$ for all $l \in E^\star$, where $E^\star$ is the space of all continuous linear functions on $E$. If, for every $l \in E^\star$, the sequence $l(x_n)$ is fundamental,
then $\{x_n\}$ is called weakly fundamental. This convergence can be described by means of the so-called weak topology on $E$, in which the open sets are all possible unions of sets of the form
\begin{align}
U (a, l_1 , . . . , l_n , ε_1 , . . . , ε_n ) = \{x : |l_1 (x − a)| < ε_1 , . . . , |l_n (x − a)| < ε_n\},\\
a \in E, l_i \in E^\star, ε_i > 0, n \in \mathbb{N},
\end{align}
and also the empty set.
5. I've searched the book regarding the properties of the measurable space but unfortunately all I could found is rather general: $(X,\mathcal{A})$ a measurable space, i.e. a pair consisting of a set $X$ and a $\sigma$-algebra
$\mathcal{A}$ of its subsets.