I posted the following question also here, but thought that I can get more answers in MO. Let $(\Omega,\Sigma)$ be a measurable space and $\nu_1$, $\nu_2$ two probability measures on it. For $i=1,2$, let $L_{\nu_i}:L^{1}(\nu_i)\rightarrow\mathbb R$ be the map defined by $L_{\nu_i}(g)=\int_\Omega gd\nu_i$. Denote by $[1_E]_{\nu_i}$ the equivalence class in $L^1(\nu_i)$ of the characteristic function $1_E$. Does there always exist a linear map $M:L^1(\nu_1)\rightarrow L^1(\nu_2)$ such that $$L_{\nu_2}\circ M([1_E]_{\nu_1})=L_{\nu_1}([1_E]_{\nu_1})$$ for all $E\in\Sigma$? When $\nu_1$ is absolutely continuous to $\nu_2$ then, by the RadonNykodim Theorem, the map $M$ can be given explicitly, i.e. there exists a $\Sigma$measurable function $f:\Omega\rightarrow\mathbb R$, such that $M(g):=f\cdot g$ will satisfy the equation given above. So the question becomes interesting when $\nu_1$ and $\nu_2$ are singular to each other, and the spaces $L^1(\nu_1)$ and $L^1(\nu_2)$ are not isometric. The way I think of it is, that it could characterize in some/all cases a RadonNykodim derivative when the measures are singular to each other...comments on this are welcome!
1 Answer
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How about $M \colon L^1(\nu_1)\rightarrow L^1(\nu_2) \colon g \mapsto (x \mapsto \int g~d\nu_1)$?

$\begingroup$ ok, you are right. so the above equality does not really charachterize the radonnikodym derivative! $\endgroup$ Nov 1, 2012 at 13:23

$\begingroup$ one should impose that $M$ is injective (???) on the set of all characterisitc functions in $L^1(\nu_1)$ in order to make my question reasonable... $\endgroup$ Nov 1, 2012 at 14:47

$\begingroup$ Then the answer would be no (in general): take $\nu_1 = \mathcal{L}_{[0,1]}$ and $\nu_2 = \delta_0$ (with $\Omega = [0,1]$). Then the $M$ above is the only possible mapping with the desired properties. $\endgroup$ Nov 1, 2012 at 14:54