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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 Radon-Nykodim 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. I posted this question also on MO.

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up vote 2 down vote accepted

Here is an example where $M$ exists. Let $(E,\mathcal E)$ denote any measurable space and $\mu$ and $\pi$ two probability measures on $(E,\mathcal E)$. Let $(\Omega,\Sigma)=(E\times E,\mathcal E\otimes\mathcal E)$, $\nu_1=\mu\times\pi$ and $\nu_2=\pi\times\mu$. Then $M(f):(x,y)\mapsto f(y,x)$ defines a suitable linear map $M:L^1(\nu_1)\to L^1(\nu_2)$.

Note that if $\mu$ and $\pi$ are singular to each other, such are $\nu_1$ and $\nu_2$. But I guess the real question is to show that $M$ may not exist, or that $M$ always exists...

Edit: As mentioned on MO, $M=L_{\nu_1}$ always works, since this $M$ sends functions to constants and $L_{\nu_2}(c)=c$ for every constant function $c$.

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It seems that the existence of such a map is trivial! See the following link –  Andy Teich Nov 1 '12 at 13:45
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