Let's have two measurable spaces $(X,\Sigma_X),(Y,\Sigma_Y)$ and measure $\mu$ on $(X,\Sigma_X)$ and measure $\nu$ on product space $(X\times Y,\Sigma_{X\times Y})$.

Does analogy of Lebesgue decomposition theorem still holds? By this I mean:

Are there measures $\nu_1, \nu_2$ such that $\nu = \nu_1 + \nu_2$ and $\nu_1 \ll \mu, \nu_2 \perp \mu.$

$\nu_1 \ll \mu \overset{\text{def}}{\iff} \left(\forall A \in \Sigma_X, \,\, \mu(A) = 0 \Rightarrow \nu_1(A\times Y) = 0 \right)$

$\nu_2 \perp \mu \,\overset{\text{def}}{\iff} \left( \exists A \in \Sigma_X, \, \mu(X\setminus A) = 0 , \, \nu_2(A\times Y)=0\right)$

Can we have Radon-Nikodym derivative of $\nu_1$ too?

Is there function $f: X\times \Sigma_Y \rightarrow \mathbb{R}$? Such that $\forall A\in \Sigma_{X\times Y} \,\, \nu_1(A) = \int_X f(x,A^x) \,d\mu(x)$.

$A^x \overset{\text{def}}{=} \{ y\in Y : (x,y)\in A\}$

My approach using vector measures. Unfortunately my knowledge of vector measures is very sloppy, so I'm unable to fill all those gaps.

My idea is to take vector measure $\tilde \nu$ on $(X,\Sigma_X)$ and as vector space the space of all measures with finite variations on $(Y,\Sigma_Y)$. Measure $\tilde \nu$ is defined as $\tilde \nu(A)(B) = \nu(A\times B), A\in \Sigma_X, B\in \Sigma_Y$.

I guess that from measure $\tilde \nu$ we can uniquely reconstruct measure $\nu$, and we can use Lebesgue decomposition theorem for $\tilde \nu$ to obtain $\tilde \nu_1, \tilde \nu_2$, from these we can uniquely construct $\nu_1,\nu_2$

  • $\begingroup$ Lebesgue Decomposition Theorem and Radon-Nikodym Theorem apply to measures defined in the SAME measure spaces. So, you may try to use them to compare/study $\nu$ to the product measure, but not to compare $\nu$ to only one factor. Please, note that Lebesgue Decomposition Theorem and Radon-Nikodym Theorem have also additional conditions (regarding $\sigma$-finiteness). $\endgroup$ – Ramiro Sep 6 '15 at 14:28
  • $\begingroup$ It seams to me that you missed the point of the question. I try to state analogy of Lebesgue Decomposition Theorem in the setting where you have one measure on product space and one measure only on one factor. $\endgroup$ – tom Sep 6 '15 at 15:00
  • $\begingroup$ Your generalisation of $\perp$ seems to be somewhat broken. If $\mu \neq 0$, then $\nu_2\perp \mu$ can only hold for $\nu_2 =0$. $\endgroup$ – PhoemueX Sep 6 '15 at 18:06
  • $\begingroup$ Ahh you are right, fixed. $\endgroup$ – tom Sep 6 '15 at 19:40
  • $\begingroup$ The first part of the "extension" seems trivial. For any $A \in \Sigma_X$ define $\tilde \nu(A) =\nu(A\times Y)$. Then $\nu \ll \mu$ (in your sense) if and only if $\tilde \nu \ll \mu$ (in the usual sense). Also $\nu \perp \mu$ (in your sense) if and only if $\tilde \nu \perp \mu$ (in the usual sense). So the "Lebesgue decomposition" (in your sense) of $\nu$ w.r.t $\mu$ actually reduces to the Lebesgue decomposition (in the usual sense) of $\tilde \nu$ w.r.t $\mu$. Note that, anyway, the condition regarding $\sigma$-finetess need to be fulfilled. $\endgroup$ – Ramiro Sep 6 '15 at 21:19

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