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Let $\mu$ be a (finite if necessary) measure on the product $\sigma$-algebra $\mathcal A_1 \otimes \mathcal A_2$ of two measurable spaces $(\Sigma_1,\mathcal A_1)$, $(\Sigma_2, \mathcal A_2)$. The marginal measure of $\mu$ on $\mathcal A_i$ for $i\in\{1,2\}$ is defined as $$ \mu_1(A) := \mu(A\times \Sigma_2), \quad \mu_2(B) := \mu(\Sigma_1 \times B) \quad \text{for $A\in\mathcal A_1, B\in\mathcal A_2$}.$$

I wanted to know whether the Radon-Nikodym derivative $\frac{d\mu}{d(\mu_1\otimes \mu_2)}$ exists. For the existence, we have to show that $\mu$ is absolutely continuous w.r.t. the product measure $\mu_1\otimes\mu_2$. I've tried several approaches without success, one of them is described in the following:

Given $M\in\mathcal A_1\otimes \mathcal A_2$ with $(\mu_1\otimes\mu_2)(M)=0$ show $\mu(M)=0$. Using Fubini, we compute $$ 0 = (\mu_1\otimes\mu_2)(M) = \int_{\Sigma_1}\int_{\Sigma_2} \mathbf 1_M(x,y) ~d\mu_2(y) ~d\mu_1(x) = \int_{\Sigma_1} \mu_2(\{ y \mid (x,y)\in M \}) ~d\mu_1(x) $$ This implies $\mu_2(\{y\mid(x,y)\in M\})=0$ for $\mu_1$-a.e. $x_1\in\Sigma_1$ (and a similar statement for $\mu_2$-a.e. $x_2\in \Sigma_2$).

At this point, I am unsure whether this statement is true after all (and if so whether we need additional assumptions?).

Michael
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1 Answers1

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It is not in general.

Take $A_1 = A_2 = [0,1]$ and let $\mu$ be the unique Borel measure supported on $S=\{(x,x):x\in[0,1]\}$ which assigns $\{(x,x) : x\in[a,b]\}$ the measure $b-a$ (alternatively identify $S$ with $[0,1]$ using the map $x\mapsto (x,x)$ and then assign the Lebesuge measure).

The projection of $\mu$ to $A_1$ and $A_2$ is the Lebesgue measure on $[0,1]$, however the Lebesgue measure assigns measure zero on the diagonal, while $\mu$ assigns the measure $1$.

Note also that Lebesgue measure assigns positive measure to sets outside the diagonal while $\mu$ does not, so no measure is absolutely continuous with respect to the other.

Yanko
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