# Radon-Nikodym derivative of product measure

For $j=1,2$, let $\nu_{j},\mu_{j}$ be $\sigma$-finite measures on $(X_{j},\mathcal{M}_{j})$ such that $\nu_{j}\ll\mu_{j}$. I want to show that $\nu_{1}\times\nu_{2}\ll\mu_{1}\times\mu_{2}$ and that $\frac{d(\nu_{1}\times\nu_{2})}{d(\mu_{1}\times\mu_{2})}(x_{1},x_{2})=\frac{d\nu_{1}}{d\mu_{1}}(x_{1})\frac{d\nu_{2}}{d\mu_{2}}(x_{2})$.

I tried to show that if $E\in\mathcal{M}_{1}\otimes\mathcal{M}_{2}$ and $\mu_{1}\times\mu_{2}(E)=0$, then $\nu_{1}\times\nu_{2}(E)=0$. This criterion holds for measurable rectangles so then I tried considering $\{E\in\mathcal{M}_{1}\otimes\mathcal{M}_{2}|\mu_{1}\times\mu_{2}(E)>0\;\text{or}\;\nu_{1}\times\nu_{2}(E)=0\}$ and show that it is a $\sigma$-algebra containing all measurable rectangles but I guess it didn't quite work.

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Maybe you could use Tonelli? –  copper.hat Nov 4 '12 at 16:45

Denote $X=X_1\times X_2$, $\mu=\mu_1\times\mu_2$, $\nu=\nu_1\times\nu_2$, $\mathcal{M}=\mathcal{M}_1\times\mathcal{M}_2$ and $$f(x_1,x_2)=\dfrac{d\nu_1}{d\mu_1}(x_1)\dfrac{d\nu_2}{d\mu_2}(x_2).$$ Evidently $f$ is non-negative, $\mathcal{M}$ measurable and $\mu$ locally integrable. Then as @copper.hat commented, by Fubini's theorem(or Tonelli's theorem instead), we have:

$$\nu(E_1\times E_2)=\nu_1(E_1)\cdot\nu_2(E_2)=\int_{E_1}\dfrac{d\nu_1}{d\mu_1}d\mu_1\times\int_{E_2}\dfrac{d\nu_2}{d\mu_2}d\mu_2=\int_{E_1\times E_2}f~d\mu.$$

Consider the collection

$$\mathcal{C}:=\{E\in \mathcal{M}:\nu(E)=\int_E f~d\mu\}$$ and we want to show $\mathcal{C}=\mathcal{M}$. We have shown $\Pi:=\{E\in\mathcal{M}:E=E_1\times E_2\}\subset \mathcal{C}$, it suffices to show $\mathcal{C}$ is a $\sigma$-algebra. First let us assume that $\mu(X)<\infty$. Since $\Pi$ is a $\pi$-system, we only need to show $\mathcal{C}$ is a $\lambda$-system: (i) $X\in\mathcal{C}$; (ii) if $D,E\in\mathcal{C}$ with $D\subset E$, then $E\setminus D\in\mathcal{C}$; and if $\{E_n:n\ge 1\}\subset \mathcal{C}$ is an increasing sequence, then $\cup_n E_n\in\mathcal{C}$. The first two items are obvious and the last one follows from monotone convergence theorem. Now we have proved $\mathcal{C}=\mathcal{M}$ with the additional assumption $\mu(X)<\infty$. For general $X$, let $\{F_n=F_{1,n}\times F_{2,n}:n\ge 1\}\subset\Pi\subset\mathcal{C}$ be such that $\mu(F_n)<\infty$ for every $n$ and $\cup_n F_n=X$. Then the argument before implies that for every $n$, $\{E\in \mathcal{C}:E\subset F_n\}=\{E\in \mathcal{M}:E\subset F_n\}$. Then given $E\in\mathcal{M}$, $E\subset F_n\in\mathcal{C}$ for every $n$, so again by monotone convergence theorem, $E\in\mathcal{C}$, which completes the proof.

Therefore, when $E\in\mathcal{M}$, $\nu(E)=\int_E f~d\mu$. The conclusion follows.

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When you use the monotone convergence theorem to show that $\bigcup_{n}E_{n}\in\mathcal{C}$, are you considering the sequence $(\bigcup_{n=1}^{k}E_{n})_{k=1}^{\infty}$? Do we need to first show that $\mathcal{C}$ is closed under finite unions? –  user44532 Nov 4 '12 at 22:23
@CYC: Yes, you are correct, and it seems not so easy to prove that directly. Thank you for pointing it out. Maybe to save words, I should use $\pi-\lambda$ theorem to fix this problem. –  23rd Nov 5 '12 at 0:10
What is this $\pi-\lambda$ theorem that you mentioned? –  user44532 Nov 5 '12 at 0:44
@CYC: See the updated answer for details. –  23rd Nov 5 '12 at 0:46
I tried showing that $\mathcal{C}$ is a monotone class containing all finite disjoint unions of rectangles, then apply the monotone class lemma. It seems to work as well. –  user44532 Nov 5 '12 at 4:45

We use the following lemma:

Let $(S,\mathcal A)$ a measurable space, and $\mu$, $\nu$ two finite measures. We have $\nu\ll\mu$ if and only if for all $\varepsilon>0$, we can find $\delta>0$ such that if $A\in\mathcal A$ and $\mu(A)\leq\delta$ then $\nu(A)\leq \varepsilon$.

Let $\{E_i\}\subset\mathcal M_1$, $\{F_j\}\subset \mathcal M_2$, where $\mu_1(E_i)+\nu_1(E_i)$ and $\mu_2(F_j)+\nu_2(F_j)$ are finite for all $i,j$ and $\bigcup_{i=1}^{+\infty}E_i=X_1$, $\bigcup_{j=1}^{+\infty}F_j=X_2$. Define the finite measures $\mu^{(i,j)}(S):=\mu_1\otimes \mu_2(E_i\times F_j\cap S)$, and $\nu^{(i,j)}(S):=\nu_1\otimes \nu_2(E_i\times F_j\cap S)$. We just have to show that $\nu^{(i,j)}\ll\mu^{(i,j)}$.

Let $\mu_1^{i}(S):=\mu_1(E_i\cap S)$ and similarly for the other measures.

Let $\varepsilon>0$, $i,j$ fixed integers and $\sqrt\delta$ working for $\varepsilon$.

As $\nu^{(i,j)}+\mu^{(i,j)}$ is finite, if $\mu^{(i,j)}(S)\leq \delta/2$, we can find, by this result, $S':=\bigsqcup_{k=1}^NA_k\times B_k$ where $$\sum_{k=1}^N\nu_1(E_i\cap A_k)\nu_2(F_j\cap B_k)+\sum_{k=1}^N\mu_1(E_i\cap A_k)\mu_2(F_j\cap B_k)\leq \delta.$$ Indeed, we take $S'$ of the previous form such that $(\nu^{(i,j)}+\mu^{(i,j)})(S\Delta S')\lt\delta/2$. This implies that $$(\nu^{(i,j)}+\mu^{(i,j)})(S')\leqslant (\nu^{(i,j)}+\mu^{(i,j)})(S)+\delta/2\leqslant \delta.$$

Let $I:=\{k\in [N],\mu_1(E_i\cap A_k)\leq \sqrt \delta\}$, and $I':=\{k\in [N],\mu_2(F_j\cap B_k)\leq \sqrt \delta\}$. Then $I\cup I'=[N]$, which gives \begin{align} \nu^{(i,j)}(S)&\leq \delta+\nu^{(i,j)}(S')\\ &=\delta+\varepsilon\sum_{k\in I}\nu_2(F_j\cap B_k)+\varepsilon\sum_{k\in I'}\nu_1(E_i\cap A_k)\\ &\leq \varepsilon(1+\mu_1(E_i)+\mu_2(F_j)), \end{align} as we can assume WLOG $\delta\leq\varepsilon$.

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I do not understand how you arrive at the first main equation, ending with ''$\leq \delta$''. Can you explain more? –  Dan Douglas Nov 7 '14 at 1:15
@DanDouglas I've edited. Is it clearer? –  Davide Giraudo Nov 7 '14 at 10:09