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Fall 2019 - Problem 1

Lebesgue-Radon-Nikodym derivative, measure theory

Given $\sigma$ -finite measures $\mu_1, \mu_2, \nu_1, \nu_2$ on a measurable space $\p{X, \mathcal{X}}$ , suppose that $\mu_1 \ll \nu_1$ and $\mu_2 \ll \nu_2$ . Prove that the product measures $\mu_1 \otimes \mu_2$ and $\nu_1 \otimes \nu_2$ on $\p{X \times X, \mathcal{X} \otimes \mathcal{X}}$ satisfy $\mu_1 \otimes \mu_2 \ll \nu_1 \otimes \nu_2$ and the Radon-Nikodym derivatives obey

\deriv{\p{\mu_1 \otimes \mu_2}}{\p{\nu_1 \otimes \nu_2}}\p{x, y} = \deriv{\mu_1}{\nu_1}\p{x} \deriv{\mu_2}{\nu_2}\p{y}

for $\nu_1 \otimes \nu_2$ -almost every $\p{x, y} \in X \times X$ .

Solution.

Let $E \in \mathcal{X} \otimes \mathcal{X}$ and suppose $\p{\nu_1 \otimes \nu_2}\p{E}$ . Then by Fubini's theorem (all measures are $\sigma$ -finite)

\p{\nu_1 \otimes \nu_2}\p{E} = \int \chi_E\p{x, y} \,\diff\nu_1\p{x} \,\diff\nu_2\p{y} = \int \nu_1\p{E_y} \,\diff\nu_2\p{y} = 0.

where $E_y = \set{x \in X \mid \p{x, y} \in E}$ . Since $\nu_1 \geq 0$ , it follows that $\nu_1\p{E_y} = 0$ for $\nu_2$ -almost every $y \in X$ , so by absolute continuity, it follows that $\mu_1\p{E_y} = 0$ for $\nu_2$ -almost every $y \in X$ , and by absolute continuity again,

\nu_2\p{\set{y \in X \mid \mu_1\p{E_y} > 0}} = 0 \implies \mu_2\p{\set{y \in X \mid \mu_1\p{E_y} > 0}} = 0,

i.e., $\mu_1\p{E_y} = 0$ for $\mu_2$ -almost every $y$ . Hence,

\p{\mu_1 \otimes \mu_2}\p{E} = \int \chi_E\p{x, y} \,\diff\mu_1\p{x} \,\diff\mu_2\p{y} = \int \mu_1\p{E_y} \,\diff\mu_2\p{y} = 0,

so $\mu_1 \otimes \mu_2 \ll \nu_1 \otimes \nu_2$ . For the second claim, by absolute continuity, we have

\begin{gathered} \diff\p{\mu_1 \otimes \mu_2} = \deriv{\p{\mu_1 \otimes \mu_2}}{\p{\nu_1 \otimes \nu_2}} \,\diff\p{\nu_1 \otimes \nu_2}, \\ \diff\mu_1 = \deriv{\mu_1}{\nu_1} \,\diff\nu_1 \quad\text{and}\quad \diff\mu_2 = \deriv{\mu_2}{\nu_2} \,\diff\nu_2. \end{gathered}

Observe that if $A \times B \in \mathcal{X} \otimes \mathcal{X}$ ,

\begin{aligned} \int_{A \times B} \deriv{\p{\mu_1 \otimes \mu_2}}{\p{\nu_1 \otimes \nu_2}} \,\diff\p{\nu_1 \otimes \nu_2} &= \p{\mu_1 \otimes \mu_2}\p{A \times B} \\ &= \int \chi_{A \times B} \,\diff\mu_1 \,\diff\mu_2 \\ &= \p{\int_A \diff\mu_1}\p{\int_B \diff\mu_2} \\ &= \p{\int_A \deriv{\mu_1}{\nu_1} \,\diff\nu_1} \p{\int_B \deriv{\mu_2}{\nu_2} \,\diff\nu_2} \\ &= \int_{A \times B} \deriv{\mu_1}{\nu_1}\p{x} \deriv{\mu_2}{\nu_2}\p{y} \,\diff\nu_1\p{x} \,\diff\nu_2\p{y}. \end{aligned}

It follows that equality holds when we replace $A \times B$ with countable unions of disjoint measurable rectangles. Hence, by Carathéodory's theorem, we see that $\deriv{\p{\mu_1 \otimes \mu_2}}{\p{\nu_1 \otimes \nu_2}} \,\diff\p{\nu_1 \otimes \nu_2}$ and $\deriv{\mu_1}{\nu_1}\p{x} \deriv{\mu_2}{\nu_2}\p{y} \,\diff\nu_1\p{x} \,\diff\nu_2\p{y}$ agree on all measurable sets. Thus,

\deriv{\p{\mu_1 \otimes \mu_2}}{\p{\nu_1 \otimes \nu_2}}\p{x, y} = \deriv{\mu_1}{\nu_1}\p{x} \deriv{\mu_2}{\nu_2}\p{y}

$\nu_1 \otimes \nu_2$ -almost everywhere.